In what ratio is the line segment joining the points (-2,-3) and (3, 7) divided by the y-axis? Also, find the coordinates of the point of division. The ratio in which the y-axis divides two points (x1 , y1) and (x2 , y2) is \[\lambda: 1\] The co-ordinates of the point dividing two points (x1 , y1) and (x2 , y2) in the ratio m : n is given as, `(x , y) = ((lambdax_2 + x_1)/(lambda + 1 )) ,((lambday_2 + y_1)/(lamda + 1))` where, `lambda = m/n` Here the two given points are A(5,−6) and B(−1,−4). \[(x, y) = \left( \frac{- \lambda + 5}{\lambda + 1}, \frac{- 4\lambda - 6}{\lambda + 1} \right)\] Since, the y-axis divided the given line, so the x coordinate will be 0. \[\frac{- \lambda + 5}{\lambda + 1} = 0\]
Thus the given points are divided by the y-axis in the ratio 5:1. The co-ordinates of this point (x, y) can be found by using the earlier mentioned formula. `(x , y ) = ((5/1 (-1) + (5) )/(5/1 + 1)) , ((5/1(-4)+(-6))/(5/1 +1))` `(x , y) = (0/6) , (-26/6)` `(x , y ) = ( 0 , - 26/6)`
Thus the co-ordinates of the point which divides the given points in the required ratio are `(0,-26/6)`.
Solution: The coordinates of the point P(x, y) which divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂), internally, in the ratio m₁: m₂ is given by the Section Formula: P(x, y) = [(mx₂ + nx₁) / m + n, (my₂ + ny₁) / m + n] Let the ratio in which the line segment joining A(- 3, 10) and B(6, - 8) be divided by point C(- 1, 6) be k : 1. By Section formula, C(x, y) = [(mx₂ + nx₁) / m + n, (my₂ + ny₁) / m + n] m = k, n = 1 Therefore, - 1 = (6k - 3) / (k + 1) - k - 1 = 6k - 3 7k = 2 k = 2 / 7 Hence, the point C divides line segment AB in the ratio 2 : 7. ☛ Check: NCERT Solutions for Class 10 Maths Chapter 7 Video Solution: NCERT Class 10 Maths Solutions Chapter 7 Exercise 7.2 Question 4 Summary: The ratio in which the line segment joining the points (- 3, 10) and (6, - 8) is divided by (- 1, 6) is 2 : 7. ☛ Related Questions:
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