Please answer this question (10) GIVING BRAINLIEST!!

Explanation:

For any rectangle, the diagonals are always the same length. We can use congruent triangles to prove this.

This means AC = BD.

Also, the diagonals of a rectangle cut each other in half (bisect). This indicates the following two equations

We'll use that second equation along with BP = -2x+23 and DP = 3y-6 to form the equation -2x+23 = 3y-6. This will be used later.

By the segment addition postulate, we know that

BP+DP = BD

(-2x+23)+(3y-6) = BD

BD = -2x+3y+17

Since the diagonals are equal, we also know that AC = -2x+3y+17

We are given that AC = 2x+4

Equating the two right hand sides leads to the equation 2x+4 = -2x+3y+17

The conclusion of each the last two sections was the following two equations

We have two equations and two unknowns. We have enough info to be able to find x and y.

Let's isolate 3y in the first equation

-2x+23 = 3y-6

3y-6 = -2x+23

3y = -2x+23+6

3y = -2x+29

Then we can plug this into the second equation

2x+4 = -2x+3y+17

2x+4 = -2x+(3y)+17

2x+4 = -2x+(-2x+29)+17 replace 3y with -2x+29

Now solve for x

2x+4 = -2x+(-2x+29)+17

2x+4 = -2x-2x+29+17

2x+4 = -4x+46

2x+4x = 46-4

6x = 42

x = 42/6

x = 7

We then use this to find y

3y = -2x+29

3y = -2(7)+29

3y = -14+29

3y = 15

y = 15/3

y = 5

I used the method of replacement on the sistem of equations.

The final answer I got was y=5

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.