Question

Find the equation of two straight lines through the point $\left( {4,5} \right)$ which makes an acute angle of ${45^ \circ }$ with $2x - y + 7 = 0$.

Answer 1

We will use the formula of angle between two lines to find the slope of the line which makes an acute angle of ${45^ \circ }$ with $2x - y + 7 = 0$. Then using the given point $\left( {4,5} \right)$ and the obtained slopes, we will find the the equation of the two straight lines through the point $\left( {4,5} \right)$ which makes an acute angle of ${45^ \circ }$ with $2x - y + 7 = 0$.

Complete step by step answer:
We have to find the equation of two straight lines through the point $\left( {4,5} \right)$ which makes an acute angle of ${45^ \circ }$ with $2x - y + 7 = 0$. Let the slope of the straight lines which makes an acute angle of ${45^ \circ }$ with $2x - y + 7 = 0$ be $m$.In slope intercept form we can write the line $2x - y + 7 = 0$ as $y = 2x + 7$. Slope of this line is ${m_1} = 2$. As we know, the acute angle $\theta$ between the two lines whose slopes are ${m_1}$ and ${m_2}$ are given by $\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$.

Using the given conditions, we can write
$\Rightarrow \tan {45^ \circ } = \left| {\dfrac{{2 - m}}{{1 + 2m}}} \right|$
Using, $\tan {45^ \circ } = 1$ and on simplifying, we get
$\Rightarrow 1 = \pm \left( {\dfrac{{2 - m}}{{1 + 2m}}} \right)$
First taking positive sign, we get
$\Rightarrow 1 = \dfrac{{2 - m}}{{1 + 2m}}$
$\Rightarrow 1 + 2m = 2 - m$
On simplifying, we get
$\Rightarrow 3m = 1$

Dividing both the sides by $3$, we get
$\Rightarrow m = \dfrac{1}{3}$
Now taking negative sign, we get
$\Rightarrow 1 = - \dfrac{{2 - m}}{{1 + 2m}}$
$\Rightarrow 1 + 2m = - \left( {2 - m} \right)$
On simplifying, we get
$\Rightarrow 1 + 2m = - 2 + m$
$\Rightarrow m = - 3$
We can write the equation of the line through the point $\left( {4,5} \right)$ and having slope $m = \dfrac{1}{3}$,
$\Rightarrow y - 5 = \dfrac{1}{3}\left( {x - 4} \right)$

On cross multiplication, we get
$\Rightarrow 3\left( {y - 5} \right) = \left( {x - 4} \right)$
$\Rightarrow 3y - 15 = x - 4$
On simplifying, we get
$\Rightarrow x - 3y + 11 = 0$
Now, we can write the equation of the line through the point $\left( {4,5} \right)$ and having slope $m = - 3$,
$\Rightarrow y - 5 = - 3\left( {x - 4} \right)$
$\Rightarrow y - 5 = - 3x + 12$
On simplifying, we get
$\therefore 3x + y - 17 = 0$

Therefore, $x - 3y + 11 = 0$ and $3x + y - 17 = 0$ is the equation of two straight lines through the point $\left( {4,5} \right)$ which makes an acute angle of ${45^ \circ }$ with $2x - y + 7 = 0$.

Note: When two lines intersect at a point, two angles are formed. One is acute angle and other is obtuse angle. These angles can be calculated by substituting values of the slopes of the lines. The formula is $\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$. Here, the positive sign stands for acute angle and the negative sign stands for obtuse angle formed between the given lines.