Question

In an isosceles triangle ABC, the coordinates of the point B and C on the base BC are respectively (1, 2) and (2, 1). If the equation of the line AB is $y = 2 x$ , then the equation of the line AC is

• A. $y = \frac { 1 } { 2 } ( x - 1 )$
• B. $y = \frac { x } { 2 }$
• C. $y = x - 1$
• D. $2 y = x + 3$

Answer 1

Answer: (B)

$y = \frac { x } { 2 }$

Answer 2

Slope of BC = $\frac { 1 - 2 } { 2 - 1 } = - 1$

$\therefore$ $\angle A B C = \angle A C B$

⇒ $\left| \frac { 2 + 1 } { 1 + 2 ( - 1 ) } \right| = \frac { m + 1 } { 1 + m ( - 1 ) }$ ⇒ $\frac { m + 1 } { 1 - m } = | - 3 |$ ⇒ $\frac { m + 1 } { 1 - m } = \pm 3$

⇒ $m = 2 , \frac { 1 } { 2 }$ .

But slope of AB is 2; $\therefore$ $m = \frac { 1 } { 2 }$

(Here m is the gradient of the line AC)

Equation of the line AC is $y - 1 = \frac { 1 } { 2 } ( x - 2 )$

⇒ $x - 2 y = 0$ or $y = \frac { x } { 2 }$ .