Question Bank Distance Formula, Section Formula and Equation of a Straight Line

Transcription

1 Question Bank Distance Formula, Section Formula and Equation of a Straight Line 1. A ( 3, ), B ( 5, 5), C (, 3) and D (4, 4) are the four points in a plane. Show that ABCD is a rhombus but not a square. Solution. The given points are A ( 3, ), B ( 5, 5), C (, 3) and D (4, 4). AB = [ ] 5 ( 3) + ( 5 ) = ( 5 + 3) + ( 7) = ( ) + ( 7) = = 53 units BC = [ ( 5) ] + [ 3 ( 5) ] = ( + 5) + ( 3 + 5) = 7 + = = 53 units CD = [ ] (4 ) + 4 ( 3 = + (4 + 3) = DA = + 7 = = 53 units ( 3 4) + ( 4) = (7) + ( ) = = 53 units AB = BC = CD = DA. ABCD is either a rhombus or a square. Diag. AC = [ ( 3)] + ( 3 ) = = Diag. BD = ( + 3) + ( 5) = = 50 = 5 units. ( + 3) + ( 5) Math Class X 1 Question Bank

2 = = [4 ( 5)] + [4 ( 5)] (4 + 5) + (4 + 5) = = = 16 = 9 units Diag. AC Diag. BD ABCD is a rhombus but not a square. Proved. Find the co-ordinates of the circumcentre of ΔABC with vertices at A(3, 0), B( 1, 6) and C(4, 1). Also, find its circum-radius. Solution. The vertices of the given triangle are A (3, 0), B ( 1, 6) and C (4, 1). Let O (x, y) be the circumcentre of ΔABC. Then, OA = OB = OC OA = OB = OC Now, OA = OB (x 3) + (y 0) = [x ( 1)] + [y ( 6)] (x 3) + y = (x + 1) + (y + 6) x + y 6x + 9 = x + y + x + 1y x + 1y + 8 = 0 x + 3y = 7...(i) And, OB = OC [ x ( 1)] + [y ( 6)] = (x 4) + [y ( 1)] (x + 1) + (y + 6) = (x 4) + (y + 1) x + y + x + 1y + 37 = x + y 8x + y x + 10y + 0 = 0 x + y =...(ii) Solving (i) and (ii), we get x = 1 and y = 3. The circumcentre of ΔABC is O (1, 3). Math Class X Question Bank

3 Circum-radius = OA = (1 3) + ( 3 0) = ( ) + ( 3) = = 13 units. 3. Find the co-ordinates of the centre of a circle which passes through the points A (0, 0), B ( 3, 3) and C (5, 1). Also, find the radius of the circle. Solution. Let P (x, y) be the centre of the circle passing through the points A (0, 0), B ( 3, 3) and C (5, 1). Then, PA = PB = PC PA = PB = PC Now, PA = PB (x 0) + (y 0) = (x + 3) + (y 3) x + y = x x + y + 9 6y 6x 6y = 18 x y = 3...(i) And, PB = PC (x + 3) + (y 3) = (i 5) + (y + 1) x x + y + 9 6y = x x + y y 16x 8y = 8 x y = 1...(ii) Subtracting (ii) from (i), we get x = 4 x = 4 Substituting x = 4 in (i), we get 4 y = 3 y = 7 Hence, centre of the circle is P (4, 7). Radius of the circle = PA = (4 0) + (7 0) = = 65 units. Math Class X 3 Question Bank

4 4. KM is a straight line of 13 units. If K has the co-ordinates (, 5) and M has the co-ordinates (x, 7), find the possible values of x. Solution. We have, KM = (x ) + ( 7 5) KM = x + 4 4x KM = x 4x (i) But KM = 13 (given) KM = 169 (ii) From (i) and (ii), we get 169 = x 4x x 4x 1 = 0 x 7x + 3x 1 = 0 x(x 7) + 3(x 7) = 0 (x 7) (x + 3) = 0 x = 7 or x = 3 Hence, the possible values of x are 7 and In what ratio does the point P (p, 1) divide the line segment joining the points A (1, 3) and B (6, )? Hence, find the value of p. Solution. Let P (p, 1) divide the line segment joining the points A (1, 3) and B (6, ) in the ratio k : 1, i.e., AP : PB = k : 1 k k 1 ( 3), k + 1 k + 1 But P is (p, 1) k 3 = 1 k + 1 k 3 = k 1 3k = k = 3 Math Class X 4 Question Bank

5 The required ratio is 3 : 1, i.e., : 3 (internally). Also, Putting 6 k + 1 = p..(i) k + 1 k = in (i), we get ρ = 3 = = = Hence, p = The centre of a circle is C ( 1, 6) and one end of a diameter is A (5, 9). Find the co-ordinates of the other end. Solution. Let the other end of the diameter of the circle be B (x, y) whose one end is the point A (5, 9). The mid-point of AB is. 5 + x 9 + y, The centre of the circle is C ( 1, 6). Since the centre of the circle is the mid-point of AB, 5 + x 9 + y = 1 and = x = and 9 + y = 1 x = 7 and y = 3 The co-ordinates of the other end of the diameter are ( 7, 3). 7. Two vertices of a ΔABC are A (6, ) and B (4, 3). If the coordinates of its centroid be (3, 1), find the co-ordinates of the third vertex of the triangle. Solution. Let C (a, b) be the third vertex of ΔABC. Math Class X 5 Question Bank

6 Then, the co-ordinates of its centroid are : G a, b 3 3 i.e., 10 + a 1 + b, 3 3 But, the co-ordinates of its centroid are G (3, 1) a = 3 3 and 1 + b 3 = a = 9 and 1 + b = 3 a = 1 and b = 4 Hence, third vertex of the triangle is C ( 1, 4). 8. The mid points of the sides BC, CA and AB of ΔABC are D (, 1), E ( 1, 3) and F (4, 5) respectively. Find the coordinates of A, B and C. Solution. Let the coordinates of A, B and C be (x 1, y 1 ), (x, y ) and (x 3, y 3 ) respectively. D is the mid-point of BC x + x3 y + y3 = and = 1 x + x 3 = 4...(i) and y + y 3 =... (ii) E is the mid point of AC x1 + x3 y1 + y3 = 1 and = 3 x 1 + x 3 =...(iii) and y 1 + y 3 = 6... (iv) F is the mid point of AB x1 + x y1 + y = 4 and = 5 x 1 + x = 8...(v) and y 1 + y = (vi) From (i) and (iii), we get x 1 x = 6... (vii) From (ii) and (iv), we get y 1 y = 8... (viii) Math Class X 6 Question Bank

7 From (vii) and (v), we get x 1 = x 1 = 1... (vii) From (viii) and (vi), we get y 1 = y 1 = 1 From (vii) and (viii), we get x = 7 and y = 9 From (i) and (ii), we get x 3 = 3 and y 3 = 7 Hence, coordinates of A, B and C are (1, 1), (7, 9) and ( 3, 7) respectively. 9. Show that the line segment joining the points A ( 5, 8) and B (10, 4) is trisected by the coordinate axes. Also, find the points of trisection of AB. Solution. Let the x-axis cut the line segment AB at P(x, 0) which divide it in the ratio p : 1. Also, let the y-axis cut the line segment AB at Q(0, y) which divide it in the ratio q : 1. P (x, 0) divides AB in the ratio p : 1. p ( 4) = p + 1 4p = 8 p = P (x, 0) divides AB in the ratio : (i) ( 5) 15 x = = = Coordinates of P are (5, 0). Q (0, y) divides AB in the ratio q : 1. y ( 5) 0 = q q = 5 q = 1 Q (0, y) divides AB in the ratio 1 : 1 i.e., 1 :... (ii) y = 1 ( 4) + 8 = Coordinates of Q are (0, 4). From (i), AP PB = 1 and from (ii) AQ 1, = QB Math Class X 7 Question Bank

8 AP = PB and AQ = QB... (iii) Now, AP + PB = AB and AQ + QB = AB 3PB = AB and 3AQ = AB [From (iii)] PB = 1 3 AB and AQ = 1 3 AB PB = AQ...(iv) Now, QB = QP + PB AQ = QP + AQ [From (iii) and (iv) AQ = QP... (v) From (iv) and (v), we get AQ = QP = PB. Proved. The points of trisection of AB are (5, 0) and (0, 4). 10. G is the centroid of a triangle with vertices A( a, 0), B(0, a) and C(a, b). Prove that GA + GB + GC = (AB + BC + CA ) Solution. Coordinates of the Solution. Coordinates of the centroid G of ΔABC are a α 0 + a + β, 3 3 α a β + a i.e.,, 3 3 Now, GA α a β + a = GA = + a α + a β + a = +... (i) 3 3 GB α a β + a = GB = 0 + a 3 3 α a β a = +... (ii) 3 3 Math Class X 8 Question Bank

9 GC α a β + a = α + β 3 3 α a a β = From (i), (ii) and (iii), we get... (iii) GA + GB + GC = 1 9 [(α + a) + (β + a) + (α a) + (β a) + ( α a) + (a β) ] = 1 9 [α + 4a + 4aα + β + a + aβ + α + a aα + β + 4a 4aβ + 4α + a + 4aα + a + 4β 4aβ] = 1 9 [6α + 6β + 6aα 6aβ + 1a ] = 3 [α + β + a + aα aβ)...(vii) AB = (0 + a) + (a 0) = a BC = (0 a) + (a β) = α + (α β) CA = (a + α) + (β 0) = (α + a) + β From (iv), (v) and (vi), we get AB + BC + CA (iv)... (v)... (vi) = a + α + a + β aβ + α + a + aα + β = α + β + 4α + aα aβ = (α + β + a + aα aβ)... (viii) From (vii) and (viii), we get GA + GB + GC = 1 3 (AB + BC + CA ). Math Class X 9 Question Bank

10 11. A straight line passes through the points P ( 1, 4) and Q (5, ). It intersects the co-ordinate axes at points A and B. M is the mid-point of the segment AB. Find : (i) the equation of the line. (ii) the co-ordinates of A and B. (iii) the co-ordinates of M. Solution. Given points are P ( 1, 4) and Q (5, ). (i) The slope of the line 4 PQ = 5 ( 1) = 1. The line passes through P ( 1, 4) and has slope 1. Its equation is y 4 = 1 [x ( 1)] [ y y 1 = m (x x 1 )] x + y 3 = 0. (ii) The line PQ meets x-axis, i.e., y = 0 where x = 0 x = 3 The co-ordinates of A are (3, 0). The line PQ meets y-axis, i.e., x = 0 where 0 + y 3 = 0 y = 3 The co-ordinates of B are (0, 3). (iii) Since M is mid-point of the segment AB, its co-ordinates are ,, i.e, 3 3,. 1. In the adjoining figure, AB and CD are the lines x y + 6 = 0 and x y = 4 respectively. (i) Write down the co-ordinates of A, B, C and D; (ii) Prove that ΔOAB and ΔODC are similar; (iii) Is figure ABCD cyclic? Math Class X 10 Question Bank

11 Give reasons for your answer. Solution. (i) Equation of AB is x y + 6 = 0 AB cuts y-axis at the point, where x = 0. Putting x = 0, we get y = 6. Co-ordinates of A are A (0, 6). Again, AB cuts x-axis at the point, where y = 0. Putting y = 0, we get x = 3. Co-ordinates of B are B ( 3, 0). Equation of CD is x y = 4. CD cuts y-axis at the point, where x = 0. Putting x = 0 in its equation, we get y = Co-ordinates of C are C (0, ). Again, CD cuts x-axis at the point, where y = 0. Putting y = 0 in its equation, we get x = 4. Co-ordinates of D are D (4, 0). (ii) Thus, OA = 6 units, OB = 3 units, OC = units and OD = 4 units. AB = OA + OB = = 45 = 3 5 units CD = OC + OD = + 4 = 0 = 5 units OA 6 3 OB 3 AB = =, = and =. OD 4 OC CD 5 Thus, in ΔOAB and ΔODC, we have = OA OB AB = = OD OC CD. ΔOAB ~ ΔODC. Proved (iii) Let P (h, k) be a point equidistant from A, B, C and D, then, PA = PB = PC = PD PA = PB = PC = PD. (h 0) + (k 6) = (h + 3) + (k 0) = (h 0) + (k + ) = (h 4) + (k 0) h + k 1k + 36 = h + 6h k = h + k + 4k + 4 = h + k 8h + 16 h + 4k = 9 and k + h = 3 Math Class X 11 Question Bank

12 k = and h = 1. 1 Thus, a circle can be drawn with centre P, and radius PA to pass through A, B, C and D. Hence, the points A, B, C, D are cyclic. 13. P (3, 4), Q (7, ) and R (, 1) are the vertices of Δ PQR. Write down the equation of the median of the triangle through R. Solution. Let M be the mid-point of PQ Then, co-ordinates of M are, = (5, 1) Equation of the median RM is given by : 1 ( 1) y ( 1) = [x ()] 5 ( ) y + 1 = 7 (x + ) 7y + 7 = x + 4 x 7y = A line passes through the point ( 1, ) and cuts off negative intercept a on the x-axis and positive intercept b on the y-axis such that a : b = : 1. Find the slope and the equation of the line. Solution. Let the line cut the x-axis at A and the y-axis at B. a : b = : 1 Let OA = a and OB = a So, the coordinates of A are ( a, 0) and that of B are (0, a). Now equation of the line through the points ( a, 0) and (0, a) is given by a 0 y 0 = ( x + a) 0 + a y = x + a x + y = a... (i) Also, (i) passes through ( 1, ) Math Class X 1 Question Bank

13 1 + 4 = a a = 5 Equation of the line is given by x + y = 5 x 5 x + y = 5 y = + Slope of the line 1 and equation of the line is y x = The equations of the lines PQ and PR are 3x 4y = 1 and x + y = respectively. Find the equation of the line PM, if the coordinates of M are (, 5). Solution. To get the coordinates of the point P, we solve the given equations simultaneously. Multiplying x + y = by 4 and adding to 3x 4y = 1, We get 4 x + 4 y = 8 3 x 4 y = 1 7 x = 7 x = 1 Substituting x = 1 in x + y =, we get 1 + = x = 1 Coordinates of P are (1, 1). y y1 Now equation of PM is given by y y 1 = x x (x x 1) 1 y 1 = 5 1 (x 1) 1 (4 x 4) y 1 = 3 3y + 3 = 4x 4 4x + 3y = 7 Math Class X 13 Question Bank

14 16. Find the equation of the perpendicular from the point P ( 1, ) on the line 3x + 4y 1 = 0. Also, find the co-ordinates of the foot of perpendicular. Solution. The given line is 3x + 4y 1 = 0... (i) 4y = 3x + 1 y = 3 4 x The slope of the line (i) = 4 From P ( 1, ), draw PN perpendicular to the given line. 4 1 The slope of the line PN = m = 3 m1 The equation of the line through P ( 1, ) and having slope 4 3 is y ( ) = 4 3 [x ( 1)] 3y + 6 = 4x + 4 4x 3y = 0... (ii) which is the required equation of the perpendicular from P to the given line. To find the co-ordinates of N (the foot of perpendicular), solve (i) and (ii) simultaneously. Multiplying (i) by 3 and (ii) by 4, and on adding, we get 5x 44 = 0 x = 44 5 Multiplying (i) by 4 and (ii) by 3, and on subtracting, we get 5y 4 = 0 y = Hence, the co-ordinates of the foot of perpendicular are, 5 5. Math Class X 14 Question Bank

15 17. ABCD is a rhombus. The co-ordinates of the points A and C are (3, 6) and ( 1, ) respectively. Write down the equation of the diagonal BD. Solution. We know that the diagonals of a rhombus bisect each other at right angles. Let the diagonals AC and BD of the given rhombus ABCD intersect at M. 3 + ( 1) 6 + So, M = mid-point of AC =, = (1, 4) Now, the diagonal BD is a line passing through (1, 4) and perpendicular to AC. y1 y 6 4 The slope of AC = = = = 1 x1 x 3 ( 1) 4 1 The slope of BD = [ m 1 m = 1] 1 So, by the point-slope form, the equation of BD is y y 1 = m (x x 1 ) y 4 = ( 1) (x 1) y 4 = x + 1 or x + y = Find the equation of a straight line parallel to the line x 5y + 1 = 0 and passing through the point which divides the line joining ( 5, 4) and (3, 1) in the ratio : 1. Solution. Let P(x, y) be the point which divide the join of ( 5, 4) and (3, 1) in the ratio : ( 5) Then, x = and y = x = 1 3 = and y = Also, equation of the given line is x 5y + 1 = 0 y = x + 1 = Math Class X 15 Question Bank

16 Gradient of the given line, m 1 = 5. The required line p is parallel to the given line. Gradient of the required line = 5. 1 Now, the required line passes through P, 3 and have gradient 5 Its equation is given by y = 1 x y 30 = 6x 6x 15y + 8 = The points A (, 3), B (3, 5) and C ( 1, 1) are the vertices of the triangle ABC. Find the equation of the altitude of the triangle through A. Solution. Let the slope of the altitude AD be m. y y The slope of BC = 1 = = = x1 x Since AD BC, m. 3 = 1 m = 3 Thus, AD is a line of slope and passing through the point A (, 3). By the point-slope form, the equation of AD is y 3 = (x ) or 3 (y 3) = (x ) 3 3y 9 = x + 4 x + 3y = The points A (1, ), B (3, 4) and C (5, 6) are the vertices of a ΔABC. Find the equation of the right bisectors of the sides BC and CA. Hence, find the circumcentre of ΔABC. Solution. Let the right bisector of BC intersect it at D and the right bisector of AC intersect it at E. Math Class X 16 Question Bank

17 Then, coordinates of D are , i.e., (4, 5) And coordinates of E are , i.e., (3, ) Gradient m 1 of BC = = 1 Gradient of right bisector of BC, m = 1 [ m 1 m = 1] Equation of the right bisector of BC, which passes through (4, 5) and has gradient 1 is given by y + 5 = 1 (x 4) or x y = 9 Gradient m 3 of BC = = 1 Gradient m 4 of the right bisector of AC = 1 [ m 3 m 4 = 1] Equation of the right bisector of AC, which passes through (3, ) and has gradient 1 is given by y + = 1 (x 3) or y + 4 = x 3 or x y = 7 To find the coordinates of the circumcentre of ΔABC, we solve x y = 9 and x y = 7 simultaneously. x y = 9 x y = 7 + y = Substituting y = in x y = 9, we get x = 9 + = 11 Hence, the equations of the right bisectors of the sides BC and AC are x y = 9 and x y = 7 respectively. The coordinates of the circumcentre of ΔABC are (11, ) Math Class X 17 Question Bank

18 1. Using coordinate geometry, prove that the diagonals of a square bisect each other perpendicularly. Solution. Let OABC be the square, where coordinates of O, A, B and C are (0, 0), (a, 0), (a, a) and (0, a) respectively. Coordinates of the mid-point of AC are a , a i,e., a, a Also the coordinates of the mid-point OB are a , a i,e., a, a i.e.,. a a, are coordinates of the mid-point of AC as well as OB. a a Hence, AC and OB bisect each other at,... (i) a 0 Slope of AC (m 1 ) = 0 a = 1 a 0 Slope of OB (m ) = = 1 a 0 m 1 m = 1 1 = 1 Hence, AC and OB are perpendicular to each other.... (ii) From (i) and (ii), we conclude that the diagonals of a square bisect each other perpendicularly Proved.. Three sides of a parallelogram have the equations x y =, x + y = 3 and y = x + 3. Find the equation of the fourth side if it passes through the origin. Solution. We have x y = y = x y = x + 3 and x + y = 3 y = 3 x We see that the gradient of y = x is and that of y = x + 3 is also. y = x and y = x + 3 are parallel sides of the parallelogram. Math Class X 18 Question Bank

19 Hence, the required line will be parallel to y = 3 x. Gradient of y = 3 x is 1 Gradient of the required line is also 1. The required line also passes through the origin i.e., through (0, 0). Equation of the fourth side of the parallelogram is given by y 0 = 1 (x 0) y + x = 0 Math Class X 19 Question Bank