Let [tex](x_A,y_A),\ (x_B,y_B)[/tex] be the coordinates of the points A and B and [tex](x_0,y_0)[/tex] be the coordinates of the point O that divides the segment AB in ratio 2:3.
Consider vectors
[tex]\overrightarrow{AO}=(x_0-x_A,y_0-y_A),\\ \\\overrightarrow{OB}=(x_B-x_0,y_B-y_0).[/tex]
These vectors are collinear and
[tex]\dfrac{\overrightarrow{AO}}{\overrightarrow{OB}}=\dfrac{2}{3}.[/tex]
Then
[tex]\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..[/tex]
This means that
[tex]\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..[/tex]
Thus,
[tex]x_0=\dfrac{2x_B+3x_A}{5},\ y_0=\dfrac{2y_B+3y_A}{5}.[/tex]
Answer: [tex]\left(\dfrac{2x_B+3x_A}{5},\dfrac{2y_B+3y_A}{5}\right).[/tex]
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
Hope this helped
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