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

Hint: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.