Question

A (-1, 1), B (1, 3) and C (3, a) are points and if AB = BC, then find ‘a’

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

The correct option is A 5

Given,
$A(-1,1),B(1,3)$ and $C(3,a)$

We know, distance
$=\sqrt{\left[\right(x2-x1)2+(y2-y1\left)2\right]}$

For distance AB,
$A(x_1,y_1)=(-1,1)\text{and}B(x_2,y_2)=(1,3)$
AB $=\sqrt{\left[\right(1+1{)}^2+(3-1{)}^2]}$
$=\sqrt{[2^2+2^2]}$
$=\sqrt{8}$

For distance BC,
$B(x_1,y_1)=(1,3)\text{and}C(x_2,y_2)=(3,a)$
$BC=\sqrt{\left[\right(3-1{)}^2+(a-3{)}^2]}$
$=\sqrt{[2^2+(a-3{)}^2]}$
$=\sqrt{[4+(a-3{)}^2]}$

Given, AB = BC
Hence,
$\Rightarrow \sqrt{8}=\sqrt{[4+(a-3{)}^2]}$
$\Rightarrow 4+(a-3{)}^2=8$
$\Rightarrow (a-3{)}^2=4$
$\Rightarrow a-3=2$
$\Rightarrow a=5$