SOLUTION: Write the equation of the line that passes through the given point and is parallel or perpendicular to the given line. Write the answer in slope-intercept form.
(−8, 1),
pe
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Question 513698: Write the equation of the line that passes through the given point and is parallel or perpendicular to the given line. Write the answer in slope-intercept form.
(−8, 1),
perpendicular to
y + 8x = −14
y =
Found 2 solutions by umail08, Maths68:
Answer by umail08(87) (Show Source): You can put this solution on YOUR website!
Write the answer in slope-intercept form.
(−8, 1),
y + 8x = -14
y = -8x -14 … slope = -8
The slope of a perpendicular line would be 1/8 … the negative reciprocal
So the equation of a line perpendicular to y + 8x = -14 and passing thru (-8,1) would be:
(-8,1) … (x1,y1)
y-y1 = m(x-x1)
y – 1 = 1/8(x-(-8))
y-1=1/8(x+8)
y – 1 = 1/8x + 1
y = 1/8x + 2
Answer by Maths68(1474) (Show Source): You can put this solution on YOUR website!
If lines are Perpendicular
=========================
Standard Form of Equation of the line:
y=mx+b
Given
y + 8x = −14
rearrage the above equation according to the standard form
y=-8x-14
Compare above equation with the standard form equation
m=-8 and b=-14
Since lines are perpendicular multiplicatin of their slope will be (-1)
So slope of the required line will be (1/8)
Now we have a point(-8,1) and slope (1/8)of the line we can easily find required lines by putting these values in the equation of the straight line poin-slope form.
m=(y2-y1)/(x2-x1)
1/8=(y-1)/(x-(-8))
1/8=(y-1)/(x+8)
1(x+8)=8(y-1)
x+8=8y-8
8y-8=x+8
8y=x+8+8
8y=x+16
y=(x+16)/8
y=x/8+16/8
y=x/8+2
The standard form of the equation of the line
y=(1/8)x+2
================================================================================================================
If lines are Parallel
=========================
Standard Form of Equation of the line:
y=mx+b
Given
y + 8x = −14
rearrage the above equation according to the standard form
y=-8x-14
Compare above equation with the standard form equation
m=-8 and b=-14
Since lines are parallel their slope will be same
So slope of the required line will be (-8)
Now we have a point(-8,1) and slope (-8)of the line we can easily find required lines by putting these values in the equation of the straight line poin-slope form.
m=(y2-y1)/(x2-x1)
-8=(y-1)/(x-(-8))
-8=(y-1)/(x+8)
-8(x+8)=(y-1)
-8x-64=y-1
-8x-64+1=y
-8x-63=y
y=-8x-63
Above equatoin is the required equation of the line in standard form.
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