2 Marks Questions

Question 12 Marks

Prove that lines $7 x-5 y+20=0$ and $5 x+7 y+$ $19=0$ are mutually perpendicular to each other.

Answer

Given equation of lines
$\begin{array}{ll}
& 7 x-5 y+20=0 ......(1)\\
\text { and } & 5 x+7 y+19=0
\end{array}$
Their slopes are $\quad m_1=\frac{7}{5}$ and $m_1=\frac{-5}{7}$
We know that if slopes of two lines be $m_1$ and $m_2$ then lines will be perpendicular to each other if product of their slopes is -1 . i.e. $m_1 m_2=-1$
Putting values of $m_1$ and $m_2$
$m_1=2$ and $m_2=\frac{-1}{2}$
$m_1 m_2=2 \times\left(\frac{-1}{2}\right)=-1$
So, both given lines are mutually perpendicular.

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Question 22 Marks

Find the equation of a line passing through point of intersection of lines $2 x+y=5$ and $x-2 y=$ 0 which makes an angle $45^{\circ}$ with $x$-axis.

Answer

Equation of line passing through point of intersection of lines $2 x+y-5=0$ and $x-2 y=0$ is given by
$\begin{array}{r}
2 x+y-5+\lambda(x-2 y)=0........(1) \\
(2+\lambda) x+(1-2 \lambda) y-5=0
\end{array}$
Slope of line $m=-\frac{(2+\lambda)}{(1-2 \lambda)}$
According to question,
$\begin{aligned}
& =\frac{-(2+\lambda)}{1-2 \lambda}=\tan 45^{\circ}=1 \\
-2-\lambda & =1-2 \lambda \\
\lambda & =3
\end{aligned}$
$\begin{array}{l}\text { From equaion (1) : } \\ \Rightarrow \quad 2 x+y-5+3(x-2 y)=0 \\ \Rightarrow \quad 2 x+y-5+3 x-6 y=0\end{array}$
$5 x-5 y-5=0 \Rightarrow x-y=1$

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