PHYSICS-15-10- 11th (PQRS) SOLUTION

https://docs.google.com/document/d/13DpSs9CbvFPXGbk9Qeb7IV1KnhL-AzvL/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true The Straight Line Recall of Cartesian system of rectangular co-ordinates in a plane, distance formula, area of a triangle, condition for collinearity of 3 points and section formula. Centroid & incentre of a triangle, locus and its equation, translation of axes, slope of a line, parallel and perpendicular lines, intercepts of a line on the co-ordinate axes. Coordinate geometry is the study of geometry using algebra. The plane curves which would be considered are : straight line, circle, parabola, ellipse and hyperbola. The system of coordinates used here is cartesian system of coordinates, introduced by philosopher Descartes. It is by far the most important system. Cartesian System of Rectangular Co-ordinates To identify a point P in the two dimensional plane, we associate with it, an ordered pair of numbers, in the following way: Let OX and OY be two fixed perpendicular straight lines in the plane of the paper. X The line OX is called, the axis of x ; the line OY is called, the axis of y ; the two together are called the axes of coordinates. The point O is called the origin (of reference) and OXY is referred to as the Cartesian system. We shall restrict ourselves to rectangular Cartesian system (where OX and OY are perpendicular). From any point P in the plane, draw lines PM and PN parallel to Y and X axes respectively. The distance OM = x is called as abscissa of P ; the distance ON = y is called as ordinate of P ; the two together, i.e. (x, y), are referred to as the coordinates of P. Distances measured along OX are positive and those measured opposite to it are negative. Distances measured in the direction of OY are positive and those measured opposite to it are negative. OXY = 1st quadrant Abscissa & ordinate both +ve OX′Y = 2nd quadrant Abscissa –ve & ordinate +ve OX′Y′ = 3rd quadrant Abscissa –ve & ordinate –ve OXY′ = 4th quadrant Abscissa +ve & ordinate –ve C H A P T E R CHAPTER INCLUDES : Ca rtesia n system of rectangular co-ordinates Distance formula Are a of triangle and condition for collinearity of 3-points Area of a plane polygon Section formula (mid point) C entroid, inc entre & ex-centres of a triangle Locus & its equation Shifting of origin (Translation of axes) Rotation of axes Slope of a line Intercepts of a line General equation of a straight line Various forms of equation of straight line Conditions for two lines to be intersecting, parallel or perpendicular Point of Intersection of two lines Angle between two lines Concurrency of lines Family of lines Angle bise ctors between two lines Area of parallelogram or rhombus Tricks for brief solutions Solved examples In a triangle ABC, the maximum of the three abscissa is –2 and minimum of three ordinates is +1. The triangle shall lie in the quadrant : (1) OXY (2) OX′ Y (3) OX′ Y′ (4) OXY′ Solution : Answer (2) Since maximum of abscissa is –2, it is clear that x1, x2, x3 all are less than zero i.e. on the left of Y-axis, so either OX′Y or OX′Y′ quadrants are possible. Since minimum of ordinates is +1, hence all ordinates are positive i.e. above X-axis, so OX′Y is the natural choice. DISTANCE FORMULA Distance between points P (x1, y1) and Q (x2, y2) i.e. length of line PQ is given by PQ = Distance of point P(x1, y1) from origin is given by OP = Three points A(2, 3), B(2, 6), C(5, 3) form a triangle. The triangle ABC is (1) Equilateral (2) Scalene (3) Right angled & non-isosceles (4) Right angled isosceles Solution : Answer (4) AB = = 3 AC = = 3 BC = = 3 AB2 + AC2 = BC2 hence triangle is right angled isosceles. AREA OF TRIANGLE If A (x1, y1), B (x2, y2) and C (x3, y3) form a triangle then area of ΔABC is given by x1 Δ = x2 x3 y1 1 y 2 1 y3 1 x1 or Δ = 1 | x2 2 x3 x1 y1 y2 | = 1 {x (y − y ) + x (y − y ) + x (y − y )} 'Stair-method' y3 2 y1 Area of ΔOAB, where O is origin of co-ordinate system is given by Δ = 1 (x y − x y ) 2 1 2 2 1 Cor. : If a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 and a3x + b3y + c3 = 0 are the sides of a triangle, then area of triangle a1 Δ = a2 a3 b1 c1 b2 c2 b3 c3 where C1, C2, C3 are cofactors of c1, c2, c3 in the determinant above. Condition for Collinearity of 3-points If A, B & C are collinear, the area of triangle ABC has to be zero i.e. x1 y1 x2 y 2 x3 y 3 1 1 = 0 1 or x1(y 2 − y 3 ) + x2 (y3 − y1) + x3 (y1 − y 2 ) = 0 AREA OF A PLANE POLYGON Let A1, A2 An are the vertices of a n sided plane polygon, its area is given by Stair method Area of Polygon = x1 y1 x y2 y3 xn yn x1 y1 i.e. = [x1y 2 x2 y1) + (x2 y 3 x3 y 2 ) + + (xn y1 x1yn )] Important results To prove that a given four sided figure is : Square : Prove that four sides are equal & diagonals are equal Rhombus : Four sides equal, diagonals unequal. Rectangleyes : Opposite sides equal & diagonals equal. Parallelogram : Opposite sides equal & diagonals unequal. The area of pentagon ABCDE, where A(–2, 3), B(3, 3), C(5, 4), D(2, 6) and E(–2, 5) is (1) 29 units (2) Solution : Answer (4). By Stair method 47 units (3) 2 29 units (4) 15 units 2 − 2 3 3 3 Area of given pentagon = 1 | 5 4 | 2 2 6 − 2 5 − 2 3 = (−6 − 9 + 12 − 15 + 30 − 8 + 10 + 12 − 6 + 10) = (30) = 15 units The area of polygonal figure A(–2, 3), B(4, 3), C(2, 5), D(2, 3), E(0, 3) and F(0, 1) is Zero (2) 4 (3) 8 (4) None of these Solution : Answer (2) By Stair method area comes out to be zero, whereas area comes out to be 4 units. Here ABCD movement is anticlockwise where as EFAE is clockwise, Hence +ve and –ve area have cancelled. Care should be taken to roughly plot the figure on co-ordinate axes. Given five points O(0, 0), A(0, –2), B(3, 0), C(5, 1) and D(6, 2) and a line 2x – 3y = 6. Choose the wrong statement. Area ΔOAB = Area of triangle formed by given line with co-ordinate axes. Area ΔABC < Area ΔOAB Area ΔABD is zero Area ΔABC > Area ΔBDC Solution : Answer (4) Area ΔOAB = (x1y 2 − x2 y1 = 0 + 6 = 3 units Area bound by x = 0, y = 0 and 2x – 3y – 6 = 0 is 1 0 1 | 0 1 2 2 − 3 0 0 |= 1 | −6 | − 6 2 = 3 units ΔABC = 0 1 | 3 − 2 1 0 1 |= 1 | 2(−2) + 1(3) |= 1 units 2 5 1 1 2 2 ΔABD = 0 − 2 1 3 0 1 = 6 2 1 [+2(3 − 6) + 1(6)] = 1 | 0 |= 0 2 ΔBDC = 3 0 1 6 2 1 = 1 [3(2 − 1) + 1(6 − 10)] = 1 | −1|= 1 units 5 1 1 2 2 2 Hence answer (4) is wrong SECTION FORMULA Coordinates of the point R (x, y ) which divides the join of the points P(x1, y1) and Q(x2, y2). internally, in the ratio m1 : m2, are x = m1x2 + m2 x1 m1 + m2 y = m1y 2 + m2 y1 m1 + m2 m1 m2 P R Q externally, in the ratio m1 : m2, are m1 x = m1x2 − m2 x1 m1 − m2 y = m1y 2 − m2 y1 m1 − m2 R Q P m2 where m1 ≠ m2 ⎛ x1 + x2 , y1 + y 2 ⎞ Cor 1 : Mid point (m1 = m2) of PQ is ⎜ 2 2 ⎟ Cor 2 : If R ( x, y ) divides P (x1, y1) and Q (x2, y2) in the ratio λ : 1 (λ > 0) then x = λx2 ± x1 ; λ ± 1 y = λy 2 ± y1 λ ± 1 Here '+' sign is taken for internal division & '–' sign for external division. Cor 3 : For finding the ratio of division, use λ : 1. If λ comes out to be positive, it indicates internal division otherwise external if λ is negative. ⎡− (Ax1 + By1 + C) ⎤ Cor 4 : Line Ax + By + C = 0 divides join of P (x1, y1) and Q (x2, y2) in the ratio ⎢ (Ax2 By 2 + C )⎥ Diagonal AC & BD of quadrilateral ABCD are divided in the ratio of λ : 1 and t : 1 respectively by their point ⎛ 6 , 46 ⎞ ⎛ 1 , 37 ⎞ of intersection, where A(0, 2), B(3, 3), C ⎜ ⎝ ⎟ and D ⎜ 5 ⎠ ⎝ ⎟ are the vertices. Then λ and t respectively ⎠ are (1) 4, 5 (2) 5, 4 (3) 3, 2 (4) 6, 5 Solution : Answer (2) Equating the x coordinate of intersection point, we get 6λ t + 3 5 = 2 …(i) λ + 1 t + 1 equating y coordinate of intersection point, we get 46λ + 2 37t + 3 5 = 4 …(ii) λ + 1 t + 1 solving equations (i) & (ii), we get λ = 5 t = 4 CENTROID, INCENTRE & EX-CENTRES OF A TRIANGLE If vertices of ΔABC have coordinates A(x1, y1), B(x2, y2) and C(x3, y3), then ⎛ x1 + x2 + x3 , y1 + y 2 + y 3 ⎞ Coordinates of its centroid are ⎜ ⎟ ⎝ 3 3 ⎠ ⎛ ax1 + bx2 + cx3 , ay1 + by 2 + cy3 ⎞ Coordinates of its incentre are ⎜ a + b + c a + b + c ⎟ where a = BC, b = AC and c = AB ⎝ ⎠ A Coordinates of excentre Ia are ⎛ − ax1 + bx2 + cx3 , − ay1 + by 2 + cy3 ⎞ ⎜ − a + b + c − a + b + c ⎟ If G, C and H denote the centroid, circumcentre and orthocentre respectively of ΔABC, then, G, C and H are collinear and G divides CH internally in the ratio 1 : 2. Circumcentre of Triangle It is the point of intersection of perpendicular bisectors of sides, so its distance from all three vertices is same. If O (x, y) be circumcentre of ΔABC, [A(x1, y1), B(x2, y2), and C(x3, y3)] then OA2 = OB2 = OC2 i.e. (x − x1)2 + (y − y1)2 = (x − x 2 )2 + (y − y 2 )2 = (x − x3 )2 + (y − y 3 )2 Alternative method I If D, E, F are mid points of BC, AC and AB respectively, then when solved shall give (x, y). Slope of BC × Slope of OD = –1 ⎫ Slope of CD × Slope of OE = –1 ⎪ Slope of AB × Slope of OF = –1 ⎪ Alternative method II Circumcentre is given by Solving any two we get x & y A ⎡ x sin 2A + x sin 2B + x sin 2C y sin 2A + y sin 2B + y sin 2C ⎤ ⎢ 1 2 3 , 1 2 3 ⎥ ⎣ sin 2A + sin 2B + sin 2C sin 2A + sin 2B + sin 2C ⎦ B C Orthocentre It is the point of intersection of the perpendicular drawn from the vertices of the triangle on the opposite sides. When vertices and the angles of the triangle are given, then orthocentre is given by, ⎛ x1 tan A + x2 tan B + x3 tanC , y1 tan A + y 2 tan B + y 3 tanC ⎞ ⎝ tan A + tan B + tanC tan A + tan B + tanC ⎠ 3. The orthocentre of the triangle with vertices (0, 0) (x1, y1) and (x2, y2) is ⎛ x1x2 – y1y 2 ⎞ ⎛ x1x2 + y1y2 ⎞⎫⎪ ⎨(y1 – y 2 ) ⎜ x y – x y ⎟ , (x1 – x2 ) ⎜ x y ⎟⎬ – x y ⎩⎪ ⎝ 2 1 1 2 ⎠ ⎝ 1 2 2 1 ⎠⎪⎭ Triangle ABC has its vertices as (5 from origin. 2,5), (3 2,9) and (− 2,2) respectively. The centroid is at a distance (1) Solution : Answer (2) ⎛ 7 2 16 ⎞ (2) (3) 118 (4) 3 108 5 G, Centroid = ⎜ , ⎟ ⎜ ⎟ ⎝ ⎠ OG = = 1 98 + 256 = 3 354 = 3 LOCUS AND ITS EQUATION When a point moves, so as always to satisfy a given condition or conditions, the path it traces out is called as its locus under these conditions. Equation to the locus (or curve) is the relation between coordinates of an arbitrarily chosen point on the curve and which relation holds for no other points except those lying on the curve. Standard method for writing equation of a locus Assume the point whose locus (path) is to be found is (h, k). Make the equation involving (h, k) as per the conditions given. Simplify this equation. Substitute h with x and k with y in the simplified form of equation & you get the equation of locus. Find the equation of the locus of a point so that sum of its distance from two given point P(3, 2) and Q(4, 3) is 4. Solution : Let the required variable point be R (h, k) PR + QR = 4, Hence + = 4 3. h2 + k 2 + 13 − 6h − 4k = 16 + h2 + k 2 + 25 − 8h − 6k − 8 ⇒ 2h + 2k − 28 = −8 ⇒ h + k − 14 = −4 ⇒ h2 + k 2 + 196 − 28h − 28k + 2hk = 16h2 + 16k 2 + 400 − 32h − 24k ⇒ 15h2 + 15k 2 − 2hk − 4h + 4k + 204 = 0 4. Hence locus is Shifting of Origin Change of axes, by changing origin, the direction of axes remaining the same. Let OXY and O′X′Y′ be two rectangular Cartesian system of axes. Let P be any point in the plane of the axes and let P and O′ have coordinates (x, y) and (h, k) respectively with respect to OXY system. X Then the coordinates (x′, y′) of P with respect to the system OX′Y′ are given by x = x′ + h ; y = y ′ + k Rotation of axes Change of axes (without changing the origin), by changing the direction of axes, both systems of coordinates being rectangular. If a point P in the plane of OXY has coordinates (x, y) and Y′ (x′, y′) with respect to the system OXY and OX′Y′ respectively, then x = x′cos θ − y ′sinθ y = x′sin θ + y ′cos θ O X SLOPE OF A LINE One of the simplest loci we come across in geometry is a straight line ; the geometrical property which is true in case of this locus is that slope between any two points on a line, is a constant m. This constant is known as slope of the line. If A(x1, y1) and B(x2, y2) are any two points then slope between A and B is defined as : m = y1 − y 2 x1 − x2 = tan θ : 0 ≤ θ < π ; θ ≠ π 2 where θ is the angle of inclination of the line joining A and B with the positive direction of x-axis. If x1 = x2 then slope is not defined. Points A(x1, y1), B(x2, y2) and C(x3, y3) are collinear (i.e. lie on a straight line) if slope between A and B is equal to slope between B and C. ⎛− a ⎞ Slope of line ax + by + c = 0 is ⎜ ⎟ ⎝ b ⎠ Two lines with slopes m1 and m2 are Parallel if m1 = m2 Perpendicular if m1m2 = –1 Intercepts of a Line Let a line L ≡ ax + by + c = 0, intersects OX - axes at A and OY - axes at B, then OA and OB are called x-intercept and y-intercept of line respectively. For x - intercept, substitute y = 0 in the equation i.e. ax + c = 0 ∴ x = – c/a is the x - intercept Similarly for y - intercept put x = 0 by + c = 0 ∴ y = – c/b is the y - intercept GENERAL EQUATION OF A STRAIGHT LINE An equation of the form ax + by + c = 0 where a & b both are not zero simultaneously, represents a straight line. Slope of a line is denoted by m = tanθ; where θ is the inclination of line to the positive direction of x-axis. (θ measured anticlockwise from +ve x-axis is taken as positive). Slope of x-axis – zero Slope of y-axis – Infinite (undefined) Slope of line equally inclined with both axes is either 1 or –1 (θ = 45° or 135°). Slope of a line joining two points (x1, y1) and (x2, y2) is m = y2 − y1 = y1 − y2 x2 − x1 x1 − x2 Slope of line ax + by + c = 0 is Collinearity of three points A, B & C can be checked by The three vertices of a triangle are A(5, 5), B(7, 3), C(1, 2). Find the slope of median from A. Solution : D = ⎛ 1+ 7 , 2 + 3 ⎞ A(5, 5) ⎜ ⎟ ⎝ 2 2 ⎠ ⎛ 4, 5 ⎞ = ⎜ ⎟ ⎝ 2 ⎠ A = (5, 5) C D Hence slope of AD = 5 − 5 / 2 5 − 4 5/ 2 = 1 = 5/2 (1, 2) B(7, 3) The intercepts of a line ax + by + c = 0 on x and y-axes can be found by substituting y = 0 and x = 0 respectively in the equation i.e., x intercept ⇒ ax + c = 0 i.e., x = –c/a y intercept ⇒ by + c = 0 i.e., y = –c/b Hence length of x intercept = |–c/a| and length of y intercept is |–c/b| Equation (i) x-axis is y = 0 (ii) y-axis is x = 0 Equation of line (a is a non-zero constant) parallel to x-axis is y = a parallel to y-axis is x = a VARIOUS FORMS OF EQUATION OF STRAIGHT LINE Line with slope m and a given point (x1, y1) on it y − y1 = m (x − x1) (Slope point form) Line with two given points (x1, y1) and (x2, y2) on it y − y1 y1 − y2 = x − x1 x1 − x2 (Two point form) Line with given slope m and intercept c on y-axis y = mx + c (Slope intercept form) Line with given intercepts a and b on x and y axes respectively x + y = 1 (Double intercept form) a b Line at perpendicular distance p from the origin and where, the perpendicular makes angle α with OX x cos α + y sin α = p (Normal or perpendicular form) Line making an angle α with OX and passing through (x1, y1) and r as the directed distance of any point P(x, y) on the line x = x1 + r cos α ; y = y1 + r sin α (Symmetric or parametric form) Find the co-ordinates of two points which are 3 distance from the point (1, 3) and lie on a straight line 2π passing from this point and inclined at 3 angle to x-axis. Solution : Clearly symmetric or parametric form shall be most helpful in such questions x = x1 + r cos α; y = y1 + r sin α α = 2π/3 (given) x1 = 1 (given) y1 = 3 (given) r = ±3 (given) So points are x = 1± 3 2 cos 2π / 3 = 1± 3 2⎛− 1 ⎞ = 1μ 3 ⎜ ⎟ ⎝ 2 ⎠ ⎛ 3 ⎞ y = 3 ± 3 2 sin2π / 3 = 3 ± 3 2⎜ ⎟ = 3 ± ⎛ ⎜1+ ⎝ 3 , 3 − ⎞ ⎛ ⎟ and ⎜1+ ⎠ ⎝ ⎜ ⎝ 3 , 3 − ⎟ ⎠ ⎞ ⎟ are the required points. ⎠ Reduction of General Equation to Different Standard Forms ax + by + c = 0 ⇒ y = −ax − c Slope - intercept form (y = mx + c) b b m = −a b −c & y intercept = b Hence, slope = coefficient of x coefficient of y ax + by + c = 0 ⇒ ax + by = –c ⇒ (−c / a) + = 1 (−c / b) intercept Here x intercept = –c/a and intercept on y-axis is –c/b. ax + by + c = 0 ax + by = – c a x + b y = − c Normal or perpendicular form (x cosα + y sinα = p) If c < 0 then –c > 0 Hence, p = −c and cosα = a ,sinα = b If c > 0 then −a x − = p = c ,cosα −a and sin α = −b Write the equation 3x – 4y + 5 = 0 in normal form. Solution : 3x – 4y + 5 = 0 ⇒ –3x + 4y = 5 −3 x + ∴ y = 1 ∴ x cos α + y sin α = 1 where cos α = − 3 5 and sin α = 4 5 Conditions for Two Lines to be Intersecting, Parallel or Perpendicular Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 Coincident if a1 = b1 = c1 a2 b2 c2 Parallel if a1 = b1 ≠ c1 a2 b2 c2 Perpendicular if a1a2 + b1b2 = 0 Intersecting if a1 ≠ b1 a2 b2 Give the equation of lines parallel and perpendicular to line 2x + 3y + 5 = 0, which pass through origin. Solution : Line parallel to given line is 2x + 3y + k = 0. k = 0 if it passed through origin, hence 2x + 3y = 0 is the required parallel line. Any line perpendicular to 2x + 3y + 5 = 0 is 3x – 2y + λ = 0 (λ = 0, since it passes through origin) ∴ 3x = 2y is the required perpendicular line POINT OF INTERSECTION OF TWO LINES By solving the two equations of straight lines simultaneously, the point of intersection can be found for lines l1 = a1 x + b1 y + c1 = 0 & l2 = a2 x + b2 y + c2 = 0 Point of intersection is ⎛ b1c2 − b2c1 c1a2 − c2a1 ⎞ ⎜ ; a b − a b a b a b ⎟ ; a1b2 ≠ a2b1 ⎝ 1 2 2 1 1 2 2 1 ⎠ Angle between Two Lines tanφ = where φ is acute angle and m1 & m2 are slopes of two lines Otherwise tanφ = ± ; one sign gives acute angle and the other gives obtuse angle Here m1 or m2 or both should not be infinite. 1. The two lines which make angles φ with the given line y = mx + c and passing through the point (x1, y1) are given by y − y1 = m1(x − x1) and y − y1 = m2 (x − x1) where m1 = m − tan φ 1+ m tanφ ; m2 = m + tan φ 1− m tanφ i.e. m1 = tan (θ − φ) and m2 = tan (θ + φ) if m = tan θ. 2. The length r of the line segment drawn through a given point (x1, y1) and making an angle θ with x-axis, to meet the line ax + by + c = 0 is given by r = − ax1 + by1 + c a cos θ + b sin θ Points in Relation to a Line ax + by + c = 0 The points P(x1, y1) and Q(x2, y2) lie on the same side or on the opposite side of the line ax + by + c = 0 according as ax1 + by1 + c and ax2 + by2 + c have same sign or opposite sign. The ratio in which the line ax + by + c = 0 divides the line segment joining P(x1, y1) and Q(x2, y2) is ax1 + by1 + c ax2 + by 2 + c The length of the perpendicular from P(x , y ) to ax + by + c = 0 is p = ax1 + by1 + c 1 1 1 a2 + b2 Coordinates of the foot of the perpendicular drawn from P(x1, y1) to the line ax + by + c = 0 are given x − x1 = by a y − y1 b = − ax1 + by1 + c Coordinates of the image of P(x1, y1) in the line ax + by + c = 0 are given by x − x1 = a y − y1 b = − 2(ax1 + by1 + c) CONCURRENCY OF LINES The lines a1x + b1y + c1 = 0 ; a2 x + b2 y + c2 = 0 and a3 x + b3 y + c3 = 0 are concurrent if they pass through a1 the same point ; the condition for concurrency is a2 a3 b1 c1 b2 c2 = 0 b3 c3 Show that the line 3x + 4y – 6 = 0 does not divide the triangle ABC whose vertices are (0.0), (1,–1) & (–3, 2). Solution : 3x + 4y – 6 = 0 x = 0,y = 0 ⇒ LHS = −6⎤ x = 1, y = −1 ⇒ LHS = −7 ⎥ x = −3,y = 2⇒ LHS = −7 ⎥⎦ Hence all the three vertices fall on the same side of line, hence the line does not divide the triangle. The three line 2x + 3y – 5 = 0; 2x + 4y – 6 = 0 and x + 7y – 8 = 0 are (1) Parallel (2) Concurrent Form a triangle (4) Neither parallel nor intersecting but skew lines Solution : 2 3 − 5 2 4 − 6 1 7 − 8 = 2(–32 + 42) – 3(–16 + 6) – 5(14 – 4) = 20 + 30 – 50 = 0 Hence lines are concurrent FAMILY OF LINES Lines through the point of intersection of two given lines a1x + b1y + c1 = 0 and a2 x + b2 y + c2 = 0 are given by : (a1x + b1y + c1) + λ (a2 x + b2 y + c2 ) = 0 where λ is a parameter. It represents a family of lines. Any particular line (member of the family) can be found from the additional condition stated about the required line. If A(x1, y1), B(x2, y2), C(x3, y3) are the vertices of a triangle ABC then x Equation of median through A is x1 x2 y 1 x y1 1 + x1 y2 1 x3 y 1 y1 1 = 0 y3 1 x Equation of internal bisector of angle A is b x1 x2 where b = AC and c = AB. y 1 x y1 1 + c x1 y 2 1 x3 y 1 y1 1 = 0 y3 1 ANGLE BISECTORS BETWEEN TWO LINES The equations of the bisectors of the angles between two intersecting lines a x + b y + c = 0 and a x + b y + c = 0 are given by a1x + b1y + c1 = ± a2 x + b2 y + c2 1 1 1 2 2 2 Let φ be the angle between one of the bisectors and one of the lines a1x + b1y + c1 = 0 . If tan φ < 1 i.e. φ < 45°, then that bisector is the acute angle bisector of the two given lines. The other equation represents the obtuse angle bisector. Rule for writing a particular bisector Write the equations of the lines so that constant terms are positive = If a1a2 + b1b2 > 0 then …(1) gives the obtuse angle bisector. If a1a2 + b1b2 < 0 then (1) gives the acute angle bisector. For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0, find the equation of the bisector of the angle which contains the origin. Solution : Re-writing the given equations so that constant terms are positive, we have –4x – 3y + 6 = 0 …(i) and 5x + 12y + 9 = 0 …(ii) Now a1a2 + b1b2 = (–4) (5) + (–3) (12) = –20 – 36 = –56 < 0 So, the origin lies in acute angle. ∴ The equation of the bisector of the acute angle between the lines (i) and (ii) is −4x − 3y + 6 = + 5x + 12y + 9 52 + (12)2 ⇒ –52x – 39y + 78 = 25x + 60y + 45 ⇒ 7x + 9y – 3 = 0 AREA OF PARALLELOGRAM OR RHOMBUS Area of a parallelogram or a rhombus, equations of whose sides are given, can be obtained by using the following formula Area = p1p2 sin θ Where p1 = DL = distance between lines AB and CD, p2 = BM = distance between lines AD and BC, θ = angle between adjacent sides AB and AD. In the case of a rhombus, p1 = p2. Thus, Area of rhombus = 1 C sinθ Also, area of rhombus = 1 d d 2 1 2 where d1 and d2 are the lengths of two perpendicular diagonals of a rhombus. A TRICKS FOR BRIEF SOLUTIONS If (x1, y1) and (x2, y2) are the two vertices of an equilateral triangle, then the third vertex is given by, ⎡(x1 + x2 ) ± 3(y1 – y 2 ) , y1 + y 2 μ 3(x1 – x2 ) ⎤ ⎢⎢ 2 ⎥ 2 ⎥⎦ If (x1, y1) and (x2, y2) are the vertices of the hypotenuse of a right angle triangle, then the third vertex is given by ⎡(x1 + x2 ) ± (y1 – y2 ) , (y1 + y 2 ) ± (x1 – x2 ) ⎤ ⎢ ⎥ ⎣ ⎦ Area of the triangle, whose sides are y = m1x + c1, y = m2x + c2 and y = m3x + c3 is given by A = 1 ∑(c1 – c2 )2 2 m1 – m2 Area of rhombus, formed by ax ± by ± c = 0, is given by, A = Area of the parallelogram formed by the lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, a1x + b1y + d1 = 0, a2x + b2y + d2 = 0 is given by A = SOLVED EXAMPLES Example 1 : If O be the origin and if points Q1 and Q2 have coordinates (x1, y1) and (x2, y2) respectively show that OQ1 ⋅ OQ2 cos ∠Q1OQ2 = x1x2 + y1y 2 . Solution : OQ2 + OQ2 − Q Q2 The cosine formula applied to triangle Q1OQ2 gives cos ∠Q2OQ1 = 1 2 1 2 2 ⋅OQ1 ⋅OQ2 = (x1 − 0)2 + (y1 − 0)2 + (x2 − 0)2 + (y 2 − 0)2 − [(x1 − x2 )2 + (y1 − y 2 )2 ] 2 ⋅OQ1 ⋅ OQ2 ∴ OQ1 ⋅ OQ2 cos ∠Q1OQ2 = x1x2 + y1y 2 = 2 (x1x2 ) + 2(y1y 2 ) 2 ⋅ OQ1 ⋅ OQ2 Example 2 : Find the points which divide the line joining the points (–3, –4) and (–8, –7) internally in the ratio 7 : 5 externally in the ratio 7 : 5 Solution : Let A(–3, –4) ; B(–8, –7) If P(x, y) divides AB internally in the ratio 7 : 5 then x = 7 ⋅ (−8) + 5 (−3) = 7 + 5 y = 7 ⋅ (−7) + 5 (−4) = 7 + 5 − 71 12 − 69 12 A (–3, –4) P (x, y) B (–8, –7) Let Q(α, β) divide AB externally in the ratio of 7 : 5. α = 7 ⋅ (−8) − 5 (−3) = 7 − 5 β = 7 ⋅ (−7) − 5 (−4) = 7 − 5 − 41 2 − 29 2 A (–3, –4) B (–8, –7) Q (α, β) Example 3 : Coordinates of the points A, B, C and P are (6, 3), (–3, 5), (4, –2) and (x, y) respectively. Show that Area of Δ PBC = Area of Δ ABC 7 Solution : Area of ΔABC = − 3 5 4 − 2 6 3 1 1 = 49 1 2 Area of ΔPBC is modulus of the determinant x y − 3 5 4 − 2 Area of Δ PBC = Area of Δ ABC 1 1 = 7 (x + y − 2) 1 2 7 value of k are the points (k, 2 – 2k), (1 – k, 2k) (–4 –k, 6 – 2k) collinear ? Solution : If points are collinear then k 1 − k − 1 − k 2 − 2k 2k 6 − 2k 1 1 = 0 1 k ⇒ 1 − 2k − 4 − 2k 2 − 2k 4k − 2 4 1 0 = 0 0 ⇒ 4(1 – 2 k) – (4k – 2) (–4 – 2k) = 0 ⇒ 4 – 8k + 16k – 8 + 8k2 – 4k = 0 ⇒ 8k2 + 4k – 4 = 0 ⇒ 2k2 + k – 1 = 0 ⇒ k = 4 = − 1 ± 3 ⇒ 4 k = 1 , − 1 2 Example 5 : The ends of a rod of length l move on two mutually perpendicular lines. Find the locus of the point on the rod which divides it in the ratio 1 : 2. Solution : Let the two mutually perpendicular lines be the x and y axes and let P(x0, y0) be any point on the locus. Let AB denote the corresponding position of the rod such that l 2 = AB2 = a2 + b2 AP = 2 BP and where A(a, 0) and B(0, b). Then, 2b + 0 0 3 ; x0 = ⎛ 3y ⎞2 ∴ ⎜ 0 ⎟ + (3x0 )2 = l 2 ⎝ 2 ⎠ 2 2 4l 2 y 0 + 4x0 = 9 x 2 2 4l 2 Equation to locus is 4x + y = . 9 Example 6 : ABCD is a variable rectangle having its sides parallel to fixed directions. The vertices B and D lie on x = a and x = –a and A lies on the line y = 0. Find the locus of C. Solution : Let A be (x1 , 0), B be (a , y2) and D be (–a, y3). We are given that AB and AD have fixed directions and hence their slopes are constant, say, m1 and m2. y 2 = m and y3 = m ∴ a − x1 a − x1 Further, m1m2 = –1, since ABCD is rectangle y 2 = m and y 3 = 1 a − x1 a + x1 m1 …(1) Let the coordinates of C be (α, β). Now mid point of BD ≡ mid point of AC ⇒ x1 + α = 0 and y 2 + y 3 = β 2 2 2 ∴ a = –x1 and β = y2 + y3 …(2) By (1) and (2) we have : − (m2 − 1) α + m β = (m2 + 1) a 1 1 1 Locus of C is given by m1y = (m2 + 1) a + (m2 − 1) x Example 7 : Vertices of a triangle are (2 cosθ, sinθ), (sinθ, cosθ) (–cosθ, –2 sinθ). Find the locus of its centroid, as θ varies. Solution : Let centroid of the Δ be (h, k), then h = 2 cos θ – cos θ + sin θ ; 3 k = sin θ – 2 sin θ + cos θ 3 ⇒ 3h = cos θ + sin θ ; 3k = cos θ – sin θ squaring and adding, ⇒ 9h2 + 9k2 = 2 ∴ Required locus is, 9(x2 + y2) = 2. Example 8 : The extremities of a diagonal of a square are (a, 0) and (0, b), where a > 0, b > 0. Find the coordinates of the extremities of the other diagonal. Solution : Let ACBD be the given square with A(a, 0) and B(0, b) AB ≡ ⎛ a , b ⎞ Coordinates of the mid-point M of ⎜ ⎟ ⎝ 2 2 ⎠ AB = ⇒ AM = BM = CM = DM = 2 Slope of AB = − b a ⇒ slope of CD = a = tan θ b (say) ⇒ cos θ = , sin θ = x − a y − b Equation of CD in parametric form is 2 = b a2 + b2 2 = r a a2 + b2 To find coordinates of C & D put r = ± 2 C ≡ ⎛ a + b , a + b ⎞ and D = ⎛ a − b , b − a ⎞ ⇒ ⎜ ⎟ ⎝ ⎠ ⎜ ⎟ ⎝ 2 2 ⎠ Example 9 : Find the coordinates of the orthocentre of the triangle whose vertices are (0, 0), (2, –1) and (–1, 3). Solution : Let OAB be the triangle as shown in the figure. Slope AB = 3 − (−1) = − 4 − 1− 2 3 D Slope of altitude OD = 3 4 Slope OB = –3 and its equation is y = 3 x 4 1 …(1) O (0, 0) A(2, –1) ∴ Slope of altitude through A = 3 Eq. of altitude from A is y + 1 = 1 (x − 2) 3 3y – x + 5 = 0 …(2) Solving (1) and (2), the coordinates of orthocentre are x = –4, y = –3 : p1 be the lengths of the perpendiculars drawn from the origin upon the straight lines x sin θ + y cos θ = 1 a sin 2θ 2 and x cos θ – y cos θ = a cos 2θ, prove that 4p2 + p 2 = a2 . Solution : We have p = ⇒ p2 = 1 a2 sin2 2θ and p = ⇒ p2 = a2 cos2 2θ 4 1 1 Now 4p2 + p 2 = a2sin22θ + a2cos22θ = a2 Example 11: Find the equation of the line passing through (2, 3) and making intercepts of length 2 units between the lines y + 2x = 3 and y + 2x = 5. Solution : y + 2x = 3 …(1) y + 2x = 5 …(2) Equation of the line passing through P(2, 3) is x − 2 = y − 3 = r …(3) cos θ sin θ The points of intersection of (3) with (2) and (1) are A(2 + r cos θ, 3 + r sin θ) and B(2 + (r – 2) cos θ, 3 + (r – 2) sin θ) respectively for some r. ∴ (3 + r sin θ) + 2 (2 + r cos θ) = 3 (3 + (r − 2) sin θ) + 2 (2 + (r − 2) cos θ) = 5 …(4) …(5) By (4) and (5) sin θ + 2 cos θ = –1 sin θ = –1 – 2 cos θ sin2θ = (–1 – 2 cos θ)2 cos θ (5 cos θ + 4) = 0 ∴ cos θ = 0 or cos θ = − 4 5 and accordingly sin θ = 1 or 3 . 5 (3) now gives equations of required lines. Example 12 : Find the coordinates of those points on the line 3x + 2y = 5 which are equidistant from the lines 4x + 3y = 7 and 2y = 5. Solution : It is obvious that the desired points lie on 3x + 2y = 5 as well as on the bisectors of angles formed by the lines 4x + 3y = 7 …(1) 2y – 5 = 0 …(2) Equations to the angle bisectors are 4x + 3y − 7 = ± 2y − 5 5 2 8x + 6y − 14 = ± (10y − 25) 8x + 16y – 39 = 0 ; 8x – 4y + 11 = 0 Solving 3x + 2y = 5 and 8x + 16y = 39 we get x = 1 , y = 77 Solving 3x + 2y = 5 and 8x – 4y + 11 = 0 we get 16 x = − 32 1 , y = 73 14 28 Example 13 : The line L has intercepts a and b on the coordinate axes. While keeping the origin fixed, if the coordinate axes are rotated through a fixed angle, the same line has intercepts p and q on the rotated axes. Show that 1 + 1 a2 b2 = 1 + 1 p2 q 2 Solution : Suppose that the axes are rotated in the anticlockwise direction through an angle α. The equation of the line L with respect to the old axes is given by x + y = 1. To find the equation of L with respect to the new a b rotated axes, we replace x by x cos α – y sin α and y by x sin α + y cos α so that the equation of L with respect to the new axes is 1 (x cos α − y sin α) + 1 (x sin α + y cos α) = 1 a b ⎛ 1 cos α + 1 sin α⎞ x + ⎛ 1 cos α − 1 sin α⎞y = 1 ⇒ ⎜ a b ⎟ ⎜ ⎟ ⎠ ⎝ b a ⎠ .....(1) Since p and q are the intercepts made by this line on coordinate axes we have 1 = 1 cos α + 1 sinα and 1 = 1 sinα + 1 cos α p a b q a b Squaring and adding we get 1 + 1 = 1 + 1 p2 q 2 a2 b2 Example 14 : A straight line through the origin ‘O’ meets the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at the points P and Q respectively. Then find the ratio in which the point ‘O’ divides the segment PQ. Solution : Taking PQ in any direction, ratio of OP and OQ will be same, so taking PQ ⊥ to the given parallel lines OP = ⊥ from (0, 0) on 2x + y – 9 = 0 . 2 = = OQ = ⊥ from (0, 0) on 2x + y + = = 6 ∴ OP = OQ = 9 = 3 12 4 ❑ ❑ ❑