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(B) The slope of the given line $x - 2y = 3$ is $m_1 = \frac{1}{2}$.Let the slope of the required lines be $m$.The angle between the lines is $45^{\ci

Vedclass · Accepted Answer

$3x - y - 7 = 0$ and $x + 3y - 9 = 0$

Vedclass · Answer

(B) The slope of the given line $x - 2y = 3$ is $m_1 = \frac{1}{2}$.Let the slope of the required lines be $m$.The angle between the lines is $45^{\circ}$,so $	an 45^{\circ} = \left| \frac{m - m_1}{1 + m \cdot m_1} ight|$.$1 = \left| \frac{m - 1/2}{1 + m/2} ight| = \left| \frac{2m - 1}{2 + m} ight|$.This gives two cases: $2m - 1 = m + 2$ or $2m - 1 = -(m + 2)$.Case $1$: $m = 3$. The equation is $y - 2 = 3(x - 3)$ $\Rightarrow y - 2 = 3x - 9$ $\Rightarrow 3x - y - 7 = 0$.Case $2$: $2m - 1 = -m - 2$ $\Rightarrow 3m = -1$ $\Rightarrow m = -1/3$. The equation is $y - 2 = -\frac{1}{3}(x - 3)$ $\Rightarrow 3y - 6 = -x + 3$ $\Rightarrow x + 3y - 9 = 0$.Thus,the equations are $3x - y - 7 = 0$ and $x + 3y - 9 = 0$.

The 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:

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