SATHEE: Straight Line and Pair of Straight Lines 1 Question 61

Straight Line and Pair of Straight Lines 1 Question 61

61. The equations of the perpendicular bisectors of the sides $A B$ and $A C$ of a $△ A B C$ are $x - y + 5 = 0$ and $x + 2 y = 0$ , respectively. If the point $A$ is $(1, - 2)$ , find the equation of the line $B C$ .

(1986, 5M)

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Solution:

Let the coordinates of $B$ and $C$ be $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ respectively. Let $m_{1}$ and $m_{2}$ be the slopes of $A B$ and $A C$ , respectively. Then,

$m_{1} = slope of A B = \frac{y_{1} + 2}{x_{1} - 1}$

and

$m_{2} = slope of A C = \frac{y_{2} + 2}{x_{2} - 1}$

Let $F$ and $E$ be the mid point of $A B$ and $A C$ , respectively. Then, the coordinates of $E$ and $F$ are $E \frac{x_{2} + 1}{2}, \frac{y_{2} - 2}{2}$ and $F \frac{x_{1} + 1}{2}, \frac{y_{1} - 2}{2}$ , respectively.

Now, $F$ lies on $x - y + 5 = 0$ .

$\begin{array}{ll} \Rightarrow & \frac{x_{1} + 1}{2} - \frac{y_{1} - 2}{2} = - 5 \\ \Rightarrow & x_{1} - y_{1} + 13 = 0 \end{array}$

Since, $A B$ is perpendicular to $x - y + 5 = 0$ .

$∴ ($ slope of $A B) \cdot ($ slope of $x - y + 5 = 0) = - 1$ .

$\begin{array}{lc} \Rightarrow & \frac{y_{1} + 2}{x_{1} - 1} \cdot (1) = - 1 \\ \Rightarrow & y_{1} + 2 = - x_{1} + 1 \\ \Rightarrow & x_{1} + y_{1} + 1 = 0 \end{array}$

On solving Eqs. (i) and (ii), we get

$x_{1} = - 7, y_{1} = 6$

So, the coordinates of $B$ are $(- 7, 6)$ .

Now, $E$ lies on $x + 2 y = 0$ .

$∴ \frac{x_{2} + 1}{2} + 2 \frac{y_{2} - 2}{2} = 0$

$\Rightarrow x_{2} + 2 y_{2} - 3 = 0 .$

Since, $A C$ is perpendicular to $x + 2 y = 0$

$∴$ (slope of $A C) \cdot ($ slope of $x + 2 y = 0) = - 1$

$\begin{array}{ll} \Rightarrow & \frac{y_{2} + 2}{x_{2} - 1} \cdot - \frac{1}{2} = - 1 \\ \Rightarrow & 2 x_{2} - y_{2} = 4 \end{array}$

On solving Eqs. (iii) and (iv), we get

$x_{2} = \frac{11}{5} and y_{2} = \frac{2}{5}$

So, the coordinates of $C$ are $\frac{11}{5}, \frac{2}{5}$ .

Thus, the equation of $B C$ is

$\begin{aligned} y - 6 & = \frac{2 / 5 - 6}{11 / 5 + 7} (x + 7) \\ \Rightarrow & - 23 (y - 6) & = 14 (x + 7) \\ \Rightarrow 14 x + 23 y - 40 & = 0 \end{aligned}$