To review these concepts, go to The Slope and Equation of a Line Help
Slope Practice Problems
Directions: When asked to find an equation for a line, put your answer in the form y = mx + b unless the line is horizontal ( y = a ) or vertical ( x = a ).
Practice

Find the slope of the line containing the points (4, 12) and (−6, 1).

Find the slope of the line with x intercept 5 and y intercept −3.

Find an equation of the line containing the point (−10,4) with slope .

Find an equation of the line with y intercept −5 and slope 2.

Find an equation of the line in Figure 1.13.

Find an equation of the line containing the points and (−2, −1).

Determine whether the lines 3 x − 7 y = 28 and 7 x + 3 y = 3 are parallel or perpendicular or neither.

Find an equation of the line containing (2, 3) and perpendicular to the line x − y = 5.

Find an equation of the line parallel to the line x = 6 containing the point (−3, 2).

Determine whether the lines 2 x − 3 y = 1 and − 4 x + 6 y = 5 are parallel or perpendicular or neither.
Solutions


The x intercept is 5 and the y intercept is −3 mean that the points (5, 0) and (0, −3) are on the line.

Put x = −10, y = 4, and in y = mx + b to find b .

m = 2, b = −5, so the line is y = 2 x − 5.

From the graph, we can see that the y intercept is 3. We can use the indicated points (0, 3) and (2, 0) to find the slope in two ways. One way is to put these numbers in the slope formula.
The other way is to move from (0, 3) to (2, 0) by going down 3 (so the numerator of the slope is −3) and right 2 (so the denominator is 2). Either way, we have the slope . Because the y intercept is 3, the equation is .

We will use x = −2 and y = −1 in y = mx + b.

We will solve for y in each equation so that we can compare their slopes.
The slopes are negative reciprocals of each other, so these lines are perpendicular.

Once we have found the slope for the line x − y = 5, we will use its negative reciprocal as the slope of the line we want.
x − y = 5
y = x − 5
The slope of this line is 1. The negative reciprocal of 1 is − 1. We will use x = 2, y = 3, and m = −1 in y = mx + b.
3 = −1(2) + b 5 = b
The equation is y = −1 x + 5, or simply y = −x + 5.

The line x = 6 is vertical, so the line we want is also vertical. The vertical line that goes through (−3, 2), is x = −3.

We will solve for y in each equation and compare their slopes.
The slopes are the same, so these lines are parallel.
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