Question

Equations of the two straight lines passing through the point $(3,2)$ and making an angle of ${45}^{\circ }$ with the line $x-2y=3$, are

• A. $3x+y-7=0\text{and}x+3y+9=0$
• B. $3x-y-7=0\text{and}x+3y-9=0$
• C. $x+3y-7=0\text{and}x+3y-9=0$
• D. none of these

Answer 1

Answer: (B)

$3x-y-7=0\text{and}x+3y-9=0$

The slope of the given line $=\frac{-1}{-2}=\frac{1}{2}$

Angle between the lines = ${45}^{\circ }$

$\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=tan{45}^{\circ }=1$

$\left|\frac{m_1-1/2}{1+\frac{m_1}{2}}\right|=1$

$|m_1-1/2|=|1+\frac{m_1}{2}|$

Here,$m_1-1/2=1+\frac{m_1}{2}$ or, $m_1-\frac{1}{2}=-1-m_1/2$

$\frac{m_1}{2}=1+\frac{1}{2}=\frac{3}{2}$ $m_1+\frac{m_1}{2}=\frac{1}{2}-1$

$m_1=3$ $\frac{3m_1}{2}=\frac{-1}{2}$

$m_1=\frac{-1}{3}$

Line passes through $(3,2)$

$y-y_1=m(x-x_1)$

$y-2=3(x-3)$

$3x-9=y-2$

$3x-y=7.$

$m=\frac{-1}{3}$

$\Rightarrow y-y_1=m(x-x_1)$

$y-2=\frac{-1}{3}(x-3)$

$3y-6=-x+3$

$x+3y-6-3=0$

$x+3y-9=0$