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agasfer [191]
2 years ago
8

Find an equation for the perpendicular bisector of the line segment whose endpoints

Geography
1 answer:
zzz [600]2 years ago
8 0

Answer:

y= -7/2x-26/2 or y= -7/2x-13

Explanation:

To solve this question you need all of these formulas:

gradient: \frac{y2-y1}{x2-x1} , Point-Slope y-y1=m(x-x1) ,

perpendicular m (m1)x(m2)=-1 , Midpoint (\frac{x1+x2}{2},\frac{y1+y2}{2} )

and equation of a line y=mx+c

where m stands for gradient

First things first.

To start you have to know the data of the line that includes the two endpoints, so you calculate its gradient (m) of this line with the gradient fromula: m=\frac{-4-(-8)}{5-(-9)} , which equals 4/14 or 2/7

(It can also be called rise/run) (Remember the rule of signs where - and - equal +)

with that information you can proceed with the point-slope or point-gradient formula, so you plug the values: y-(-8) = 2/7 (x-(-9)), which results in y+8=2/7(x+9) and then y+8=2/7x+18/7.

To finish the equation you move eight to the other side. To simplify things you can change it into a fraction as I did, and remember to change signs.

y=2/7x+18/7 -8 -> y=2/7x+18/7-56/7 . This gives us the number of y=2/7x-38/7, which is the equation of the first line.

Now to know the gradient of the second line you apply the formula of perpendicular bisector where m1 x m2 = -1. We know m1 (gradient of the first line) is 2/7, so m2 = \frac{-1}{2/7} = -7/2. m2 is therefore -7/2

Now you have to know the midpoint between the two endpoints, which will act as the start point of the perpendicular bisector

M (midpoint) = (\frac{-9+5}{2} ,\frac{8+(-4)}{2} ), which give us the coordinates of (-2, -6)

(remember, x coordinate is always first)

with this point we can apply again the point-slope formula to know the equation of the line:

y-(-6)=-7/2(x-(-2)) -> y+6=-7/2(x+2) -> y+6=-7/2x - 14/2

Move the 6 to isolate the y

y=-7/2x -14/2 - 12/2

which equals y= -7/2x -26/2

You can check the results in this page: GraphPlotter

https://www.transum.org/Maths/Activity/Graph/Desmos.asp

To make sure the answer is correct.

Hope it helps :)

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