Long Questions (Question 3)

Posted on April 21, 2020 by user

Question 3:

The diagram shows a triangle LMN where L is on the y-axis. The equation of the straight line LKN and MK are 2y – 3x + 6 = 0 and 3y + x– 13 = 0 respectively. Find

(a) the coordinates of K

(b) the ratio LK:KN

Solution:
(a)

2y – 3x + 6 = 0 ----(1)

3y + x – 13 = 0 ----(2)

x = 13 – 3y ----(3)

Substitute equation (3) into (1),

2y – 3 (13 – 3y) + 6 = 0

2y – 39 + 9y + 6 = 0

11y = 33

y = 3

Substitute y = 3 into equation (3),

x = 13 – 3 (3)

x = 4

Coordinates of K = (4, 3).

(b)

Given equation of LKN is 2y – 3x + 6 = 0

At y – axis, x = 0,

x coordinates of point L = 0.

$\begin{array}{l} Ratio L K : K N \\ Equating the x coordinates, \\ \frac{L K (10) + K N (0)}{L K + K N} = 4 \\ 10 L K = 4 L K + 4 K N \\ 6 L K = 4 K N \\ \frac{L K}{K N} = \frac{4}{6} \\ \frac{L K}{K N} = \frac{2}{3} \\ Ratio L K : K N = 2 : 3 \end{array}$