Question

Show that the perpendicular let fall from any point on the straight line $2x+11y-5=0$ upon the two straight lines $24x+7y=20$ and $4x-3y-2=0$ are equal to each other.

Answer 1

Let (h, k) be the point on the line $2x+11y-5=0$

$\Rightarrow 2h+11k-5=0\dots \dots \left(1\right)$

Let p and q be length of perpendicular from (h, k) on lines $24x+7y-20=0$ and $4x-3y-2=0$ so,

p = q

$\frac{24h+7k-20}{\sqrt{(24{)}^2+(7{)}^2}}=\frac{4h-3k-2}{\sqrt{(4{)}^2+(-3{)}^2}}$

$\frac{24h+7k-20}{\sqrt{576+49}}=\frac{4h-3k-2}{\sqrt{25}}$

$\frac{24h+7k-20}{25}=\frac{4h-3k-2}{5}$

$2h+7k-20=20h-15k-10$

$4h=-22k+10$

$4(\frac{5-11k}{2}=-22k+10)$ [Using equation (1)]

$10-22k=-22k+10$

LHS = RHS

So,

Distance $24x+7y=20$ and $4x-3y-2=0$ from any point on the line $2x+11y-5=0$ is equal.