Question

The line which passes through the intersection point of the lines $x-2y+2=0$ and $x+3y=13$ and parallel to the line $3x+4y=0$ formed a triangle with the coordinates axes then perimeter of the triangle is

Answer 1

Given lines are: $x-2y+2=0$ and $x+3y=13$ intersect at $(4,3)$

Required line passes through $(4,3)$ and is parallel to $3x+4y=0$

$\therefore$ Equation of required line: $y-3=\frac{-3}{4}(x-4)$

$\to 4y-12=-3x+12$

$\to 3x+4y=24$

Intersects x-axis at $(8,0)$ and y-axis at $(6,0)$

$\therefore$ Perimeter$=8+6+\sqrt{8^2+6^2}$

$=8+6+10=24$ units