Mathematics, 27.01.2022 19:40 manny2822

Find the equation of a line passing through the point (2,3), and perpendicular to the line with equation 3y-6x=4? The answer in the book is 2y- x-8=0 and I don't know how I am always getting this wrong. My working out has been rearranging the first equation and finding that the gradient is -2.

Having a -2 gradient means, m1 (in this case, -2) x m2 = -1, thus m2 must be 1/2.

Then doing, y-3= 1/2(x-2)

thus giving me 2(y-3)=1(x-2)

2y - 6 = x -2

rearranging to give me: 2y = x + 4

I understand if rearranged to other form it would be, 2y -x -4 = 0 but I don't know how it is -8?

Answers: 2

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Find the equation of a line passing through the point (2,3), and perpendicular to the line with equa...