Question

$L1:x-2y+10=0$

$L2:x+2y-6=0$

What is the ratio in which the point of intersection of line L1 and L2 divides the line segment AB of L1?

Answer 1

$\text{Let the point of intersection of L1 and L2 be C(x,y)}$

$x-2y+10=0$ -------- (1)

$x+2y-6=0$----------(2)

Solving (1) and (2) simultaneously we get

$x=-2$

$y=4$

$C\equiv (-2,4)$

$\text{Let the ratio in which C divides AB be k: 1}$

$-2=\frac{8k-6}{k+1}$

$-2(k+1)=(8k-6)$

$-2k-8k=2-6$

$-10k=-4$

$5k=2$

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

$\text{The required ratio is 2 : 5}$