Question

Find the x-value for point C such that AC and BC from a 2:3 ratio.

Answer 1

Given:

Point C divides AB such that AC:BC=2:3.

To find:

The x-value for point C.

Solution:

Section formula: If a point divide a line segment in m:n, then

$Point=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)$

Form the given graph it is clear that the endpoints of the line segment AB are A(-3,5) and B(3,0).

Point C divides AB such that AC:BC=2:3. Using section formula, the coordinates of point C are

$C=\left(\dfrac{2(3)+3(-3)}{2+3},\dfrac{2(0)+3(5)}{2+3}\right)$

$C=\left(\dfrac{6-9}{5},\dfrac{0+15}{5}\right)$

$C=\left(\dfrac{-3}{5},\dfrac{15}{5}\right)$

$C=\left(-0.6,3\right)$

The x-value of C is -0.6.

Therefore, the correct option is B.