
a.
Calculate the slope − intercept forms of the equations of the lines through the given points which are parallel to the given line.
a.

Answer to Problem 68E
The slope intercept form for the parallel line is y=−x−1 .
Explanation of Solution
Given:
It is given in the question that the coordinates and equation are (−3,2),x+y=7 .
Concept Used:
In this , use the concept that the slope intercept form is y=mx+c and the parallel line have the same slope.
Calculation: The equation is x+y=7 ,make this in slope intercept form,
y=−x+7
A parallel line has the same slope as the line we are comparing it to. Then, plug in the given point into the slope intercept form to get the equation of the parallel line.
(2)=1(−3)+b2+3=bb=5
The parallel line will be : y=−x−1 .
Conclusion:
The equation is y=−x−1 .
b.
Calculate the slope − intercept forms of the equations of the lines through the given points which are perpendicular to the given line.
b.

Answer to Problem 68E
The slope intercept form for the perpendicular line is y=x+5 .
Explanation of Solution
Given:
It is given in the question that the coordinates and equation are (−3,2),x+y=7 .
Concept Used:
In this , use the concept that the slope intercept form is y=mx+c and the perpendicular line has a slope that is the negative reciprocal of the line.
Calculation: The equation is x+y=7 ,make this in slope intercept form,
y=−x+7
A perpendicular line has a slope that is the negative reciprocal of the line we are comparing it to. Then plug in the given point into the slope intercept form to get the equation of the perpendicular line.
(2)=1(−3)+b2+3=bb=5
Now,the perpendicular line will be y=x+5 .
Conclusion:
The equation is y=x+5 .
Chapter 1 Solutions
Precalculus with Limits: A Graphing Approach
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