Question

find the equation of parallel line and perpendicular line .

Answer 1

Part 1)

The slope-intercept form of the line equation

$y = mx+b$

where m is the slope and b is the y-intercept

Given the equation

$y = 3x-4$

comparing with the slope-intercept form of the line equation y = 3x-4

Thus, the slope of the line: m = 3

We know that the parallel lines have the same slopes.

Thus, the slope of the paralellel line is also: 3

substituting the slope m = 3 and the point (5, -3) in the slope-intercept form

$y = mx+b$

-3 = 3(5) + b

-3 = 15 + b

b = -3-15

b = -18

Therefore, the value of y-intercept b = -18

now substituting the slope m = 3 and the y-intercept b = -18 in the slope-intercept form

$y = mx+b$

y = 3x + (-18)

y = 3x - 18

Therefore, the equation of line parallel to the given line $y = 3x-4$ will be:

$y = 3x - 18$

Part 2)

The slope-intercept form of the line equation

$y = mx+b$

where m is the slope and b is the y-intercept

Given the equation

$y = 3x-4$

comparing with the slope-intercept form of the line equation y = 3x-4

Thus, the slope of the line: m = 3

We know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line, such as:  

slope = m = 3

Thus, the slope of the the new perpendicular line = – 1/m = -1/3 = -1/3

substituting m = -1/3 and the point   (5, -3)  in the slope-intercept form

$y = mx+b$

$-3\:=\:-\frac{1}{3}\left(5\right)+b$

$-\frac{5}{3}+b=-3$

$b=-\frac{4}{3}$

now substituting the slope m = -1/3 and the y-intercept b = -4/3 in the slope-intercept form

$y = mx+b$

$y=\:-\frac{1}{3}x+\left(-\frac{4}{3}\right)$

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

Therefore, the equation of the line perpendicular to the given line will be:

• $y=\:-\frac{1}{3}x-\frac{4}{3}$