Mathematics-19.Unit-15 STRAIGHT LINES AND PAIR OF LINES

1.1 DISTANCE BETWEEN TWO POINTS 3 1.2 SECTION FORMULA 3 1.3 SOME BASIC DEFINITIONS 4 1.4 AREA OF A TRIANGLE 5 2. LOCUS 6 3. STRAIGHT LINE 7 3.1 GENERAL EQUATION OF A STRAIGHT LINE 7 3.2 INTERCEPT OF A STRAIGHT LINE ON THE AXES 7 3.3 DIFFERENT FORMS OF THE STRAIGHT LINES 7 3.4 POSITION OF A POINT WITH RESPECT TO A GIVEN LINE 10 3.5 POSITION OF TWO POINTS WITH RESPECT TO A LINE 11 3.6 ANGLE BETWEEN TWO STRAIGHT LINES 12 3.7 EQUATION OF THE STRAIGHT LINE PASSING THROUGH A POINT AND INCLINED TO A GIVE LINE 13 3.8 THE DISTANCE BETWEEN TWO PARALLEL LINES 13 3.9 PERPENDICULAR DISTANCE OF A POINT FROM A LINE 14 3.10 TO FIND THE IMAGE OF A POINT IN A LINE 14 3.11 CO-LINEARITY OF THREE POINTS 15 3.12 CONCURRENCY OF THREE LINES 15 3.13 BISECTORS OF ANGLE BETWEEN TWO GIVEN LINES 15 3.14 FAMILY OF LINES 18 3.15 PAIR OF STRAIGHT LINES 19 3.16 ROTATION OF CO-ORDINATE SAXES 21 4. OBJECTIVE ASSIGNMENT 21 1.1 DISTANCE BETWEEN TWO POINTS Let P and Q are two points whose coordinates are (x1, y1) and (x2, y2), then PQ = = Note: Distance is always positive. Therefore we often write PQ instead of PQ. KEY TOOLS In order to prove that a given figure is a 1. Square: Prove that the four sides are equal and the diagonals are equal. 2. Rhombus (but not a square): Prove that the four sides are equal but the diagonals are not equal. 3. Rectangle: Prove that the opposite sides are equal and the diagonals are also equal. 4. Parallelogram (but not a rectangle): Prove that the opposite sides are equal but diagonals are not equal. Note: That in each of these cases diagonals bisects each other. 5. Equilateral Triangle: Prove all three sides are equal. 6. Isosceles Triangle: Prove two opposite side are equal. (Opposite angle also equal) 1.2 SECTION FORMULA The coordinates of a point which divides the line segment joining two given points A(x1, y1) and B(x2, y2) internally in the given ratio m : n are The coordinates of the point P(x, y) which divides A (x1, y1) and B (x2, y2) externally in the ratio m:n are Illustration 1: Find the ratio in which x-axis and y-axis divides the line segment joining (–2, – 3) and (–1, 2). Key concept: If the ratio, in which a given line segment is divided, is to be determined, then for convenience instead of taking m:n, we take the ratio k:1. If the value of k turns out to be positive, it is an internal division and if k is negative it is an external division. Solution: Let x-axis divides the line segment in k : 1. Then the coordinate of point on x-axis which divides PQ in K : 1 is . But if any point is lying on the x-axis, then its y-coordinate will be zero Hence x – axis divides PQ in 3:2 internally. Similarly coordinates of any point on y-axis which divides PQ in k :1 is . But if any point is lying on the y-axis, then it’s x-coordinate will always zero Hence y-axis divides PQ in 2:1 externally 1.3 SOME BASIC DEFINITIONS Centroid (G): The centroid of a triangle divides each median in the ratio 2:1(2 from the vertex and 1 from the opposite side). The coordinates of centroid are given by G  . Incentre (I): The coordinates of the in-centre of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) are given by I  where a, b and c are length of the sides BC, CA and AB respectively. Orthocentre (H): The co-ordinates of the orthocentre of the triangle A(x1, y1), B(x2, y2), C(x3,y3) are . Circumcentre (C): • The coordinates of the circum-centre of the triangle with vertices A(x1, y1), B(x2, y2), C(x3, y3) is given by O = . • Note: The circum centre of a right angled triangle is the mid point of its hypotenuse. • In an equilateral triangle centroid, in-centre, orthocentre and circum centre all coincides. • Centroid divides the line joining orthocentre and circum centre in 2:1 internally. Illustration 2: If the vertices of a triangle be (2 , 6), (3, – ) and (0, 0). Then find its orthocentre and circumcentre. Key concept: In a right angled triangle orthocentre is the right angled vertex, and circumcentre is the mid point of the hypotenuse. Solution: Clearly, the given triangle is right angled at vertex(0, 0), so its orthocentre is (0, 0) and circumcentre will be the mid point of hypotenuse which is . Illustration 3: Find the orthocentre of the triangle whose vertex are (0, 0), (1, 4) and (3, 0). Solution: Let the orthocentre be H. Clearly x co-ordinate of orthocentre is 1 and let the y co-ordinate be k. Then from  OB’H and BB’A Hence the orthocentre is 1.4 AREA OF A TRIANGLE Let (x1, y1), (x2, y2) and (x3, y3) respectively be the coordinates of the vertices A, B, C of a triangle ABC. Then the area of triangle ABC, is 2. LOCUS The equation to a locus is the relation which exists between the coordinates of any point on the path, and which holds for no other point except those lying on the path. Working rule to find the locus of a point: Step 1: Let the coordinates of the moving point be P(h, k). Step 2: Write down the given geometrical condition and express these conditions in terms of h and k. Step 3: Eliminate the variable to get the relation in h and k i.e. this relation must contain only h, k and known quantities. Step 4: Express the given relation in h and k in the simplest form and then put x for h and y for k. The relation thus obtained, will be the required equation of the locus of P(h, k). Illustration 4: Find the locus of a point P which divides the line joining (1, 0) and (2 cosq, 2 sinq) internally in the ratio 2 : 3 for all q. Solution: Let the coordinate of point P which divides the line joining (1, 0) and (2cosq, 2sinq) in the ratio 2 : 3 be (h, k) then cosq = , sinq = = (5h-3)2 + (5k)2 = 16 Hence locus is (5x – 3)2 + (5y)2 = 16 Illustration 5: The ends of a rod of length move on two mutually perpendicular lines. Find the locus of the point on the rod, which divides it in the ratio 1 : 2. Solution: Suppose the two perpendicular lines are x = 0 and y = 0 and intercepts a and b are cut respectively on the two lines. Then the two points on these lines are (0, a) and (b, 0). The point P has coordinates given by h = , k = Also = a2 + b2 Thus the required locus is x2 + . 3. STRAIGHT LINE 3.1 GENERAL EQUATION OF A STRAIGHT LINE Any equation of first degree of the form ax + by + c = 0, where a, b, c, are constants always represents a straight line (at least one out of a and b is non-zero). Slope or gradient of a line: If  is the angle at which a straight line is inclined to the positive direction of x-axis, then tan is called the slope or gradient of the line, where 0   < 180 (  90). Note: (i) If a line is parallel to x-axis, then its slope is equal to tan0 = 0. (ii) Slope of a line perpendicular to x-axis is not defined. Whenever we say that the slope of a line is not defined, we mean that the line is perpendicular to x-axis. 3.2 INTERCEPT OF A STRAIGHT LINE ON THE AXES Intercept of a line on x-axis: If a line cut the axis at (a, 0), then ‘a’ is called the intercept of a line on x-axis or the x-intercept. If a line intersect the x-axis at (–2, 0), then the length of intercept is –2 = 2. Intercept of a line on y-axis: If a line cuts y-axis at (0, b) then b is called the intercept of the line on y-axis or y intercept. 3.3 DIFFERENT FORMS OF THE STRAIGHT LINES Slope intercept form: Equation of a line whose slope is m (m = tan) and which cut an intercept c on the y- axis (i.e. which passes through the point (0, c) is given by y = mx + c. Intercept form: The equation of the line which cuts off intercepts a and b on x-axis and y-axis respectively is given by . Thus intercept of a straight line on x-axis can be found by putting y = 0 in the equation of the line and then finding the value of x. Similarly intercept on y-axis can be found by putting x = 0 in the equation of the line and then finding the value of y. Normal Form: The equation of a straight line upon which the length of perpendicular from the origin is p and the perpendicular makes an angle with the positive direction of x-axis is given by x cos + y sin = p Note: In normal form of equation of a straight line p is always taken as positive and is measured from positive direction of x-axis in anticlockwise direction between 0 and 2 . Note: In the normal form x cos +y sin =p, p is always taken as positive. Point-slope form: The equation of a line passing through the point (x1, y1) and having slope m is given by . Two points form: The equation of a straight line passing through two given points (x1, y1) and (x2, y2) is given by Note: Slope of the line passing through two points (x1, y1) and (x2, y2) is given by Parametric form: The equation of line passes through a given point P(x1, y1) and makes a given angle  with the positive direction of the x-axis is given by where r is the distance between the variable point Q (x, y) and the fixed point P(x1, y1). • Any point on the line will be of the form (x1 + r cos, y1 + r sin). For different values of r, we will get different points on the line • Here r will gives the distance of the point Q from the fixed point P(x1, y1). • If P(x1, y1) is any point on the line which makes an angle with the positive direction of x axis, then there will be two points on the line at a distance r from P(x1, y1). One will be relatively upward if r is taken positive and other will be relatively downward if r is taken negative. Illustration 6: Find the points on the line y = x – 2, which are lying at a unit distance from the point (5, 3). Key concept 1: In general case if in any question there is something related to distance, we use the parametric form of line. Solution: Clearly (5, 3) is lying on the given line. Now we will write the parametric equation of the given line passing through (5, 3), which is Here r will represent the distance between (5, 3) and (x, y) Any point on the line is . Now if we take r = 1, we get the coordinate of the point which is at a unit distance from (5, 3) and relatively upward, and if we take r = –1, we will get the point which is also at a unit distance from (5, 3) but relatively downward. Hence the points are Key concept 2: Take any point on the given line in terms of a single variable and use the distance formula. Solution: Any point on the given line can be taken as (,  - 2). Given that the distance between the points (,  - 2) and (5, 3) is 1. Hence the required points are Illustration 7: If the straight line through the point P (3, 4) makes an angle /6 with x-axis and meets the line 12x + 5y + 10 = 0 at Q, find the length of PQ. Solution: The equation of a line passing through P (3, 4) and making an angle /6 with the x-axis is , where r represents the distance of any point on this line from the given point P (3, 4). The co-ordinates of any point Q on this line are If Q lies on 12x + 5y + 10 = 0, then   length PQ = 3.4 POSITION OF A POINT WITH RESPECT TO A GIVEN LINE Let the given line be ax + by + c = 0, and we have to check whether the point (, ) lies above the line or below the line. To check first draw a perpendicular line from P(, ) on the x-axis which intersect the given line at Q. x-coordinate of Q will be  and corresponding to , y-coordinate of Q will be . Now if , the point will lie above the line if point will lie on the line and if point will be lie below the line. Illustration 8: If the point (, 2) lies between the acute angle region formed between the lines y = 2x and y = 3x, then find the range of . Key concept 1: Since y co-ordinate of (, 2) is always non negative, point (, 2) will always lie above the x-axis. Now for acute angle region the point (, ) should lie below the line y = 3x and above the line y = 2x. Solution: (, ) should lie below the lie y = 3x  3 < 2 and (, )should also lie above the line y = 2x  2 > 2  2 < 2 < 3  2 <  < 3 Key concept 2: Join P with the origin now slope of OP should lie between the slope of the lines y = 2x and y = 3x. Solution:  Slope of OP  (slope of the line y = 2x, slope of the line y = 3x)    (2, 3): Key concept 3: First identity the locus of point , that means curve on which P lies for different values of . Then find the portion of the curve lying between the acute angle region between the lines. From the graph it is clear that Solution: First we will find the locus of P x =  and y = 2. Now on eliminating  we get y = x2. Hence the point P will always lie on the curve y = x2. 3.5 POSITION OF TWO POINTS WITH RESPECT TO A LINE Let the given line be ax + by + c = 0 and P(x1, y1), Q (x2, y2) be two given points. Now if  both ax1 + by1 + c and ax2 + by2 + c are of opposite sign, then the points P and Q will lie on the opposite side of the line ax + by + c = 0 and, if  both ax1 + by1 + c and ax2 + by2 + c are of the same sign, then the points P and Q will lie on the same side of the given line. Illustration 9: Find the range of  such that the point (2, 2) and (2, 6) lies on the same side of the line x + y –1 = 0. Key concept: Sign of x + y –1 should be same for both the points. Solution: For (2, 6) x + y – 1 = 2 + 6 –1 is + ve. Hence 2 + 2 –1 > 0  ( + 1)2 – 2 > 0 Common mistake: Generally students solve the inequality ( + 1)2 > 2 like in this way ( + 1)2 > 2   + 1 >   > –1 but this is wrong 3.6 ANGLE BETWEEN TWO STRAIGHT LINES If  is the acute angle between two lines, then tan = where m1 and m2 are the slopes of the two lines and are finite. Notes: • If the two lines are perpendicular to each other then m1m2 = -1. • Any line perpendicular to ax + by + c = 0 is of the form bx – ay + k = 0. • If the two lines are parallel or are coincident, then m1 = m2. • Any line parallel to ax + by + c = 0 is of the form ax + by + k = 0. • If any of the two lines is perpendicular to x-axis, then the slope of that line is not define (infinite). Let m1 = , then = or  = |90 - |, where tan = m2 i.e. angle  is the complimentary to the angle which the oblique line makes with the x-axis. Illustration 10: Find the equation to the sides of an isosceles right-angled triangle, the equation of whose hypotenuse is x - 2y = 3 and the opposite vertex is the point (2, 2). Solution: Clearly other two sides of the triangle will be making an angle of 45o with the given line. That means we have to find the equation of line passing through the point (3, 2) and making an angle of 45o with the given line. Slope of the given line x – 2y = 3 is m1 = 1/2.  tan 45° =  Hence the required equations of the two lines are 3x – y – 7 = 0 and x + 3y – 9 = 0 3.7 EQUATION OF THE STRAIGHT LINE PASSING THROUGH A POINT AND INCLINED TO A GIVE LINE The equations of the lines through the point (x1, y1) and making equal angles  with the given line are y – y1 = mA(x – x1), y – y1 = mB(x - x1). 3.8 THE DISTANCE BETWEEN TWO PARALLEL LINES The distance between two parallel lines: ax + by + c1 = 0 and ax + by + c2 = 0 is . 3.9 PERPENDICULAR DISTANCE OF A POINT FROM A LINE The length of the perpendicular from P(x1, y1) on ax + by + c = 0 is . 3.10 TO FIND THE IMAGE OF A POINT IN A LINE Let the given line be ax + by + c = 0 and we have to find the image of the point P(, ). Working rule: Step 1: Find the equation of line PP’ (where P’ is image of P). Slope of PP’ will be b/a, because PP’ is perpendicular to given line. Hence equation of PP’ is Step 2: Find the point of intersection of line PP’ with the given line and let it be R(p, q). Clearly R is the mid point of PP’. Step 3: Now Also Hence the image is (2p – , 2q – ) Illustration 11: Find the image of (5, 7) about line x –y + 6 = 0 Solution: slope of line PP’ = –1 Hence equation of PP’ y – 7 = – 1 (x – 5)  x + y – 12 = 0 Point of intersect of x + y – 12 = 0 with x – y + 6 = 0 is (3, 9). Hence 3.11 CO-LINEARITY OF THREE POINTS Following methods can be used to prove that three given points P(x1, y1), Q(x2, y2), R(x3, y3) are collinear. Method 1: If P, Q and R are collinear, then area of triangle formed by these three points should be equal to zero Method 2: Any one point of the given three points should divide other two points either internally or externally. for some real m and n. Method 3: Slope of PQ = Slope of QR 3.12 CONCURRENCY OF THREE LINES The condition for 3 lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, a3x + b3y + c3 = 0 to be concurrent is . 3.13 BISECTORS OF ANGLE BETWEEN TWO GIVEN LINES Let a1x+b1y+c1 = 0……..(1) a2x + b2y + c2 = 0….(2) are two intersecting lines. Let any point p(x, y) be any point on either of the two bisectors of angles of (1) and (2). Then p is equidistance from (1) and (2) which are the required equations of the two bisectors of angles between (1) and (2). If the two given lines are not perpendicular i.e. a1 a2 + b1 b2  0, then one of these equation is the equation of the bisector of acute angle and the other that of the obtuse angle. The equation of acute and the obtuse angle bisectors: Method 1 Step 1: Take one of the given lines and let its slope be m1 and take one of the bisectors and let it’s slope be m2. Step 2: If  be the acute angle between them, then find Step 3: If tan > 1 then the bisector taken is the bisector of the obtuse angle and the other one will be the bisector of the acute angle. If tan < 1 then the bisector taken is the bisector of the acute angle and the other one will be the bisector of the obtuse angle. Method 2: If the constant term c1 and c2 in the two equations a1x+b1y+c1 = 0 and a2x + b2y + c2 = 0 are of the same sign, then Case 1: if then will give the equation of obtuse angle bisector and will give the equation of acute angle bisector. Case 2: if then will give the equation of acute angle bisector and will give the equation of obtuse angle bisector. Note: Whether both the lines are perpendicular or not but the angle bisectors of these lines will always be mutually perpendicular. The equation of the bisector of the angle which contain a given point: The equation of the bisector of the angle between the two lines containing the point ( ) is if are of the same signs or if are of the opposite signs The equation of the bisector of the angle containing the origin: Write the equations of the two lines so that the constants c1 and c2 are positive. Then the equation is the equation of the bisector containing the origin. Note: if , then the origin will lie in the acute angle and if then origin will lie in the obtuse angle. Illustration 12: For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0 find equation of the bisector of the (a) acute angle (b) angle which contain (1, 2) (c) angle which contain origin Solution: (a) The equation of given lines are 4x + 3y – 6 = 0 …(1) and 5x + 12y + 9 = 0 …(2) equation of angle bisectors between these two line are ….(3) 9x – 7y – 41 = 0 and 7x + 9y – 3 = 0 Here we will get two equation of bisectors. Two find the acute angle bisector take one equation of the given line and one equation of bisector. Given line be 4x + 3y – 6 = 0 and one bisector be 7x + 9y – 3 = 0. Now Hence bisector 7x + 9y – 3 = 0 will be acute angle bisector. (b) Equation of bisector which contain the point (1, 2) since 4.1 + 3.2 – 6 > 0 and 5.0 + 12.2 + 9 > 0 Hence equation of bisector which bisect the angle which contains the point (1, 2) is  9x – 7y – 41 = 0 (c) Since 4.0 + 3.0 – 6 < 0 and 5.0 + 12.0 +9 > 0. Hence the equation of the bisector which contain the origin is  7x + 9y – 3 = 0 3.14 FAMILY OF LINES Suppose L1= a1x+b1y+c1 = 0 and L2=a2x + b2y + c2 = 0 are two intersecting lines and let the point of their intersection be ( ). Now if we write these two equations in this form (where is a parameter) ………(1) then for different values of , (1) will give different straight lines. Now ( ) always lies on (1) whatever be the value of . Hence (1) represent a family of straight lines passing through the point of intersection of a1x+b1y+c1 = 0 and a2x + b2y + c2 = 0. Note: • Whenever we have to show that a line always passes through a fixed point, we use the concept of family of lines. • Family of lines perpendicular to a given line ax+by+c = 0 is given by bx-ay+k=0, where k is a parameter. • Family of lines parallel to a given line ax+by+c = 0 is given by ax+by+k=0, where k is a parameter. Illustration 13: Find the equation of the straight line which belongs to both of the following family of lines 5x + 3y – 2 + l1 (3x – y – 4) = 0 and x - y + 1 + l2 (2x – y – 2) = 0 Solution. Lines of first family are concurrent at (1, -1) and that of second at (3, 4) \ Required line passes through both of these points \ Equation is 5x – 2y – 7 = 0 Illustration 14: If a, b and c are three consecutive odd integers then prove that the variable line ax + by + c = 0 always passes through (1, –2). Solution: Since a, b and c are three consecutive odd integers, these must be in form of 2n + 1, 2n +3 and 2n + 5 respectively where n I. The given line can be written as (2n + 1)x + (2n +3)y + (2n + 5) = 0  n/2 (x + y + 1) + (x + 3y + 5) = 0. Now we can say this will represent a family of line passing through the point of intersection x + y + 1 = 0 and x + 3y + 5 = 0, which is (1, -2) Illustration 15. The base BC of ABC is bisected at (p, q) and equation of sides AB and AC are px + qy = 1 and qx + py = 1. Then show that the equation of median through A is ( px + qy –1) – . Solution. Any line through A is given by (px + qy – 1) + (qx + py – 1) = 0. If this line represent the median through A then this will be passing through (p, q). Hence  = . Thus the required line is (px + qy –1) – 3.15 PAIR OF STRAIGHT LINES The general equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of straight lines if and h2  ab.  abc + 2fgh - af2 - bg2 - ch2 = 0 and h2  ab. The homogeneous second degree equation ax2 + 2hxy + by2 = 0 represents a pair of straight lines through the origin if h2  ab. If the lines through the origin whose joint equation is ax2 + 2hxy + by2 = 0, are y = m1x and y = m2x, then y2 - (m1 + m2)xy + m1m2x2 = 0 and y2 + xy + = 0 are identical, so that . If  be the angle between two lines, through the origin, then =  . The lines are perpendicular if a + b = 0 and coincident if h2 = ab. Joint Equation of Pair of Lines Joining the Origin and the Points of Intersection of a Curve and a Line: If the line lx + my + n = 0, ((n  0) i.e. the line does not pass through origin) cut the curve ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 at two points A and B, then the joint equation of straight lines passing through A and B and the origin is given by homogenizing the equation of the curve by the equation of the line. i.e. ax2 + 2hxy + by2 + (2gx + 2fy) is the equation of the lines OA and OB Illustration 16: If the lines joining origin to the points of intersection of the curve 2x2 + 3y2 + m = 0 and the line y = 2x + 3 are perpendicular, then find the value of m. Solution. Homogenise the given curve with line, we get the required line Since, the lines are perpendicular Þ Coefficient of x2 + coefficient of y2 = 0 Þ m = -9 Illustration 17: Find the value of unknowns for which the given equation represents two coincident lines x2 + 9y2 + pxy + qx + ry + 1 = 0. Solution: Let the line is ax + by + c = 0 Equation of pair of lines (ax + by + c)2 = 0 a2x2 + 2abxy + b2y2 + 2acx + 2bcy + c2 = 0 Comparing this with given equation a2 = 1 a = 1 b2 = 9 b = 3 c2 = 1 c = 1 p = 2ab = 6 q = 2ac = 2 r = 2bc = 6 3.16 ROTATION OF CO-ORDINATE AXES Let OX, OY be the original axes and OX’ and OY’ be the new axes obtained after rotating OX and OY through an angle  in the anticlockwise direction. Let P be any point in the plane having coordinates (x, y) with respect to axes OX and OY and (x’, y’) with respect to axes OX’ and OY’. Then x = x’ cos – y’ sin, y = x’ sin + y’ cos ...(1) and x’ = x cos + y sin, y’ = – x sin + y cos …(2) Note: The above transformation can also be displaced by a table. x’ y’ x cos –sin y sin  cos • If f(x, y) = 0 is the equation of a curve then it’s transformed equation is f( x’ cos – y’ sin, x’ sin + y’ cos) = 0 4. OBJECTIVE ASSIGNMENT 1. If the algebraic sum of distances of points (2, 1) (3, 2) and (¬-4, 7) from the line y = mx + c is zero, then this line will always pass through a fixed point whose coordinate is (A) (1, 10) (B) (1, 3) (C) (1, 6) (D) (1/3, 10/3) Solution: we have Hence the given line y = (10 ¬– 3c)x + c Hence the line always passes through the point of intersection of y = 10x and 3x = 1, which is (1/3, 10/3) Hence (D) is the correct answer. 2. In centre of the triangle formed by the lines y = x, y = 3x and y = 8 ¬– 3x is (A) (B) (C) (D) Solution: How students used to solve this problem. First they find the vertices and sides of the triangle. And then they use the formula coordinates of the incentre are . It takes a lot of time, a lot of scope of calculation mistake is also there. So here we should use the basic concept of incentre which is “incentre is the concurrency of angle bisectors” Equation of bisector OO1 is Here we will get two equation of the bisectors, but we have to consider the bisector which is bisecting the internal angle of the triangle. For the internal bisector the slope of the bisector should be positive. ….(1) equation of BB1 is …..(2) Hence (C) is the correct answer. 3. Consider the triangle OAB where O = (0, 0), B = (3, 4). If the orthocentre of a triangle is H(1, 4) then coordinate of A is (A) (0, 15/4) (B) (0, 17/4) (C) (0, 21/4) (D) (0, 19/4) Solution: We know that Orthocentre is concurrency of altitudes and line joining Orthocentre to vertex is perpendicular to the opposite side. Let A = (h, k), Where mAH and mOB are slopes of lines AH and OB, which are perpendicular to each other. …(1) Slope of OA Since OA and BH are mutually perpendicular it implies that h = 0. Putting h = 0 in (1) we get k = . Thus orthocentre is Hence (D) is the correct answer. 4. The sides of a triangle are x + y = 1, 7y = x and . Then which of the following is an exterior point of the triangle? (A) orthocentre (B) centroid (C) In-centre (D) none of these Solution: In any triangle centroid and in-centre always lie within the triangle. If the triangle is an obtuse angle triangle then the orthocentre and the circumcentre fall outside of the triangle. Now the angle between 7y= x and is obtuse angle because the first line is inclined at angle less than 45 with the x-axis, where the second line make 150 with the x-axis. Hence (A) is the correct answer. 5. ABC is an equilateral triangle such that the vertices B and C lie on two parallel lines at a distance 6. If A lies between the parallel lines at a distance 4 from one of them then the length of a side of the equilateral triangle is (A) 8 (B) (C) (D) 1 Solution: From the choice of the axis A = (0, 0), B = (2cot, 2), C = (4cot(60 - ), -4) Now (side of equilateral triangle)2 = 4cot2 + 4 = 16 cot2 (60 - ) + 16 Hence the required length = Hence (C) is the correct answer. 6. The equation of the bisector of the acute angle between the lines 2x – y + 4 = 0 and x – 2y = 1 (A) x + y + 5 = 0 (B) x – y + 1 = 0 (C) x – y = 5 (D) x + y + 1 = 0 Solution: Its quits clear from the figure that origin is contained in the acute angle Hence the required bisector is .  x – y + 1 = 0 Hence (B) is the correct answer. 7. The diagonals of the parallelogram whose sides are lx + my + n = 0, lx + my + n’ = 0, mx + ly + n = 0, mx + ly + n’ = 0 include an angle (A) /2 (B)  / 3 (C) (D) Solution: Since the distance between the parallel lines lx + my + n = 0 and lx + my + n’ = 0 is same as the distance between the parallel lines mx + ly + n = 0 and lx + my + n’ = 0. Therefore, the parallelogram is rhombus. Since the diagonals of a rhombus are at right angles, therefore the required angle is /2 Hence (A) is the correct answer. 8. A ray of light is sent along the line which passes through the point (2, 3). The ray is reflected from the point P on x-axis. If the reflected ray passes through the point (6, 4), then the coordinates of P are (A) (B) (C) (D) Solution: Method 1. Let the reflected ray makes an angle  with + ve direction of x-axis, then the incident ray makes angle ( ¬- ) with positive direction of x-axis. Now, the slope of the incident ray ….(1) ….(2) from (1) and (2) we get Method 2. Take the image of P(2, 3) about the x-axis, which is P’. Now P’QR will be collinear. Hence first find the equation of line P’Q and then find the point of intersection of P’Q with the x-axis to get the required point. P’ ≡ (2, -3), equation of P’Q is 7x ¬– 4y = 26. Hence the required point is Hence (A) is the correct answer. 9. The point (2t2 + 2t + 4, t2 + t + 1) lies on the line x + 2y = 1, for (A) All real values of t (B) Some real values of t (C) (D) for no value of t Solution: The given point lies on the line if (2t2 + 2t + 4) + 2(t2 + t + 1) = 1  4t2 + 4t + 5 = 0 Here discriminant = 16 – 4.4.5 = - 64 < 0 Hence for no real value of ’t’ it is possible. Hence (D) is the correct answer. 10. Without changing the direction of coordinates axes, origin is transferred to (, ) so that the linear terms in the equation x2 + y2 + 2x –4y + 6 = 0 are eliminated. The point (, ) is (A) (-1, 2) (B) (1, -2) (C) (1, 2) (D) (-1, -2) Solution: The given equation is x2 + y2 + 2x – 4y + 6 = 0 …..(1) Putting x = x’ +  and y = y’ +  in (1) we get x’2 + y’2 + x’(2 + 2) + y’(2 - 4) + (2 + 2 + 2 - 4 + 6) = 0. To eliminate linear term, we should have 2 + 2 = 0 and 2 - 4 = 0   = -1 and  = 2  (, ) ≡ (-1, 2) Hence (A) is the correct answer. 11. Equation ax2 + 2hxy + by2 = 0 represents a pairs of lines, combined equation of lines that can be obtained by taking the mirror of lines about the x-axis is (A) ax2 + 2hxy + by2 = 0 (B) bx2 + 2hxy + ay2 = 0 (C) bx2 - 2hxy + ay2 = 0 (D) none of these Solution: Let the lines represented by ax2 + by2 + 2hxy = 0 be y = m1x and y = m2x, then m1 + m2 = , m1m2 = a/b. If these lines reflected about the x-axis, there equation becomes y + m1x = 0 and y + m2x = 0 and their combined equation is (y + m1x) (y + m2x) = 0  y2 + xy (m1 + m2) + m1m2x2 = 0  by2 – 2hxy + ax2 = 0 Hence (D) is the correct answer. 12. If the line x cuts the curve x3 + ax2 + bx – 72 = 0 at A, B and C, then OA. OB.OC (Where ‘O’ is origin) is (A) 576 (B) –576 (C) a + b –c – 576 (D) a + b + c – 576 Solution: The line x is passing through the origin and slope is , hence in parametric form the equation of given line can be written as ….(1) Any point on the line (1) is . If the line cuts the given curve, then . This is a cubic equation in r. Roots of this equation r1, r2, r3 will represent OA, OB and OC. Therefore OA.OB.OC = r1r2r3 = 576 Hence (A) is the correct answer. 13. A variable line drawn through the point (1, 3) meets the x-axis at A and y-axis at B. If the rectangle OAPB is completed, where ‘O’ is the origin, then locus of ‘P’ is (A) (B) x + 3y = 1 (C) (D) 3x + y = 1 Solution: Let the line be . Since the line is passing through (1, 3), hence . Now A = (a, 0), B = (0, b)  P = (a, b) Thus locus of ‘P’ is Hence (C) is the correct answer. 14. If the straight lines ax + by + P = 0 and x cos + y sin = P are inclined at an angle /4 and concurrent with straight line x sin - ycos  = 0, then the value of a2 + b2 is (A)1 (B) 0 (C)2 (D) 13 Solution: ON = distance of origin from the line x sin + y cos = P OM = Perpendicular distance of (0, 0) from the line ax + by + P = 0 Now OMN is a right angle triangle with ONM = /4  OM = ON sin /4 =  a2 + b2 = 2 Hence (C) is the correct answer. 15. The least value of , for which the lines x =  + m, y = - 2 and y = mx are concurrent, is (A) (B) (C) 0 (D) 1 Solution: Since lines are concurrent, point of intersection of lines x =  + m and y = -2 which is will also satisfy y = mx  m2 + m + 2 = 0 Since m is real 2 -8  0    . Hence the least value of = . Hence (B) is the correct answer. 16. If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is (A) square (B) circle (C) straight line (D) two intersection lines Solution: Let the perpendicular line be x = 0 and y = 0. According to given condition Which represent a square. Hence (A) is the correct answer. 17. The straight lines L1  4x – 3y + 2 = 0, L2  3x + 4y – 4 = 0 and L3  x – 7y + 6 = 0 (A) form a right angled triangle (B) form a right angled isosceles triangle (C) are concurrent (D) none of these Solution: Since the line L1 and L2 are perpendicular and angle between L1 and L2 is 45 Hence the triangle will be right angled isosceles. Hence (B) is the correct answer. 18. If a,b,c are in H.P. then the straight line always passes through a fixed point, that point is (A) (-1,-2) (B) (-1, 2) (C) (1, -2) (D) (1,-1/2) Solution: Since a, b, c are in H.P.  Hence Hence the line always passes through (1, -2) Hence (C) is the correct answer. 19. The extremitas of the base of an isosceles triangle are (2, 0) and (0, 2). If the equation of one of the equal sides is x = 2 then equation of the other equal side is (A) x = y = 2 (B) x – y + 2 (C) y = 2 (D) 2x + y = 2 Solution: Adjacent figure represents the given isoceles triangle. Clearly the equation of other equal side is y = 2. Hence (C) is the correct answer. 20. If coordinate axes are the angle bisectors of the pair of lines ax2 + 2hxy + by2, then (A) a – b = 0 (B) a2 + b2 = 0 (C) h = 0 (D) b2 + a = 0 Solution: Equations of angle bisectors of the given lines is It should be same as xy = 0 Hence (C) is the correct answer. 21. P (x, y) is called a good point if x, y  N. Total number of good points lying inside the quadrilateral formed by the line 2x + y = 2, x = 0, y = 0 and x + y = 5, is equal to (A) 4 (B) 2 (C) 6 (D) 1 Solution: Adjacent figure indicates that there are exactly six good points inside the quadrilateral ABCD. Hence (C) is the correct answer. 22. Consider a family of straight lines (x + y) +  (2x – y + 1) = 0. Equation of the straight line belonging to this family that is farthest form (1, –3) is (A) 13y + 6x = 7 (B) 15y + 6x = 7 (C) 13y – 6x = 7 (D) 15y – 6x = 7 Solution: Given family is concurrent at .If Q = (1, –3) then mPQ . Now member of the family that is farthest from ‘Q’ will have it’s slope as . Thus equation of required line is i.e. 15y – 6x – 7 = 0 Hence (D) is the correct answer. 23. The sides of a rectangle are x = 0, y = 0, x = 4, and y = 3. The equation of straight line having slope that divides the rectangle in to two equal halves is (A) 2y = x + 1 (B) 2x = y + 1 (C) 2y + x = 1 (D) 2x + y = 1 Solution: Let the required line be 2y = x + c. We have Area of rectangle ABCD is 12 sq. units Thus area of trapezium OAA1B is 6 = Thus required line is 2y = x + 1 Hence (A) is the correct answer. 24. If the point P(a, a2) lies completely inside the triangle formed by the lines x = 0, y = 0 and x + y = 2, then exclusive range of ‘a’ is (A) a  (0, 1) (B) (C) (D) Solution: Clearly Also a2 + a – 2 < 0 Hence (A) is the correct answer. 25. A light ray emerging from the point source placed at P(2, 3) is reflected at a point ‘Q’ on the y-axis and then passes through the point R (5, 10). Coordinates of ‘Q’ is (A) (0, 3) (B) (0, 2) (C) (0, 5) (D) (0, 1) Solution: If P1 be the reflection of P in y–axis then P1 = (–2, 3) Equation of line P1R is (y – 3) = Hence (C) is the correct answer. 26. The distance of the line 2x – 3y = 4 from the point (1, 1) in the direction of the line x + y=1is (A) (B) 5 (C) (D) none of these Solution: Parametric equation of line PQ Any point on this line Now Hence the distance is Hence (A) is the correct answer. 27. The equation of the bisector of angle between the lines x + y = 1 and 7x – y = 3 that contains the point (2, 3) is (A) 3x + y – 1 = 0 (B) x – 3y + 1 = 0 (C) 2x + y – 1 = 0 (D) x – 2y + 1 = 0 Solution: Equation of bisector Now at (2, 3) x + y –1 > 0 and 7x – y – 3 > 0. Hence the equation of bisector contains the point (2, 3) is Hence (B) is the correct answer. 28. If the point (1+cos , sin ) lies between the region corresponding to the acute angle between the lines 3y=x and 6y=x then (A) (B) (C) (D) none of these Solution: Slope of OP should lie between Hence (D) is the correct answer. 29. If the distance of a point (3, k) from the line 3x + 7y –1 = 0 is then k is equal to (A) 1 (B) –1 (C) (D) none of these Solution: Hence (B) is the correct answer. 30. The number of points on the line 3 x + 4y = 5, which are at a distance of sec2q +2 cosec2q, q Î R, from the point (1, 3), is (A) 1 (B) 2 (C) 3 (D) infinite Solution: The distance of the point (1,3) from the line 3x+4y=5 is 2/5. Now the value of is greater than 2/5, hence two such lines will be possible. Hence (B) is the correct answer. 31. The locus of point of intersection of lines xcosa + ysina = a and xsina - ycosa = b is (a is variable) (A)2(x2 + y2) = a2 +b2 (B)x2-y2 = a2 – b2 (C) x2 + y2 = a2 + b2 (D)None of these Solution: Let the point of intersection be (h, k) Hence hcos + k sin = a Hence the locus is x2 + y2 = a2 + b2 Hence (C) is the correct answer. 32. If { a, b  R – {0}}, then the line , will pass through (A) (2, 1) (B) (1,1) (C) (2,3) (D) (0, 0) Solution: (B) Given a4b4 – a4– b4 = 2a2b2  a4b4= (a2+b2)2  a2 + b2 = a2b2 Given line b2x +a2y = a2b2  b2x+a2y = a2+b2  (x – 1) b2 + a2(y – 1) clearly the line will be passing through (1,1). Hence (B) is the correct answer. 33. Point P ( a, b) lie on the line y = x + 1. P is shifted in the direction perpendicular to given line so that it meets x –axis at ( , 0) , then (A)  =2a + 1 (B) = a +1 (C)  = a – 1 (D)  = a +2 Solution: parametric equation of line passing through (a, b,) and  to y = x + 1 is = any point on this line is given by If this point is lying on the X-axis, then b + = 0   = – Hence the point is (a – b, 0) Hence (A) is the correct answer. 34. If x2 – 7xy – y2 = 0 , x + y = 2 , represents the sides of a triangle, then distance between orthocenter and the side of triangle is (A) 2 (B) 3 (C) 2 (D) Solution: Clearly x2– 7xy+y2 =0 will represent two perpendicular lines  triangle will be a right angled triangle and ortho-centre will be the right angled vertex which is (0, 0). Clearly distance of (0, 0) from x + y – 2=0 is . Hence (C) is the correct answer. 35. The foot of the perpendicular on the line 3x + y =  drawn from the origin is C if the line cuts the x–axis and y–axis at A and B respectively then BC : CA is (A) 1: 3 (B) 3 : 1 (C) 1: 9 (D) 9 : 1 Solution: Clearly tan  = from  OCA tan  =  OCB tan  =  = = Hence (D) is the correct answer. 36. A straight line x=y meets the parallel lines 3x – 4y = 6 and 6x – 8y + c = 0 at points P and Q respectively such that (where O is the origin). Then value of c (c > 0) is (A) 10 (B) 14 (C) 16 (D) 9 Solution: Given parallel lines are 3x – 4y = 6 ………(i) and 3x – 4y + = 0 ……..(ii) Now the equation of line OP is = =  any point on this line is given by . If=OP, then the point will also satisfy the equation (i). Hence OP = – 6 . Similarly OQ = =  C = 9 Hence (D) is the correct answer. 37. The equation of acute angle bisector between lines and is (A) y=x (B) x+y=0 (C) y=x+1 (D) y=-x+1 Solution: Give lines are y = x and y = x Clearly inclination of both the lines with the positive direction of x–axis are 75 and 150. Clearly the inclination of bisector will be 450. Hence its equation is y = x. Hence (A) is the correct answer. 38. A straight line passing through P(3, 1) meet the coordinate axes at A and B. It is given that distance of this straight line from the origin ‘O’ is maximum. Area of triangle OAB is equal to (A) 50/3 sq. units (B) 25/3 sq. units (C) 100/3 sq. units (D) 20/3 sq. units Solution : Distance of the given line will be maximum when line is perpendicular to OP  slope of OP line should be – 3. Hence the line is 3 x + y = 10. Hence (A) is the correct answer. 39. Equation of the bisector of angle B of the triangle ABC is y = x. If A is (2, 6) and B is (1, 1); equation of side BC is (A) 2x + y – 3 = 0 (B) x – 5y + 4 = 0 (C) x – 6y + 5 = 0 (D) none of these Solution: Drop a perpendicular on y = x from A. clearly BAA’’ and BA’A’’ will be congruent  AA’’ = A’A’’. Hence image of (2,6,) w. r. t. y = x will give A’ which is (6,2). Hence (B) is the correct answer. 40. Vertex opposite to the side x + y – 2 = 0 of the equilateral triangle, with centroid at the origin; is (A) (– 1, 1) (B) (2, 2) (C) (– 2, – 2) (D) none of these Solution: Equation of line passing through (0, 0) and A. A = (–2, –2) Hence (C) is the correct answer. 41. If the line y = x cuts the curve x3 + y2 + 3x2 + 8 = 0 at the points A, B, C, then OA.OB.OC (O being origin) equals (A) -32 (B) -64 (C) 108 (D) -100 Solution: Parametric equation of line Any point on this line At point of intersection this point will also satisfy x3 + y2 + 3x2 + 8 = 0 Hence (B) is the correct answer. 42. Two points A and B move on the x –axis and the y – axis respectively such that the distance between the two points is always same. The locus of middle point of AB is (A) a straight line (B) a circle (C) a parabola (D) an ellipse Solution: Given a2 + b2 = P2 (P is constant) Hence 4h2 + 4k2 = P2 Hence (B) is the correct answer. 43. If one vertex of an equilateral triangle of side 2 is the origin and another vertex lies on the line x= y then third vertex can be (A) (0, 2) (B) (- , -1) (C) (0, -2) (D) ( ,1) Solution: Clearly the third vertex will lie on the y-axis. Hence the points are (0, 2) or (0, –2) Hence (A) is the correct answer. 44. A line passing through the point (2, 2) and the axes enclose an area . The intercepts on the axes made by the line are given by the two roots of (A) (B) (C) (D) none of these Solution: Area ……(i) Equation of line passing through (2, 2) ……(ii) 2(a + b) = ab = Here intercepts on axes made by the line are given by x2 – x + 2 = 0 Hence (C) is the correct answer. 45. In an isosceles right angled triangle, a straight line drawn from the mid-point of one of equal sides to the opposite angle. It divides the angle into two parts, and . Then tan and tan are equal to (A) ½,1/3 (B) 1/3,1/4 (C) 1/5,1/6 (D) none of these Solution: Clearly tan and Hence (A) is the correct answer. 46. Area of the triangle formed by the line x + y = 3 and angle bisector of the pair of straight lines is (A) 2 sq.units (B) 4 sq.units (C) 6 sq.units (D) 8 sq.units Solution: The equation x2 – y2 +2y = 1 represents x – y + 1 = 0 and x + y -1 = 0 Hence equation of their angle bisector are x = 0 and y = 1. Hence (A) is the correct answer. 47. Let P  (-1, 0), Q  (0, 0) and R  (3, 33) be three points. Then the equation of the bisector of the angle PQR is (A) (B) x + (C) 3x + y = 0 (D) x + Solution: Inclination as QR is 60, clearly angle bisector will be inclined at an angle of 120 with the x-axis. Hence (C) is the correct answer. 48. If a and b are real numbers between 0 and 1 such that the points (a, 1), (1,b) and (0,0) form an equilateral triangle, then 2(a + b) – ab is equal to (A) -1 (B) 0 (C) 1 (D) 2 Solution: The given vertices are P(a,1), Q(1,b) and R(0,0) and the triangle is equilateral. Hence apply the conditions that angle between PR and RQ should be 600 and QP = PR. Hence (A) is the correct answer. 49. The number of integer values of m, for which the x-coordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer, is (A) 2 (B) 0 (C) 4 (D) 1 Solution: On findings x-coordinate of point of intersection this should be an integer Here two integer values of m are possible for which x-coordinate of intersectio n is also integer. Hence (A) is the correct answer. 50. The incentre of the triangle with vertices (1, 3), (0, 0) and (2, 0) is (A) (B) (C) (D) Solution: Since the given triangle is equilateral hence in centre will contact with centroid which is Hence (D) is the correct answer. 51. If x1, x2, x3 as well as y1, y2, y3 are in G.P., with the same common ratio, then the points (x1, y1), (x2, y2) and (x3, y3) (A) lie on a st. line (B) lie on an ellipse (C) lie on a circle (D) Are vertices of a triangle Solution: Since x1, x2, x3 and y1, y2, y3 are in G.P Now, Area = Hence the points are collinear Hence (A) is the correct answer. 52. Let PQR be a right angled isosceles triangle, right angled at P(2,1). If the equation of the line QR is 2x+y=3, then the equation representing the pair of lines PQ and PR is (A) (B) (C) (D) Solution: The line PQ and PR will be inclined at an angle of 45 with the line 2x + y = 3 Hence the lines are y–1 = (x-4) and y – 1 = 3(x – 2) Find the equation of pair of lines by multiplying these two equations. Hence (B) is the correct answer. 53. If P(1, 2), Q(4, 6), R(5,7) and S(a, b) are the vertices of a parallelogram PQRS, then (A) a = 2, b = 4 (B) a = 3, b = 4 (C) a = 2, b = 3 (D) a = 3, b = 5 Solution: Here we will use the property that diagonals in parallelogram bisect each other. Hence (C) is the correct answer. 54. If P  (1, 0), Q  (-1, 0) and R  (2, 0) are 3 given points, then locus of the point S satisfying the relation SQ2 + SR2 = 2SP2 is (A) a straight line parallel to x-axis (B) a circle passing through the origin. (C) A circle with the centre at the origin (D) a straight line parallel to y-axis. Solution: (D) Let the point S be (h, k). According to given condition  2h + 3 = 0 which is a straight line parallel to y-axis. Hence (D) is the correct answer. 55. If the line x = y meets the lines x = 1, x = 2… x = n, at points A1, A2,…. An respectively then (OA1)2 + (OA2)2 …. + (OAn)2 is equal to (A) 3n2 + 3n (B) 2n3 + 3n2 + n (C) 3n3 + 3n2 + 2 (D) (3/2) (n4 + 2n3 + n2) Solution: Equation of the line in parametric form is . Any point on the given line can be taken as . Now ,if r = OA then the point will also satisfy the equation x = 1. Similarly OA2 = . Hence OA12 + OA22 + …….. + OA32 +………..+OAn2 = = = 2n3 + 3n2 + n Hence (B) is the correct answer. 56. In which ratio the line joining (2, 3) and (4, 1) divides the line segment joining (1, 2) and (4, 3)? (A) 1 : 2 (B) 2 : 1 (C) 1 : 1 (D) 2 : 3 Solution : The equation of line joining (2, 3) and (4, 1) is or x + y = 5 ......(i) Let the line joining (1, 2) and (4, 3) is divided in the ratio K : 1. The point where it is divided is But the point is lying on (i) or 4K + 1 + 3K + 2 = 5K + 5 or K = 1  Required ratio is 1 : 1 Hence (C) is the correct answer. 57. If equation 3x – 4y = 8 and 2px + 3qy + 12 = 0 represent the same lines then the values of p and q are : (A) 1, 2 (B) 4, 9 (C) 2, 3 (D) None of these Solution : Both the equation 3x – 4y = 8 and 2px + 3qy + 12 = 0 represent the same lines and q = 2 Hence (D) is the correct answer. 58. If equation y = mx + c and x cos  + ysin  = p represent the same line, then : (A) (B) (C) (D) Solution : Since the two equations represents the same line therefore  c cos  = –mp and  c cos  = p  m2p2 + p2 = c2  c2 = p2(1 + m2) Hence (A) is the correct answer. 59. The perpendicular distance between 3x + 4y – 5 = 0 and 6x + 8y – 45 = 0 is: (A) 2 (B) 3.5 (C) 4 (D) 4.5 Solution : By putting x = 0 and y = 0 in the equation we get negative values so the roots lie on one side of the two lines. The length of perpendicular from (0, 0) to 3x + 4y – 5 = 0 is The length of perpendicular from (0, 0) to 6x + 8y – 45 = 0 is Distance between the lines . Hence (B) is the correct answer. 60. The angle between the lines represented by x2 – 7xy + 12y2 = 0 is : (A) 45° (B) 60° (C) 15° (D) None of these Solution : We know that if the angle between the lines ax2 + 2hxy + by2 = 0 is  then Here a = 1, b = 12 and So Hence (D) is the correct answer.