Question

Find the x value for the point that divides the lines segment AB into ratio of 2:3

Answer 1

Let $(x_A,y_A),\ (x_B,y_B)$ be the coordinates of the points A and B and $(x_0,y_0)$ be the coordinates of the point O that divides the segment AB in ratio 2:3.

Consider vectors

$\overrightarrow{AO}=(x_0-x_A,y_0-y_A),\\ \\\overrightarrow{OB}=(x_B-x_0,y_B-y_0).$

These vectors are collinear and

$\dfrac{\overrightarrow{AO}}{\overrightarrow{OB}}=\dfrac{2}{3}.$

Then

$\left\{\begin{array}{l}\dfrac{x_0-x_A}{x_B-x_0}=\dfrac{2}{3}\\ \\\dfrac{y_0-y_A}{y_B-y_0}=\dfrac{2}{3}\end{array}\right.\Rightarrow \left\{\begin{array}{l}3(x_0-x_A)=2(x_B-x_0)\\ \\3(y_0-y_A)=2(y_B-y_0)\end{array}\right..$

This means that

$\left\{\begin{array}{l}3x_0-3x_A=2x_B-2x_0\\ \\3y_0-3y_A=2y_B-2y_0\end{array}\right.\Rightarrow \left\{\begin{array}{l}5x_0=2x_B+3x_A\\ \\5y_0=2y_B+3y_A\end{array}\right..$

Thus,

$x_0=\dfrac{2x_B+3x_A}{5},\ y_0=\dfrac{2y_B+3y_A}{5}.$

Answer: $\left(\dfrac{2x_B+3x_A}{5},\dfrac{2y_B+3y_A}{5}\right).$

Answer 2

Answer:  x value is -0.4

Step-by-step explanation:

So theres a total of 5 pieces to segment AB.

Another way to look at 2:3 is as a fraction 2/5

-Look at the x coordinates

A(-4,6)  B(5,1)

-4 to 5 = a distance of 9

-multiply 2/5 × 9 = 3.6

-Look at y coordinates

A(-4,6)  B(5,1)

-1 to 6 = distance of 5

-multiply 2/5 × 5 = 2

.Shift both coordinates with theyre corresponding #

. x axis: -4 + 3.6 = -0.4

.y axis: 6+ 2= 8

X value = -0.4

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