Question

Find the equation of the straight lines passing through the origin and making angle of ${45^\circ }$ with straight line $\sqrt 3 x + y = 11$ .

Answer 1

Hint: We will use the formula of the equations of two lines passing through a point $\left( {{x_1},{y_1}} \right)$ and making angle $\alpha $ with straight line y= mx +c which is given as-
$ \Rightarrow y - {y_1} = \dfrac{{m \pm \tan \alpha }}{{1 \mp m\tan \alpha }}\left( {x - {x_1}} \right)$ . Compare the giving straight line equation with the standard equation to find m. Put the given values in the formula and solve to get the equations.

Complete step-by-step answer:
Given the equation of straight lines passes through origin and makes ${45^\circ }$angle with straight line$\sqrt 3 x + y = 11$.
We can write $\sqrt 3 x + y = 11$ as $y = - \sqrt 3 x + 11$ - (i)
We will use the formula of the equations of two lines passing through a point $\left( {{x_1},{y_1}} \right)$ and making angle $\alpha $ with straight line y= mx +c which is given as-
$ \Rightarrow y - {y_1} = \dfrac{{m \pm \tan \alpha }}{{1 \mp m\tan \alpha }}\left( {x - {x_1}} \right)$
Here since the equation passes through origin$\left( {0,0} \right)$then ${x_1} = 0$ and ${y_1} = 0$ and also here on comparing equation (i) with standard straight line equation, we get-
$ \Rightarrow $m=$ - \sqrt 3 $ and $\alpha = {45^\circ }$
On putting the given equation in the formula, we get-
$ \Rightarrow y - 0 = \dfrac{{ - \sqrt 3 \pm \tan {{45}^ \circ }}}{{1 \mp \left( { - \sqrt 3 } \right)\tan {{45}^ \circ }}}\left( {x - 0} \right)$
We know that $\tan {45^ \circ } = 1$
On putting the value of angle, we get-
$ \Rightarrow y = \dfrac{{ - \sqrt 3 \pm 1}}{{1 \pm \left( {\sqrt 3 } \right)}}x$
We can also write the above equation as-
$ \Rightarrow y = \dfrac{{ - \sqrt 3 + 1}}{{1 + \sqrt 3 }}x$ and $y = \dfrac{{ - \sqrt 3 - 1}}{{1 - \sqrt 3 }}x$
On rationalizing, we get-
$ \Rightarrow y = \dfrac{{\left( { - \sqrt 3 + 1} \right)\left( {1 - \sqrt 3 } \right)}}{{\left( {1 + \sqrt 3 } \right)\left( {1 - \sqrt 3 } \right)}}x$ and $y = \dfrac{{\left( { - \sqrt 3 - 1} \right)\left( {1 + \sqrt 3 } \right)}}{{\left( {1 - \sqrt 3 } \right)\left( {1 + \sqrt 3 } \right)}}x$
Now, we know that $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$
On applying this formula, we get-
$ \Rightarrow y = \dfrac{{{{\left( {1 - \sqrt 3 } \right)}^2}}}{{\left( {{1^2} - {{\left( {\sqrt 3 } \right)}^2}} \right)}}x$ and $y = \dfrac{{ - {{\left( {1 + \sqrt 3 } \right)}^2}}}{{\left( {{1^2} - {{\left( {\sqrt 3 } \right)}^2}} \right)}}x$
Now, we know that ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ and${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$.
On applying both the formulae in the above equation, we get-
$ \Rightarrow y = \dfrac{{1 + 3 - 2\sqrt 3 }}{{\left( {1 - 3} \right)}}x$ and $y = \dfrac{{ - \left( {1 + 3 + 2\sqrt 3 } \right)}}{{\left( {1 - 3} \right)}}x$
On simplifying, we get-
$ \Rightarrow y = \dfrac{{4 - 2\sqrt 3 }}{{ - 2}}x$ and $y = \dfrac{{ - \left( {4 + 2\sqrt 3 } \right)}}{{ - 2}}x$
On further simplifying we get-
$ \Rightarrow y = \dfrac{{ - 4 + 2\sqrt 3 }}{2}x$ and $y = \dfrac{{4 + 2\sqrt 3 }}{2}x$
On further solving, we get-
$ \Rightarrow y = \left( { - 2 + \sqrt 3 } \right)x$ and $y = \left( {2 + \sqrt 3 } \right)x$
On rearranging, we get-
$y = \left( {\sqrt 3 - 2} \right)x$ and $y = \left( {2 + \sqrt 3 } \right)x$
These are the required equations of straight lines passing through the origin and making an angle of ${45^\circ }$ with straight line$\sqrt 3 x + y = 11$.

Note: Here the student can also directly take the different signs from starting step and solve-
$ \Rightarrow y - 0 = \dfrac{{ - \sqrt 3 + \tan {{45}^ \circ }}}{{1 - \left( { - \sqrt 3 } \right)\tan {{45}^ \circ }}}\left( {x - 0} \right)$ and $y - 0 = \dfrac{{ - \sqrt 3 - \tan {{45}^ \circ }}}{{1 + \left( { - \sqrt 3 } \right)\tan {{45}^ \circ }}}\left( {x - 0} \right)$
On solving, we get-
$ \Rightarrow y = \dfrac{{ - \sqrt 3 + \tan {{45}^ \circ }}}{{1 + \sqrt 3 \tan {{45}^ \circ }}}x$ and $y = \dfrac{{ - \sqrt 3 - \tan {{45}^ \circ }}}{{1 - \sqrt 3 \tan {{45}^ \circ }}}x$
Put the value of angle-
$ \Rightarrow y = \dfrac{{ - \sqrt 3 + 1}}{{1 + \sqrt 3 }}x$ and $y = \dfrac{{ - \sqrt 3 - 1}}{{1 - \sqrt 3 }}x$
Then solve as given in the above solution.