Question

Find the equation of a line parallel to y = x – 3 that contains the point (-2, 1). Yeah

Answer 1

Answer:

The equation of a line parallel to y = x - 3 that contains the point (-2, 1) is:

• $y = x + 3$

The graph of both the parallel equation is shown below to make you further understand the concept.

Step-by-step explanation:

The slope-intercept form of the line equation

$y = mx+b$

where

• $m$ is the slope

• $b$ is the y-intercept

Given the equation

$y = x - 3$

comparing with the slope-intercept form of the line equation

Slope m = 1

Important Tip:

• As the parallel lines never intersect, therefore, they have the same slopes.

Thus, the slope of the parallel line is also 1.

As the parallel line contains the point (-2, 1).

so substitute m = 1 and (-2, 1) in the slope-intercept form of the line equation to determine the y-intercept of the parallel line

$y = mx+b$

$1 = 1(-2) + b$        ∵ (x, y) = (-2, 1) and m = 1

$1 = -2 + b$

switch sides

$-2+b = 1$

add 2 to both sides

$-2 + b + 2 = 1 + 2$

simplify

$b = 3$

Thus, the y-intercept b of the parallel line is:  b = 3

now substitute b = 3 and m = 1 in the slope-intercept form of the line equation to determine the  equation of the parallel line

$y = mx + b$

$y = (1)x + 3$

$y = x + 3$

Therefore, the equation of a line parallel to y = x - 3 that contains the point (-2, 1) is:

• $y = x + 3$

The graph of both the parallel equation is shown below to make you further understand the concept.

From the graph:

• The green line represents the equation y = x - 3

• The black  line represents the parallel equation y = x + 3

It is clear that both lines are parallel.