Question

The equation of the line joining the point $(3,5)$to the point of intersection of the lines $4x+y-1=0$ and $7x-3y-35=0$ is equidistant from the points $(0,0)$ and $(8,34)$. State whether the statement is true or false. Justify

Answer 1

Given lines : $4x+y-1=0$

$\Rightarrow y=-4x+1$ ⋅⋅⋅(i)

and $7x-3y-35=0$ ⋅⋅⋅(ii)

Finding the point of intersection:

From (i) and (ii), we get

$\Rightarrow 7x-3(-4x+1)-35=0$

$\Rightarrow 19x=38$

$\Rightarrow x=2$

$\Rightarrow y=-8+1=-7$

The point of intersection is $(2,-7)$.

So, the equation of the line joining the points $(3,5)$ and $(2,-7)$ is

$y-5=\frac{-7-5}{2-3}(x-3)$

$[\because y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)(x-x_1\left)\right]$

$\Rightarrow y-5=12(x-3)$

$\Rightarrow 12x-y-31=0$

Distance of point $P(x_1,y_1)$ from line $ax+by+c=0$ is :

$D=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$

Distance from $(0,0)$ to the line

$12x-y-31=0$ is

$d_1=\frac{|-31|}{\sqrt{{12}^2+1^2}}=\frac{31}{\sqrt{145}}$

Distance from $(8,34)$ to the line

$12x-y-31=0$ is

$d_2=\frac{|96-34-31|}{\sqrt{{12}^2+1^2}}=\frac{31}{\sqrt{145}}$

Since, $d_1=d_2$

Hence, the line $12x-y-31=0$ is equidistant from $(0,0)$ and $(8,34)$

Hence given statement is true.