Question

Write an equation of the line containing the point (3,1) and perpendicular to the line 4x−3y=5.

Answer 1

The given line is 4x - 3y = 5.
Write the equation in standard form.
3y + 5 = 4x
3y = 4x - 5
$y= \frac{4}{3}x- \frac{5}{3}$
This is a straight line with slope = 4/3, and with y-intercept = - 5/3.

A perpendicular line should have a slope of -3/4, because the product of the slopes of two perpendicular lines is -1.
Let the perpendicular line be
$y=- \frac{3}{4}x+b$
Because the line passes through the point (3,1), therefore
$- \frac{3}{4}(3)+b=1 \\\\ - \frac{9}{4}+b=1 \\\\ b=1+ \frac{9}{4}= \frac{13}{4}$
The equation of the perpendicular line is
$y=- \frac{3}{4}x+ \frac{13}{4}$

The equation may also be written as
4y + 3x = 13.
A graph of the two lines is shown below.

Answer:
$y=- \frac{3}{4}x+ \frac{13}{4} \,\, or \,\, 4y+3x=13$