1 4. a line perpendicular to y =x+6 3 through (3, 2) -3 slope of the new line: y =-3x + 11 equation: 6. a line perpendicular to y = 5x – 2 through (-10, 6) slope of the new line: equation: 8. a line perpendicular to y = 2x + 4 through (4, 2) slope of the new line: equation:

Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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1
4. a line perpendicular to y =x+6
3
through (3, 2)
-3
slope of the new line:
y=-3x +11
equation:
6. a line perpendicular to y = 5x – 2
through (-10, 6)
slope of the new line:
equation:
8. a line perpendicular to y = 2x + 4
through (4, 2)
slope of the new line:
equation:
Transcribed Image Text:1 4. a line perpendicular to y =x+6 3 through (3, 2) -3 slope of the new line: y=-3x +11 equation: 6. a line perpendicular to y = 5x – 2 through (-10, 6) slope of the new line: equation: 8. a line perpendicular to y = 2x + 4 through (4, 2) slope of the new line: equation:
Expert Solution
Step 1 CONCEPTS USED

 A line is given to us in the form y=m1 x+c where m1 is the slope 

To find a line perpendicular to the given line we need to find a slope m2  which is in the form m1m2=-1

And then the equation of a perpendicular line becomes y=m2x+c

 

Step 2 question 6

we have given an equation of a line y=5x-2 where the slope m1=5 

so, 

m1m2=-15.m2=-1m2=-15

hence the slope is m2=-1/5

now, putting the slope in the equation we get 

                                  y=m2x+cy=-15x+c                                       (1)

now, to find the value of c we will substitute x and y by the coordinates (-10,6)

6=-15(-10)+c6=2+cc=4

putting back the value of in the equation (1) we get 

y=-15x+45y=-x+205y+x=20

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