The sides of a triangle have the equations y=-1/2x+1, y=2x-4, and y=-3x-9. Verify that the triangle is an isosceles right triangle. Algebraic solutions only!

Answer 1

Method:
solve for the vertices by solving the 3 different pairs of equations.
The take the length of sides between pairs of points using the distance formula √[(x2 - x1)^2 + (y2 - y1)^2]

I will do one of the vertices:
y = 2x-4 with y = -3x-9
2x-4 = -3x-9
5x = -5
x = -1
back in the first ...
y = 2(-1) - 4 = -6
so one vertex is (-1,-6)

let me know if it works out for you

Answer 2

how do i solve the one with the fraction?
do i multiply everything by two?
or put every thing over two?

Answer 3

Ok, I will do one more

y = -(1/2)x + 1 with y = 2x - 4

2x - 4 = -(1/2)x + 1
multiply each term my 2
4x - 16 = -x + 2
6x = 18
x = 3, then y=2(3) + 4 = 10

so another point is (3,10)

now you try the last one, let me know what you got.

Answer 4

3(x+2)=12

Answer 5

Actually in terms of method, the method you used for the second point is right

But there was a small error in calculation :)

-(1/2)x + 1 with y = 2x - 4

2x - 4 = -(1/2)x + 1

multiply each term my 2
*accidentally multiplied -4 by 4 to get 16, should be -8

4x - 8 = -x + 2
*And then over here you got 6x= 18 but bringing over that -(1)x would mean 4x +1 x =5x not 6x

5x = 8+2
5x=10
x/5=10/5
x = 2,

*Over here the formula should be y= 2x -4, not y=2x+4 which I'm sure is just a small error you did in a rush

then y=2(2) - 4
y= 4-4
y=0

so another point would be (2,0)

Hope this helped