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Write an equation in slope-intercept form for the line that passes through (-5,3) and is perpendicular to the line described by y= 5x.

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3 Answers

Hi Haley;
Write an equation in slope-intercept form for the line that passes through (-5,3) and is perpendicular to the line described by y= 5x.
The equation is in slope-intercept form...
y=mx+b
m is the slope.
b is the y-intercept, the value of y when x=0.
y=5x
Slope is 5
y-intercept is zero.  However, this will be a non-issue when establishing the perpendicular equation.
In such equation, the slope is -1/5.
y=(-1/5)x+b
Let's apply the point-slope formula...
(y-y1)=m(x-x1)
y-3=(-1/5)(x--5)
y-3=(-1/5)(x+5)
y-3=(-1/5)x-1
y=(-1/5)x+2
Given line y = 5x
compare with slope intercept line equation y = mx + b where m is slope
therefore slope of given line is 5
 
Slope of new line is perpedicular to given line
therefore its slope is - 1/5
 
equation of new line
 
y = -1/5x + b
since it passes through -5,3
 
3 = -1/5x(-5) + b
 
b = 2
 
equation of new line
 
y = -1/5 x + 2
 L1l : Y = 5X
 
 L2:   Y = -1/5 X +b    /passes through ( -5.3)
 
         3 = -1/5(-5) +b
 
         b = 3 - 1 = 2
 
      L2 : Y = -1/5 X + 2