Find the equation of the line passing through the point (4,6) and perpendicular to the line y = (3/2)x + 5.  Write your answer in general form.  Do not use any space between the letter (lower case), symbol and number, say x+y-1=0
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When two lines are perpendicular, the product of their gradients is -@. Therefore the gradient of the line we need is -⅔. This is the x coefficient, so we have y=-⅔x+a where a is the y intercept.

Plug in the point (4,6): 6=-⅔×4+a, 6+8/3=a and a=26/3. So y=-⅔x+26/3=⅔(13-x).

Get rid of the fraction and we have 3y=2(13-x)=26-2x. In general form this is: 2x+3y-26=0.

by Top Rated User (599k points)

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