Question

A line L is such that its segment between the straight lines $5x-y=4$ and $3x+4y=4$ is bisected at the point $(1,5)$. Find its equation.

Answer 1

b'Let L be $\Rightarrow y=mx+c$

$A=(x_1,y_1)$ and $B(x_2,y_2)$

C, is mid point on AB

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(1,5)$

$\Rightarrow x_1+x_2=2$ and $y_1+y_2=10$

$x_1=2-x_2$ ...(3) $y_1=10-y_2$ ..(4)

$(x_1,y_1)$ is on $5x-y=4$

$\Rightarrow 5x_1-y_1=4$ ....(1)

$x_2,y_2$ is on $3x+4y=4\left(2\right)\Rightarrow 3x_2+4y_2=4$

from (1) and (2)

$5x_1-y_1=3x_2+4y_2$

$5(2-x_2)-(10-y_2)=3x_2+4y_2$ ( using (3) and (4))

$\Rightarrow 10-5x_2-10+y_2=3x_2+4y_2$

$\Rightarrow 3y_2+8x_2=0$

$\therefore$ equation of $\angle$ is $8x+3y=0$