Find the length of the perpenducular drawn from the point of inter section of the lines $3\mathrm{x}+2\mathrm{y}+4=0,2\mathrm{x}+5\mathrm{y}-1=0$ to the straight line $7\mathrm{x}+24\mathrm{y}-15=0$. 

Given lines
$\begin{array}{l}3\mathrm{x}+2\mathrm{y}+4=0 ...\left(1\right)\\ 2\mathrm{x}+5\mathrm{y}-1=0 ...\left(2\right)\\ 7\mathrm{x}+24\mathrm{y}-15=0 ...\left(3\right)\end{array}$
Solving (1) & (2)
$$\begin{gathered} \mathrm{x} \mathrm{y} 1 \\ 2 4 3 2 \\ 5 -1 2 5 \end{gathered}$$
$\begin{array}{l}\frac{\mathrm{x}}{-2-20}=\frac{\mathrm{y}}{8+3}=\frac{1}{15-4}\\ \frac{\mathrm{x}}{-22}=\frac{\mathrm{y}}{11}=\frac{1}{11}\\ \frac{\mathrm{x}}{-22}=\frac{1}{11}\Rightarrow \mathrm{x}=-2\\ \frac{\mathrm{y}}{11}=\frac{1}{11}\Rightarrow \mathrm{y}=1\\ \mathrm{P}(-2,1)\end{array}$
d = length of the perpendicular from (–2,1) to
$\begin{array}{l}7\mathrm{x}+24\mathrm{y}-15=0\\ d=\frac{\left|{\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{c}\right|}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\\ =\frac{\left|7\right(-2)+24(1)-15|}{\sqrt{49+576}}\\ =\frac{|-5|}{25}\\ =\frac{1}{5}\end{array}$