Answer to Question #141773 in Analytic Geometry for me

Question

find the equation of the straight line that passes through the intersection point between the lines x + y = 3 and 2x - 3y = 16 and perpendicular to the straight line y = 4x - 3

Answer 1

x + y = 3 in the form y=mx+c

will be $y=-x+3$ ........(equation 1)

2x - 3y = 16 in the form y=mx+c

will be $y=\frac{2}{3}x - \frac{16}{3}$........(equation 2)

At point of intersection (equation 1) is equal to (equation 2)

$-x+3=\frac{2}{3}x - \frac{16}{3}$

$3+\frac{16}{5} = \frac{2}{3}x+x$

$\frac{25}{3} =\frac{5}{3}x$

Hence. $x = 5$

From (equation 1) $y=-x+3$

Therefore

$y = -5 +3$

$y = -2$

Hence point of intersection is (5,-2)

For perpendicular lines the product of their gradient is -1

For line $y=4x-3$ the gradient is $4$

Let the gradient of the other line be M₂

M₂ x 4 = -1

M₂ = -1/4

$\frac{y--2}{x-5} = \frac{-1}{4}$

$4y+8=-x+5$

$4y = -x-3$

The equation of the line will be,

$y= -\frac{1}{4}x -\frac{3}{4}$