Question

Find the ratio in which the line segment joining (2, -3) and (5, 6) is divided by
y-axis. Also find the point of division​

Answer 1

Answer:

5:2 externally; 12

Step-by-step explanation:

Let y-axis i.e Y(0,y) divides the line segment AB joining A(2,-3) and B(5,6) in the ratio m:n

m:n = $\frac{m}{n}$:1 = k:1 = k (Let)

So, by internal section formula

(0,y) = $(\frac{k.5 + 1.2}{k + 1}$ , $\frac{k.6 + 1.-3}{k + 1})$

(0,y) = $(\frac{5k + 2}{k + 1}$ , $\frac{6k - 3}{k + 1})$

Equating corresponding elements,

0 = $\frac{5k + 2}{k + 1}$

0 = 5k + 2

-5k = 2

k = $-\frac{5}{2}$

$\frac{m}{n}$ = $-\frac{5}{2}$

∴ m:n = 5:2 externally

Again,

y = $\frac{6k - 3}{k + 1}$

  = 12