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Here are some tips for Applied Linear Equations 2, which aligns with Florida state standards:

Applied Linear Equations 2

This topic continues the topic of Applied Linear Equations 1.
To review the Applied Linear Equations 1 topic, see here.

For two lines, where m1 is the slope of one line and m2 is the slope of the second line, the two lines are

• parallel if m1 = m2.

• perpendicular if m1 × m2 = -1.
In this case, m1 and m2 are called negative reciprocals.

Example 1: Perpendicular slope

What slope is perpendicular to slope 5/2?

To find the perpendicular slope, find the negative reciprocal of 5/2.
1. Flip the slope to get the reciprocal
 52 → 25
2. Reverse the sign to get the negative reciprocal
 25 → - 25
The negative reciprocal of 5/2 is -2/5.

Example 2: Find the equation of a line

In slope-intercept form, write the equation of the line that intersects (2, -4) and is perpendicular to x + 3y = 0
Equation:

1. Find the slope perpendicular to x + 3y = 0
 A line equation with the form ax + by = c, the slope m = - ab The line x + 3y = 0 has a slope m = - 13 Therefore, the line perpendicular to x + 3y = 0 has a slope m1 = 3

2. Knowing the slope and intersection point, find the equation of the perpendicular line

The slope m1 = 3 and a point on the line (x, y) = (2, -4)
 y = mx + b -4 = 3(2) + b -4 = 6 + b -10 = b
Plugging the values for m and b into the slope-intercept formula, we get the answer

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