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Question 84707This question is from textbook Intermediate Algebra
: Exercise #26 Find the equation of the line that contains the points (3,2)and is parallel to the line the 3x+y=-3.
Exercise #28 Find the equation of the line that contains the point (2,-5) and is perpendicular to the line y=5/2x-4.
This question is from textbook Intermediate Algebra
Answer by uma(370)   (Show Source):
You can put this solution on YOUR website! 26)given line is 3x + y = -3
==> y = - 3x - 3
This is of the form y = mx + c where m is the slope.
So slope of the given line = -3
So slope of the required line is also -3 as parallel lines have the same slope.
It passes through (3,2).
So the equation is: y - y1 = m(x - x1)
==> y - 2 = -3(x-3)
==> y-2 = -3x + 9
==> y = -3x + 9 + 2
==> y = -3x + 11
==> 3x + y = 11
==> the equation of the line that contains the points (3,2)and is parallel to the line the 3x+y=-3 is 3x + y = 11.
27)Given line is y = 5/2x - 4
==> slope of the line is 5/2
Perpendicular lines have the product of their slopes = -1
So slope of the perpendicular to the given line is -1/(5/2)
= -2/5
The line passes through (2,-5)
So the equation is: y - y1 = m(x - x1)
==> y-(-5) = (-2/5)(x-2)
==> y+5 = (-2/5)(x-2)
==> y = -2/5x + 4/5 - 5
==> y = -2/5x - 21/5
==> the equation of the line that contains the point (2,-5) and is perpendicular to the line y=5/2x-4 is y = -2/5x - 21/5
Good Luck!!!
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