SOLUTION: determine the equation of the line passing through the point (4,8) and perpendicular to the line y=2x+3

Question 625038: determine the equation of the line passing through the point (4,8) and perpendicular to the line y=2x+3
Answer by Aztec(6) (Show Source): You can put this solution on YOUR website!
The gradient of the line y=2x+3 is 2, since using the intercept formula y=mx + b where m is the gradient.
If a line "m1" is perpendicular to the line "m2", then the product of their gradients would equal to -1. (ie. m1 * m2 = -1)
So to find the line perpendicular to y=2x+3, where the gradient of the line is 2, we simply get the "negative reciprocal" of the number 2. The negative reciprocal of 2 is simply, the reciprocal of 2 which is 1/2 and the plus a minus sign in front, which would then be -1/2.
Using the line formula y-y1=m(x-x1), where we substitute the point (4,8) which is (x1,y1) and the gradient m which is -1/2.
y-8=-1/2(x-4)<===multiply by 2 to both sides.
2(y-8)=2(-1/2)(x-4)
2y-16=-2/2(x-4)
2y-16=-1(x-4)
2y-16=-x+4 <===if you are forced to write in general form, you must make x positive, so to do that, we just simply move -x and the 4 to the other side. That means that the signs change when -x and +4 move to the other side.
x+2y-20=0 or if you are made to write in the intercept form then it will be this
y=(-1/2)x+10

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