Straight Line And Pair Of Straight Lines Ques 7
- Two sides of a parallelogram are along the lines, $x+y=3$ and $x-y+3=0$. If its diagonals intersect at $(2$, $4)$, then one of its vertex is
(2019 Main, 10 Jan II)
(a) $(3,6)$
(c) $(2,1)$
(b) $(2,6)$
(d) $(3,5)$
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Solution:
Formula:
Equation Of Lines Parallel And Perpendicular To A Given Line :
- According to the given information, we have the following figure
[Note that given lines are perpendicular to each other as $\left.m _1 \times m _2=-1\right]$
Clearly, point $A$ is a point of intersection of lines
and $\quad \begin{aligned} x+y & =3 \ x-y & =-3\end{aligned}$
So, $\quad A=(0,3)$ [solving Eqs. (i) and (ii)]
Now, as point $M(2,4)$ is the midpoint of the line joining the points $A$ and $C$, so
$$ \begin{aligned} (2,4)=\frac{0+x_2}{2}, & \frac{3+y_2}{2} \\ & \because \text { mid- point }=\frac{x _1+x _2}{2}, \frac{y _1+y _2}{2} \end{aligned} $$
$$ \begin{array}{lll} \Rightarrow & 2 & =\frac{0+x _2}{2} ; 4=\frac{3+y _2}{2} \\ \Rightarrow & & x _2=4 \text { and } y _2=5 \\ \therefore \text { Thus, } C \equiv (4,5) \end{array} $$
Now, equation of line $B C$ is given by
$$ \begin{aligned} \left(y-y _1\right) & =m\left(x-x _1\right) \\ y-5 & =1(x-4) \end{aligned} $$
[line $BC$ is parallel to $x-y+3=0$ and slope
$$ \begin{aligned} & \text { of } \left.x-y+3=0 \text { is } \frac{1}{1}=1\right] \\ & \Rightarrow \quad y=x+1 \end{aligned} $$
and equation of line $DC$ is
$$ \begin{aligned} & y-5=-1(x-4) \\ & \quad[\text { line } D C \text { is parallel to } x+y=3 \text { and } \\ & \text { slope of } \left.x+y=3 \text { is } \frac{-1}{1}=-1\right] \end{aligned} $$
$\Rightarrow \quad x+y=9 \quad$…(iv)
On solving Eqs. (i) and (iii), we get $B(1,2)$ and on solving Eqs. (ii) and (iv), we get $D(3,6)$
BETA
