Can anyone me to understand how to solve coordinate proofs in

The coordinate proof is a proof of a geometric theorem which uses "generalized" points on the Cartesian Plane to make an argument.The method usually involves assigning variables to the coordinates of one or more points, and then using these variables in the midpoint or distance formulas .For example, the following is a coordinate proof of the Triangle Midsegment Theorem , which states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and exactly half the length.Without loss of generality, we can assume that one side of the triangle lies on the xx -axis with one vertex at (0,0)(0,0) and the other vertex at(a,0)(a,0).Let the third vertex have the coordinates (b,c)(b,c) . We can assume without loss of generality that this third vertex lies in the first quadrant. (If it doesn't, you can just reflect the triangle over the xx - and/or yy -axis until it does.)

Example:

To Prove: XY¯¯¯¯¯¯XY¯ is parallel to PR¯¯¯¯¯PR¯ and

XY=12PRXY=12PR

Proof:

Using the midpoint formula , the coordinates of the midpoints XX and YY are

X(0+b2,0+c2)=X(b2,c2)Y(a+b2,0+c2)=Y(a+b2,c2)X(0+b2,0+c2)=X(b2,c2)Y(a+b2,0+c2)=Y(a+b2,c2)

First note that the slope is 00 , since

m=riserun     =(c2−c2)(b2−(a+b2))     =0(b2−(a+b2))     =0m=riserun     =(c2−c2)(b2−(a+b2))     =0(b2−(a+b2))     =0

So, XY¯¯¯¯¯¯XY¯ is parallel to PR¯¯¯¯¯PR¯ .

Now, compare the lengths of the two segments.

XY=a+b2−b2        =a2PR=a−0        =aXY=a+b2−b2        =a2PR=a−0        =a

So,

XY=12PRXY=12PR

Thanks to google for example

hope this helps

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

e

step-by-step explanation:

an equation by definition has to have two or more variables (x and 2 in this case)