Question

The acute angles between the lines: - $lx+my=l+m$ and $l\left( x-y \right)+m\left( x+y \right)=2m$ is: -
(a) $\dfrac{\pi }{4}$
(b) $\dfrac{\pi }{6}$
(c) $\dfrac{\pi }{2}$
(d) $\dfrac{\pi }{3}$

Answer 1

Write the given two equations in the slope intercept form as: - $y={{m}_{1}}x+{{c}_{1}}$ and $y={{m}_{2}}x+{{c}_{2}}$, where ${{m}_{1}}$ and ${{m}_{2}}$ are the slopes of the given lines and ${{c}_{1}}$ and ${{c}_{2}}$ are the intercepts on the y – axis for the given lines. Now, assume ‘$\theta$’ as the angle between the given lines and apply the formula: - $\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$ to find the value of $\theta$.

Complete step-by-step answer:
Here, we have been provided with two lines and we have to find the measurement of acute angle between them. So, we have,
$\Rightarrow lx+my=l+m$ - (1)
$\Rightarrow l\left( x-y \right)+m\left( x+y \right)=2m$ - (2)
First, we will write both the equations in the form, $y=mx+c$ which is also called the slope intercept form. Therefore,
(i). Considering equation (1), we have,

& \Rightarrow lx+my=l+m \\\ & \Rightarrow my=-lx+l+m \\\ & \Rightarrow y=\left( \dfrac{-l}{m} \right)x+l+m \\\ \end{aligned}$$ Comparing the above equation with, $$y={{m}_{1}}x+{{c}_{1}}$$, we get, $$\Rightarrow {{m}_{1}}$$ = slope of the line 1 $$\Rightarrow {{m}_{1}}=\left( \dfrac{-l}{m} \right)$$ (ii). Considering equation (2), we have, $$\begin{aligned} & \Rightarrow l\left( x-y \right)+m\left( x+y \right)=2m \\\ & \Rightarrow lx-ly+mx+my=2m \\\ & \Rightarrow \left( l+m \right)x+\left( m-l \right)y=2m \\\ & \Rightarrow \left( m-l \right)y=-\left( l+m \right)x+2m \\\ & \Rightarrow y=\left[ \dfrac{-\left( l+m \right)}{\left( m-l \right)} \right]x+\dfrac{2m}{m-l} \\\ & \Rightarrow y=\left[ \dfrac{l+m}{l-m} \right]x+\dfrac{2m}{m-l} \\\ \end{aligned}$$ Comparing the above equation with $$y={{m}_{2}}x+{{c}_{2}}$$, we get, $$\Rightarrow {{m}_{2}}$$ = slope of line 2 $$\Rightarrow {{m}_{2}}=\left( \dfrac{l+m}{l-m} \right)$$ Now, we know that angle of two lines having slopes $${{m}_{1}}$$ and $${{m}_{2}}$$ is given as: - $$\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$$. Therefore, applying the above formula and substituting the values of $${{m}_{1}}$$ and $${{m}_{2}}$$, we get, $$\Rightarrow \tan \theta =\left| \dfrac{-\dfrac{l}{m}-\dfrac{l+m}{l-m}}{1+\left( \dfrac{-l}{m} \right)\times \left( \dfrac{l+m}{l-m} \right)} \right|$$ Here, ‘$$\theta $$’ is assumed to be the angle between the two lines. Therefore, on simplification by taking L.C.M, we get, $$\begin{aligned} & \Rightarrow \tan \theta =\left| \dfrac{\dfrac{-{{l}^{2}}+lm-lm-{{m}^{2}}}{m\left( l-m \right)}}{1-\left( \dfrac{{{l}^{2}}+lm}{lm-{{m}^{2}}} \right)} \right| \\\ & \Rightarrow \tan \theta =\left| \dfrac{\dfrac{-{{m}^{2}}-{{l}^{2}}}{ml-{{m}^{2}}}}{\dfrac{lm-{{m}^{2}}-{{l}^{2}}-lm}{ml-{{m}^{2}}}} \right| \\\ \end{aligned}$$ $$\Rightarrow \tan \theta =\left| \dfrac{\dfrac{-\left( {{m}^{2}}+{{l}^{2}} \right)}{ml-{{m}^{2}}}}{\dfrac{-\left( {{m}^{2}}+{{l}^{2}} \right)}{ml-{{m}^{2}}}} \right|$$ Cancelling the common terms, we get, $$\Rightarrow \tan \theta =\left| 1 \right|$$ Removing modulus sign we get, $$\begin{aligned} & \Rightarrow \tan \theta =1 \\\ & \Rightarrow \theta ={{\tan }^{-1}}1 \\\ \end{aligned}$$ Now, we know that the value of tangent of angle is 1 when that angle is $${{45}^{\circ }}$$ or $$\dfrac{\pi }{4}$$. Therefore, we have, $$\Rightarrow \theta =\dfrac{\pi }{4}$$ **So, the correct answer is “Option A”.** **Note:** One must not try to solve the two equations because it will give us the point of intersection which is of no use here. You may note that we can write an equation in many forms but here we need only the slope – intercept form because we have to determine the slope of the two lines so that the formula of $$\tan \theta $$ can be applied. Remember that the angle between the two lines is nothing but the angle of intersection of the lines. In the formula of $$\tan \theta $$ a modulus sign is necessary, so that we must get an acute angle.