what is an equation of the line that s perpendicullar to y-3=-4(x+2) and passes through the point (-5,7)
1 answer:
Answer:
4y=x+33
y=x/4+33/4 (slope-intercept form)
Step-by-step explanation:
y-3 = -4(x+2)
y-3 -4x-8
y= -4x-8+3
y= -4x-5
m1 = -4
For perpendicularity
m2= -1/-4 = 1/4
The equation is
y-y1 = m2(x-x1)
y-7 = 1/4(x-(-5))
y-7 = x/4+5/4
Multiply through by 4
4y-28=x+5
4y=x+5+28
4y=x+33
Divide through by 4
y=x/4+33/4 (slope-intercept form)
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Ok so you would plug in 18 for the x so it would be 6(18)-12 and then 6x18 is 108 and 108-12=96 so your answer is 96
<em>The answer is 25.12 exactly</em>
