Envision Math 2.0: Grade 8, Volume 2
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Exercise 1 Page 270

Practice makes perfect
We are given the graph of a system of linear equations that shows perpendicular lines.
Perpendicular lines

We want to find the number of solutions to this system of equations. Note that the lines described by the equations of the system intersect at a point.

Solution of the perpendicular lines

The point of intersection solves the system of linear equations. This means that the system has one solution.

We are asked to determine the slopes and y-intercepts of the lines, and then to explain the relationship between the slopes. To do so, let's express the equations of the lines in slope-intercept form.

y=3/4x+1 y=-4/3x-2 ⇒ y= 3/4x+ 1 y= -4/3x+( - 2) We can now identify the desired parts.

Equations Slope y-intercept
y= 3/4x+ 1 3/4 1
y= -4/3x+( - 2) -4/3 - 2

Notice that the slopes of two perpendicular lines are negative reciprocals of each other.

Consider the following system of equations. y= 1/2x+8 y= mx-6 We want to find the value of m that makes the lines perpendicular. In Part A, we realized that the slopes of two perpendicular lines are negative reciprocals. Therefore, the value of m is equal to the negative reciprocal of the slope of the first equation. Note that the slope of the first equation is 12.

1/2 negative reciprocal → -2/1 The value of m is -2.