Point a is located at (0, 4), and point b is located at (−2, −3). find the x value for the point that is 1 over 4 the distance from point a to point b.
−1.5
−2
−0.5
−1

The value of x is -0.5 ⇒ 3rd answer

Step-by-step explanation:

* Lets explain how to find a point which divides a line into two parts

- If point (x , y) divides a line whose endpoints are (x1 , y1) and (x2 , y2)

 at ratio m1 : m2 from (x1 , y1) to (x2 , y2), then

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* Lets solve the problem

∵ Point A located at (0 , 4)

∵ Point B located at (-2 , -3)

∵ Point (x , y) located at 1/4 the distance from A to B

- That means the distance from A to point (x , y) is 1/4 the distance

  from point A to point B

∴ The distance from A to B is divided into four equal parts, where

   the distance between point A and point (x, y) is one and the

   distance between point (x, y) and point B is three

∴ Point (x , y) divide the line AB at ratio 1 : 3 ⇒ (1 + 3 = 4)

* Now lets find the x

∵ x-coordinate of point A is 0

∴ x1 = 0

∵ x-coordinate of point B is -2

∴ x2 = -2

∵ m1 = 1 , m2 = 3

- By using the rule above

∴

* The value of x is -0.5

3

step-by-step explanation:

divide 13.5 by 9 to get the scale factor. it equals 1.5

2 * 1.5 = 3

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)