Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
2. Section 1.2
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Exercise 77 Page 42

Practice makes perfect
a We are given the line in slope-intercept form which means we can identify it's slope and y-intercept directly .
|c|c|c| Line & Slope & y-intercept [-0.8em] y= - 2/5x+ 6& - 2/5 & 6 [0.8em] If we multiply the slopes of two perpendicular lines, the product equals -1. With this information, we can find the slope of the perpendicular line.
m_1m_2=-1
m_1( -2/5)=-1
â–Ľ
Solve for m_1
- m_1(2/5)=-1
- m_1(2)=-5
m_1(2)=5
m_1=5/2
The slope of the perpendicular line is 52. |c|c|c| Perpendicular Line & Slope & y-intercept [-0.8em] y= 5/2x+ b& 5/2 & b [0.8em] To complete the equation, we also have to find the line's y-intercept. Notice that the y-intercept is a point on the y-axis which means its x-coordinate is 0. Since we know that the perpendicular line goes through (0,-3) we know that the y-intercept is at -3. Now we can complete the equation. y= 5/2x-3 Let's plot the graph of both equations. Since they are perpendicular, they will intersect at a right angle.
b Parallel lines have the same slope but different y-intercepts. To find the slope of the line, we must write it in slope-intercept form.
- 3x+2y=10
â–Ľ
Write in slope-intercept form
2y=3x+10
y=3x/2+5
y=3/2x+5
Now we can identify the line's slope and y-intercept.

|c|c|c| Line & Slope & y-intercept [-0.8em] y= 3/2x+ 5& 3/2 & 5 [0.8em] As already explained, if two lines are parallel, they have to have the same slope but different y-intercepts. Therefore, we know that the parallel line must have a slope of 32. |c|c|c| Parallel Line & Slope & y-intercept [-0.8em] y= 3/2x+ b& 3/2 & b [0.8em] Like in Part A, we see that the given point has an x-coordinate of 0. This means the points y-coordinate describes the line's y-intercept. With this information, we can write the line's equation. y=3/2x+7 If we plot both lines, we see that they are parallel.