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)

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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.


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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