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

Solution:

We have, y = mx + c

Finding the point of intersection,

Given two equations are (2x +y =5) and (x -y = 1).

$or, 2x + y =5$
$or, 2x = 5-y$
$or, x = \frac{5-y}{2}$ - (i)

$or, x - y = 1$
Put value of x from equation (i), we get,
$or, \frac{5-y}{2} - y = 1$
$or, \frac{5-y}{2} = 1+y$
$or, 5 - y = 2(1+y)$
$or, 5 -y = 2 +2y$
$or, 5 -2 = 2y +y$
$or, 3 = 3y$
$\therefore y = 1$

Put value of y in equation (i), we get,
$or, x = \frac{5-1}{2}$
$or, x = \frac{4}{2}$
$\therefore x = 2$

So, the point of intersection (x,y) = (2,1).

Given,
Equation of line 2 is 3x -4y +5 = 0
Slope of line 1 ($m_1$) = $- \frac{coefficient \; of \; x}{coefficient \; of \; y}$
$= - \frac{3}{-4}$
$= \frac{3}{4}$

Line 1 and line 2 are perpendicular. Let $m_2$ represent the slope of the line 2.
$or, m_1 × m_2 = -1$
$or, \frac{3}{4} × m_2 = -1$
$\therefore m_2 = - \frac{4}{3}$

Also, in the equation y = mx +c, m represents the slope of the line.

So, m = $m_2$ =$ - \frac{4}[3}$

To find c,
Put value of (x,y) = (2,1) and $m = - \frac{4}{3}$
$or, y = mx + c$
$or, 1 = - \frac{4}{3} 2 + c$
$or, 1 + \frac{8}{3} = c$
$or, \frac{3+8}{3} = c$
$\therefore c = \frac{11}{3}$

Now, put value of m and c in equation y = mx+c, we get,

$or, y = - \dfrac{4}{3} x + \dfrac{11}{3}$

$or, y = - \dfrac{4x +11}{3}$

$or, 3y = -(4x +11)$

$or, 3y = -4x -11$

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

Therefore, the required equation of the given lien is either $y = - \frac{4}{3} x + \frac{11}{3}$ or $4x +3y +11 = 0$.