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Question 503114: write the slope -intercept equation for the line that passes through (-3,-15) and is perpendicular to -6x+8y=3 please show all of your work
Answer by Maths68(1474)   (Show Source):
You can put this solution on YOUR website! -6x+8y=3
Arrange the above equation in slope-intercept form
8y=6x+3
y=(6x+3)/8
y=6/8(x)+3/8
y=3/4(x)+3/8
y=3/4(x)+3/8 , passing through the point (-3,-15)
Equation of the line slope-intercept form
y=mx+b
Given line
y=3/4(x)+3/8
Slope of given line m = 3/4
Since the required line is perpendicular to the given line the sign of this line will be opposite to sign of given line's slope and it will be reciprocal to the slope of the given line
Thus
slope (m) of required line will be = -4/3
Now we have a slope (-4/3) and we have a point (-3, -15)
We can easily find the equation of the line either by slope-intercept form or by Point slope-form.
Use slope-intercept
y=mx+b ..........(1)
here
y=-15, x=-3, m= -4/3 and we have to find b
Plug in the value in the equation of line
-15=(-4/3)(-3)+b
-15=4+b
-15-4=b
b=-19
Put the value m and b in (1)
y=(-4/3)x-19
Equation of the line passing through (-3, -15) and perpendicular to line y=3/4(x)+3/8) is
y=(-4/3)x-19
Use Point slope form
m=(y1-y)/(x1-x)
Plug in the values m = -4/3, x=-3, y=-15
-4/3=(-15-y)/(-3-x)
(-4/3)(-3-x)=(-15-y)
(-4)(-3-x)=(-15-y)*3
12+4x=-45-3y
12+45=-4x-3y
57+3y=-4x
3y=-4x-57
y=(-4x-57)/3
y=-4/3(x)-57/3
y=-4x/3 - 19
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