# Thread: Perpendicular Line EQ: x-2y-3=0

1. ## Perpendicular Line EQ: x-2y-3=0

I whipped through these in relatively short order until this one

Passing through (4,-7) and perpendicular to the line whose equation is $x -2y -3=0$

x -2y -3 = 0 : Subtract x from both sides and add 3 to both sides so that

-2y = -x +3 : Divide -x by -2 to get

$\frac{\1}{2}x$ The negative reciprocal of 1/2 is -2 so...

y- (-7)= -2(x-4) to get

y +7 = -2x +8 : subtract both 7 and y to get

-2x -y +1 = 0 : The answer in the book is the complete opposite. It gives

2x +y -1 = 0

But to get that requires something that I'm missing. Either that or the book is wrong. The latter is very rare and the former much more likely. Any and all help is appreciated. Thanks!

2. ## Re: Perpendicular Line EQ: x-2y-3=0

Your answer is in standard form of a line which shouldn't have a Negative x so all you do is multiply by -1. Which gives the same answer.

3. ## Re: Perpendicular Line EQ: x-2y-3=0

It's both correct, because if you multiply both sides of your equation with -1 you get the answer in the book.

4. ## Re: Perpendicular Line EQ: x-2y-3=0

Cool. Thanks guys.

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### x 2y-3=0

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