Question

the points A(-6,10),B(-4,6) and C(3,-8) are collinear,then show that AB=2/9AC

Answer 1

ab=√(-6+4)2+(10-6)2 =√4+16=√20=2√5
ac=√(-6-3)2+(10+8)2=√81+324=√405=9√5
now ab /ac=2/9
or, ab=2/9ac

Answer 2

Answer:

If the points A(-6,10),B(-4,6) and C(3,-8) are collinear, then $AB=\frac{2}{9}\times AC$

Step-by-step explanation:

It is given that points A(-6,10),B(-4,6) and C(3,-8) are collinear.

To prove: AB=2/9AC

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula we get

$AB=\sqrt{(-4-(-6))^2+(6-10)^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$

$AC=\sqrt{(3-(-6))^2+(-8-10)^2}=\sqrt{81+324}=\sqrt{405}=9\sqrt{5}$

$\frac{AB}{AC}=\frac{2\sqrt{5}}{9\sqrt{5}}$

Cancel out common factors.

$\frac{AB}{AC}=\frac{2}{9}$

Multiply both sides by AC.

$AB=\frac{2}{9}\times AC$

Hence proved.