Question: Find the equation of straight line which passes through the point of intersection of the lines 2x -3y = 1 and x+2y-4 = 0 = 0 and  perpendicular to the line 3x -4y +5 = 0.

Solution:

For line 1,
Equation of the line is 3x -4y +5 = 0
Slope of line ($m_1$) =  $- \frac{coefficient \; of \; x}{ coefficient\; of y}$
$= - \frac{3}{-4}$
$= \frac{3}{4}$

Given,
Second line passes through 2x -3y = 1 and x+2y -4 = 0.
Solving the equations simultaneously,

$or, 2x -3y = 1$
$or, 2x = 1 +3y$
$or, x = \frac{1+3y}{2}$ - (i)

$or, x +2y -4 = 0$
$or, x + 2y = 4$
$or, x = 4-2y$ - (ii)

Put value of x from equation (i) in equation (ii), we get,
$or, \frac{1 + 3y}{2} = 4 - 2y$
$or, 1 + 3y = 2(4 -2y)$
$or, 1 + 3y = 8 -4y $
$or, 3y+4y = 8-1$
$or, 7y = 7$
$\therefore y = 1$

Put value of x in equation (ii), we get,
$or,x = 4 - 2×1$
$or, x = 4 -2$
$\therefore x = 2$

(x,y) = (2,1)
So, the other line passes through the point (2,1).

Since both the lines are perpendicular, the products of their slopes result negative 1. Let $m_2$ represent the slope of line 2, we have
$m_1 × m_2 = -1$
$or, \dfrac{3}{4} × m_2 = -1$
$\therefore m_2 = - \dfrac{4}{3}$

Now, equation of line 2 when $m_2 = - \frac{4}{3}$ and (2,1) = ($x_1,y_1$) is given by;
$y - y_1 = m_2 (x -x _1)$

$or, y -1 = - \dfrac{4}{3} (x - 2)$

$or, 3(y -1) = - 4(x -2)$

$or, 3y -3 = -4x +8$

$or, 4x +3y -3-8 = 0$

$or, 4x +3y -11 = 0$ is the required equation.


Therefore, the required equation of the line is 4x +3y -11 = 0.