JEE 2024 — Mathematics PYQ

Point of intersection of the lines  
$3x+2y=14$  
$5x-y=6$  
is $A(2,4)$.

Point of intersection of the lines  
$4x+3y=8$  
$6x+y=5$  
is $B\left(\frac{1}{2},2\right)$.

Eqn of line $AB$:

$\Rightarrow y-4=\frac{4-2}{2-\frac{1}{2}}(x-2)$

$\Rightarrow y-4=\frac{4}{3}(x-2)$

$\Rightarrow 3y-12=4x-8$

$\Rightarrow 4x-3y+4=0$

Distance of the point $P(5,2)$ from the line  
$4x-3y+4=0$ is

$\left|\frac{4(5)-3(-2)+4}{\sqrt{4^2+(-3{)}^2}}\right|<br>=\frac{30}{5}<br>=6$