Question

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

Answer 1

$2x+11y=5$

Let $(h,k)$ be any point on the line

$2h+11k=52h=5-11k\Rightarrow h=\frac{5-11k}{2}$

So any point on the line will be of form $P(\frac{5-11k}{2},k)$

$24x+7y-20=\mathrm{0...........}\left(i\right)$

Let $p_1$ be the perpendicular from $P$ to $\left(i\right)$

$p_1=\frac{|24\left(\frac{5-11k}{2}\right)+7k-20|}{\sqrt{{\left(24\right)}^2+7^2}}{p}_1=\frac{|60-132k+7k-20|}{\sqrt{625}}{p}_1=\frac{|40-125k|}{25}{p}_1=\frac{|8-25k|}{5}$

$4x-3y-2=\mathrm{0.......}\left(ii\right)$

Let $p_2$ be the perpendicular from $P$ to $\left(ii\right)$

$p_2=\frac{|4\left(\frac{5-11k}{2}\right)-3k-2|}{\sqrt{4^2+3^2}}{p}_2=\frac{|10-22k-3k-2|}{\sqrt{25}}{p}_2=\frac{|8-25k|}{5}$

Clearly we get $p_1=p_2$

Hence proved