Question 34 Co-ordinate Geometry Exercise 14.3
Answer:
Solution: The coordinates of the point that splits a line segment (internally or externally) into a specific ratio are found using the Section formula. Let A(-2, -3) and B(5, 6) be the given points. Suppose x-axis divides AB in the ratio k: 1 at the point P Then, the coordinates of the point of division are \left[\frac{5 k-2}{k+1}, \frac{6 k-3}{k+1}\right] As, P lies in the x-axis, the y – coordinate is zero. So, 6k – 3/ k + 1 = 0 6k – 3 = 0 k = ½ Thus, the required ratio is 1: 2 Using k in the coordinates of P We get, P (1/3, 0)
Was This helpful? In what ratio is the line joining (2, -3) and (5, 6) divided by the x-axis. Let the line joining points A (2, −3) and B (5, 6) be divided by point P (x, 0) in the ratio k : 1. `y=(ky_2+y_1)/(k+1)` `0=(kxx6+1xx(-3))/(k+1)` `0=6k-3` `k=1/2` Thus, the required ratio is 1: 2. Concept: Co-ordinates Expressed as (x,y) Is there an error in this question or solution? Page 2In what ratio is the line joining (2, -4) and (-3, 6) divided by the y – axis. Let the line joining points A (2, -4) and B (-3, 6) be divided by point P (0, y) in the ratio k : 1. `x=(kx_2+x_1)/(k+1)` `0=(kxx(-3)+1xx2)/(k+1)` `0=-3k+2` `k=2/3` Thus, the required ratio is 2: 3. Concept: Co-ordinates Expressed as (x,y) Is there an error in this question or solution? Page 3In what ratio does the point (1, a) divide the join of (-1, 4) and (4,-1)? Also, find the value of a. Let the point P (1, a) divides the line segment AB in the ratio k: 1. `1=(4k-1)/(k+1)` `=>k+1=4k-1` `=>2=3k` `=>k=2/3` ............(1) `=>a=(-k+4)/(k+1)` `=> a = (-2/3 + 4)/(2/3 + 1)` (from 1) `=> a = 10/5 = 2` Hence, the required is 2 : 3 and the value of a is 2. Concept: Co-ordinates Expressed as (x,y) Is there an error in this question or solution? Page 4In what ratio does the point (a, 6) divide the join of (-4, 3) and (2, 8)? Also, find the value of a. Let the point P (a, 6) divides the line segment joining A (-4, 3) and B (2, 8) in the ratio k: 1. `6=(8k+3)/(k+1)` `=> 6k+6=8k+3` `=>3=2k` `=>k=3/2` .................(1) `=>a=(2k-4)/(k+1)` `=>a=(2xx3/2-4)/(3/2+1)` (from equation 1) `=>a=-2/5` Hence, the required ratio is 3:2 and the value of a is `-2/5` Is there an error in this question or solution? |