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 $\triangle A B C$ 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 $\left(x _1, y _1\right)$ and $\left(x _2, y _2\right)$ respectively. Let $m _1$ and $m _2$ be the slopes of $A B$ and $A C$, respectively. Then,
$$ m _1=\text { slope of } A B=\frac{y _1+2}{x _1-1} $$
and
$$ m _2=\text { slope of } A C=\frac{y _2+2}{x _2-1} $$
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 $F \quad \frac{x _1+1}{2}, \frac{y _1-2}{2}$, 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 $x-y+5=0$.
$\therefore \quad($ 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, $\quad E$ lies on $x+2 y=0$.
$$ \therefore \quad \frac{x _2+1}{2}+2 \frac{y _2-2}{2}=0 $$
$$ \Rightarrow \quad x _2+2 y _2-3=0 \text {. } $$
Since, $A C$ is perpendicular to $x+2 y=0$
$\therefore$ (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
$$ x _2=\frac{11}{5} \text { and } y _2=\frac{2}{5} $$
So, the coordinates of $C$ are $\frac{11}{5}, \frac{2}{5}$.
Thus, the equation of $B C$ is
$$ \begin{array}{rlrl} y-6 & =\frac{2 / 5-6}{11 / 5+7}(x+7) \\ \Rightarrow \quad & & -23(y-6) & =14(x+7) \\ \Rightarrow \quad 14 x+23 y-40 & =0 \end{array} $$
