Question

Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line $x -2y = 3.$

Answer 1

Let the slope of the required line be $m_1$.
The given line can be written as $y =\frac{ 1}{2} x – \frac{3}{2} ,$ which is of the form $y = mx + c$

∴ Slope of the given line = $m_2 = \frac{1}{2}$

It is given that the angle between the required line and line $x \- 2y = 3$ is 45°.
We know that if θisthe acute angle between lines $l_1$ and $l_2$ with slopes $m_1$ and $m_2$ respectively, then

$tanθ=\left|\frac{m_2-m_1}{1+m_1m_2}\right|$

$∴ tan45º=\left|\frac{m_2-m_1}{1+m_1m_2}\right|$

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

$⇒1=\left|\frac{\left(\frac{1-2m_1}{2}\right)}{\frac{2+m_1}{2}}\right|$

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

$⇒1=±\left(\frac{1-2m_1}{2+m_1}\right)$

$⇒1=\left(\frac{1-2m_1}{2+m_1}\right)$ or $1=-\left(\frac{1-2m_1}{2+m_1}\right)$

$⇒2+m_1=1-2m_1$ or $2+m_1=-1+2m_1$

$⇒m_1=\frac{-1}{3}$ or $m_1=3$

Case I:

$m_1 = 3$
The equation of the line passing through (3, 2) and having a slope of 3 is:
$y -2 = 3 (x \- 3)$
$y \- 2 = 3x \- 9$
$3x \- y = 7$

Case II:

$m_1 =\frac{-1}{3}$
The equation of the line passing through (3, 2) and having a slope of $\frac{-1}{3}$ is:
$y – 2 = – \frac{1}{3} (x – 3)$

$3y – 6 = – x + 3$
$x + 3y = 9$

Thus, the equations of the lines are $3x \- y = 7$ and $x + 3y = 9.$