55.  A points A(-2, -2), B$\left(\frac{-2}{7}, \frac{-20}{7}\right)$ and C(2, -4) are collinear, prove that $AB=\frac{3}{7}AC$.

Dear Student,

The given question can be solved using distance formula.

Since it is given that points A,B and C are collinear, we need to find the length of AB and AC using distance formula.


$$\begin{gathered} Dis\tan ce formula = \sqrt{(x2-x1{)}^2 +(y2-y1{)}^2} \\ For AB, \\ A = (-2,-2) , x1= -2, y1 = -2 \\ B = (\frac{-2}{7}, \frac{-20}{7}) , x2 = \frac{-2}{7} , y2 = \frac{-20}{7} \\ So, AB = \sqrt{(x2-x1{)}^2 +(y2-y1{)}^2} = \sqrt{(\frac{-2}{7} -(-2){)}^2 + (\frac{-20}{7}-(-2){)}^2} \\ =\sqrt{(\frac{-2}{7}+2{)}^2 +(\frac{-20}{7}+2{)}^2} = \sqrt{(\frac{-2+14}{7}{)}^2 +(\frac{-20+14}{7}{)}^2} \\ =\sqrt{\frac{144}{49} + \frac{36}{49} } = \frac{\sqrt{180}}{7} = \frac{\sqrt{9\times 20}}{7} = \frac{3}{7}\sqrt{20} units \\ Now, \\ For AC, \\ A = (-2,-2) , x1= -2, y1 = -2 \\ B = (2,-4 ) , x2 = 2 , y2 = -4 \\ AC = \sqrt{(x2-x1{)}^2 +(y2-y1{)}^2} = \sqrt{(2+2{)}^2 +(-4+2{)}^2} \\ =\sqrt{(4{)}^2 + (-2{)}^2} = \sqrt{16 + 4} = \sqrt{20} units \\ So, \\ AB = \frac{3}{7} \sqrt{20 } units , AC = \sqrt{20 } units \\ So, AB = \frac{3}{7} AC \end{gathered}$$


Hence proved.

Hope it helps.

Regards