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Tasya [4]
4 years ago
5

write the equation of a line that is perpendicular to the given line and that passes through the given point. y-4=5/2(x+3) ; (-7

,8)
Mathematics
2 answers:
faltersainse [42]4 years ago
7 0
The gradient of the perpendicular line would be the negative reciprocal of the original line. Therefore the gradient of the perpendicular line would be -2/5x.
Since we know y=mx+c, ∴y=-2/5x+c. Sub in the x and y values of the given point and we get that c=26/5.

The perpendicular equation would be y=-2/5x+26/5.
I hope I got this right.
Jobisdone [24]4 years ago
4 0
Comparing equation y-4 = 5/2(x+3) with slope-point form y-y1=m(x-x1), slope of given line is 5/2. Let m' be the slope of required line. product of the slope of perpendicular lines is equal to -1. Therefore, mxm' = -1. m' = -2/5. Using slope-point form to find the equation of required line, y-y1 = m(x-x1) . y-8 = -2/5(x-(-7)). y-8 = -2/5(x+7).
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