Question

The equation of the line segment AB is y=x. If A and B lie on the same side of the line mirror 2x-y=1, the image of AB has the equation

& (A)\text{ x + y =2} \\\ & \text{(B) 8x + y=9} \\\ & \text{(C) 7x - y =6} \\\ & \text{(D) None of these} \\\ \end{aligned}$$

Answer 1

Hint: We know that if two points $A\text{ (}{{\text{x}}_{1}},{{y}_{1}})$ and $\text{B (}{{\text{x}}_{2}},{{y}_{2}})$ are on the same side of the line $L=ax+by+c$, then the ratio of $a{{x}_{1}}+b{{y}_{1}}+c$ and$a{{x}_{2}}+b{{y}_{2}}+{{c}_{2}}$ must be negative. In the similar way, that if two points$A\text{ (}{{\text{x}}_{1}},{{y}_{1}})$ and $\text{B (}{{\text{x}}_{2}},{{y}_{2}})$ are on the opposite side of the line $L=ax+by+c$, then the ratio of $a{{x}_{1}}+b{{y}_{1}}+c$ and$a{{x}_{2}}+b{{y}_{2}}+{{c}_{2}}$ must be positive.
We should apply the above condition such that A and B points should lie on the same side of 2x-y=1. Now, find the image of point A with respective to 2x-y=1. Also, we needed to find the point B with respective to 2x-y=1. We needed to find the line equation passing through the images of point A and point B.

Complete step-by-step answer:
From the question, it is given that the equation of AB is y=x. Let $A\text{ (}{{\text{x}}_{1}},{{y}_{1}})$ and $\text{B (}{{\text{x}}_{2}},{{y}_{2}})$ be two points.
If $A\text{ (}{{\text{x}}_{1}},{{y}_{1}})$ and $\text{B (}{{\text{x}}_{2}},{{y}_{2}})$ are two points on the line segment AB, then the abscissa and ordinate of points A and B must be equal. Hence, the points on the line segment are $A\text{ (}{{\text{x}}_{1}},{{x}_{1}})$ and $\text{B (}{{\text{x}}_{2}},{{x}_{2}})$.
We know that if two points $A\text{ (}{{\text{x}}_{1}},{{y}_{1}})$ and $\text{B (}{{\text{x}}_{2}},{{y}_{2}})$ are on the same side of the line L=ax + by + c=0, then the ratio of $a{{x}_{1}}+b{{y}_{1}}+c$ and$a{{x}_{2}}+b{{y}_{2}}+{{c}_{2}}$ must be negative.
\dfrac{L({{x}_{1}},{{y}_{1}})}{L({{x}_{2}},{{y}_{2}})}=-k$$$$\Rightarrow \dfrac{a{{x}_{1}}+b{{y}_{1}}+c}{a{{x}_{2}}+b{{y}_{2}}+c}=-k......(1) where k is positive integer

From the question, 2x-y=1 is the mirror line. With respect to mirror line 2x-y=1, A and B should be on the same side.
By comparing 2x-y=1 with $L=ax+by+c$, we get a=2, b=-1 and c=-1.
From (1)

& \Rightarrow \dfrac{2{{x}_{1}}-{{x}_{1}}-1}{2{{x}_{2}}-{{x}_{2}}-1}=-k \\\ & \Rightarrow \dfrac{{{x}_{1}}-1}{{{x}_{2}}-1}=-k........(2) \\\ \end{aligned}$$ We know that if the image of point $$P({{x}_{a}},{{y}_{a}})$$ with respect to$$px+qy+r=0$$ is $$Q(h,k)$$ then $$\dfrac{h-{{x}_{a}}}{p}=\dfrac{k-{{y}_{a}}}{q}=\dfrac{-2(p{{x}_{a}}+q{{y}_{a}}+r)}{{{p}^{2}}+{{q}^{2}}}$$. So, the image of point $$A({{x}_{1}},{{x}_{1}})$$ with respect to 2x-y-1=0 is $$A\grave{\ }({{h}_{1}},{{k}_{1}})$$ if $$\begin{aligned} & \Rightarrow \dfrac{{{h}_{1}}-{{x}_{1}}}{2}=\dfrac{{{k}_{1}}-{{x}_{1}}}{-1}=\dfrac{-2(2{{x}_{1}}-{{x}_{1}}-1)}{{{2}^{2}}+{{1}^{2}}} \\\ & \Rightarrow \dfrac{{{h}_{1}}-{{x}_{1}}}{2}=\dfrac{{{k}_{1}}-{{x}_{1}}}{-1}=\dfrac{-2({{x}_{1}}-1)}{5}.....(3) \\\ \end{aligned}$$ From equation (3) we can find the value of $${{h}_{1}}$$. $$\dfrac{{{h}_{1}}-{{x}_{1}}}{2}=\dfrac{-2({{x}_{1}}-1)}{5}$$ By using cross multiplication, we get $$\begin{aligned} & \Rightarrow {{h}_{1}}-{{x}_{1}}=\dfrac{-4({{x}_{1}}-1)}{5} \\\ & \Rightarrow {{h}_{1}}-{{x}_{1}}=\dfrac{-4{{x}_{1}}+4}{5} \\\ & \Rightarrow {{h}_{1}}={{x}_{1}}+\left( \dfrac{-4{{x}_{1}}+4}{5} \right) \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow {{h}_{1}}=\dfrac{5{{x}_{1}}-4{{x}_{1}}+4}{5} \\\ & \Rightarrow {{h}_{1}}=\dfrac{{{x}_{1}}+4}{5}.......(4) \\\ \end{aligned}$$ From the equation (3) we can find the value of $${{k}_{1}}$$. $$\dfrac{{{k}_{1}}-{{x}_{1}}}{-1}=\dfrac{-2({{x}_{1}}-1)}{5}$$ By using cross multiplication, we get $$\begin{aligned} & \Rightarrow {{k}_{1}}-{{x}_{1}}=\dfrac{2({{x}_{1}}-1)}{5} \\\ & \Rightarrow {{k}_{1}}={{x}_{1}}+\dfrac{2({{x}_{1}}-1)}{5} \\\ & \Rightarrow {{k}_{1}}=\dfrac{5{{x}_{1}}+2{{x}_{1}}-2}{5} \\\ & \Rightarrow {{k}_{1}}=\dfrac{7{{x}_{1}}-2}{5}......(5) \\\ \end{aligned}$$ From equation (4) and equation (5), we can get the image of $$A({{x}_{1}},{{y}_{1}})$$ is $$A\grave{\ }\left( \dfrac{{{x}_{1}}+4}{5},\dfrac{7{{x}_{1}}-2}{5} \right)$$. So, the image of point $$B({{x}_{2}},{{x}_{2}})$$ with respect to 2x-y-1=0 is $$B\grave{\ }({{h}_{2}},{{k}_{2}})$$ if $$\begin{aligned} & \Rightarrow \dfrac{{{h}_{2}}-{{x}_{2}}}{2}=\dfrac{{{k}_{2}}-{{x}_{2}}}{-1}=\dfrac{-2(2{{x}_{2}}-{{x}_{2}}-1)}{{{2}^{2}}+{{1}^{2}}} \\\ & \Rightarrow \dfrac{{{h}_{2}}-{{x}_{2}}}{2}=\dfrac{{{k}_{2}}-{{x}_{2}}}{-1}=\dfrac{-2({{x}_{2}}-1)}{5}.....(6) \\\ \end{aligned}$$ From equation (6) we can find the value of $${{h}_{2}}$$. $$\dfrac{{{h}_{1}}-{{x}_{2}}}{2}=\dfrac{-2({{x}_{2}}-1)}{5}$$ By using cross multiplication, we get $$\begin{aligned} & \Rightarrow {{h}_{2}}-{{x}_{2}}=\dfrac{-4({{x}_{2}}-1)}{5} \\\ & \Rightarrow {{h}_{2}}-{{x}_{2}}=\dfrac{-4{{x}_{2}}+4}{5} \\\ & \Rightarrow {{h}_{2}}={{x}_{2}}+\left( \dfrac{-4{{x}_{2}}+4}{5} \right) \\\ & \Rightarrow {{h}_{2}}=\dfrac{5{{x}_{2}}-4{{x}_{2}}+4}{5} \\\ & \Rightarrow {{h}_{2}}=\dfrac{{{x}_{2}}+4}{5}.......(7) \\\ \end{aligned}$$ From the equation (6) we can find the value of $${{k}_{2}}$$. $$\dfrac{{{k}_{2}}-{{x}_{2}}}{-1}=\dfrac{-2({{x}_{2}}-1)}{5}$$ By using cross multiplication, we get $$\begin{aligned} & \Rightarrow {{k}_{2}}-{{x}_{2}}=\dfrac{2({{x}_{2}}-1)}{5} \\\ & \Rightarrow {{k}_{2}}={{x}_{2}}+\dfrac{2({{x}_{2}}-1)}{5} \\\ & \Rightarrow {{k}_{2}}=\dfrac{5{{x}_{2}}+2{{x}_{2}}-2}{5} \\\ & \Rightarrow {{k}_{2}}=\dfrac{7{{x}_{2}}-2}{5}......(8) \\\ \end{aligned}$$ From equation (7) and equation (8), we can get the image of $$B({{x}_{2}},{{y}_{2}})$$ is $$B\grave{\ }\left( \dfrac{{{x}_{2}}+4}{5},\dfrac{7{{x}_{2}}-2}{5} \right)$$. Now let us find the equation of A`B`. We know that if $$A({{x}_{1}},{{y}_{1}})$$ and $$B({{x}_{2}},{{y}_{2}})$$ are two points on a straight line, then the equation of line passing through A and B is $$y-{{y}_{1}}=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right)$$. So, the equation of line passing through $$A\grave{\ }\left( \dfrac{{{x}_{1}}+4}{5},\dfrac{7{{x}_{1}}-2}{5} \right)$$ and $$B\grave{\ }\left( \dfrac{{{x}_{2}}+4}{5},\dfrac{7{{x}_{2}}-2}{5} \right)$$ is $$\begin{aligned} & y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=\left( \dfrac{\left( \dfrac{7{{x}_{2}}-2}{5} \right)-\left( \dfrac{7{{x}_{1}}-2}{5} \right)}{\left( \dfrac{{{x}_{2}}+4}{5} \right)-\left( \dfrac{{{x}_{1}}+4}{5} \right)} \right)\left( x-\left( \dfrac{{{x}_{1}}+4}{5} \right) \right) \\\ & \Rightarrow y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=\left( \dfrac{\dfrac{(7{{x}_{2}}-2)-(7{{x}_{1}}-2)}{5}}{\dfrac{({{x}_{2}}+4)-({{x}_{1}}+4)}{5}} \right)\left( x-\left( \dfrac{{{x}_{1}}+4}{5} \right) \right) \\\ & \Rightarrow y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=\left( \dfrac{(7{{x}_{2}}-2)-(7{{x}_{1}}-2)}{({{x}_{2}}+4)-({{x}_{1}}+4)} \right)\left( x-\left( \dfrac{{{x}_{1}}+4}{5} \right) \right) \\\ & \Rightarrow y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=\left( \dfrac{7({{x}_{2}}-{{x}_{1}})}{({{x}_{2}}-{{x}_{1}})} \right)\left( x-\left( \dfrac{{{x}_{1}}+4}{5} \right) \right) \\\ & \Rightarrow y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=7\left( x-\left( \dfrac{{{x}_{1}}+4}{5} \right) \right) \\\ & \Rightarrow y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=7x-7\left( \dfrac{{{x}_{1}}+4}{5} \right) \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow 7x-y=7\left( \dfrac{{{x}_{1}}+4}{5} \right)-\left( \dfrac{7{{x}_{1}}-2}{5} \right) \\\ & \Rightarrow 7x-y=\dfrac{7({{x}_{1}}+4)-(7{{x}_{1}}-2)}{5} \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow 7x-y=\dfrac{7{{x}_{1}}+28-7{{x}_{1}}+2}{5} \\\ & \Rightarrow 7x-y=6 \\\ \end{aligned}$$ So, the image of the line equation of AB is 7x-y=6. Hence, option (c) is correct. Note: This sum can be solved in other ways also.  We know that if an incident ray incident on a mirror at a point, the reflected ray will pass through the same point of intersection making the same angle as the incident ray made with the mirror. Consider y=x as incident ray and 2x-y=1 as normal. Now we needed to find the intersection point of y=x and 2x-y=1. Assume $$\begin{aligned} & y=x......(1) \\\ & 2x-y=1.....(2) \\\ \end{aligned}$$ Now, add equation (1) with equation (2) $$\begin{aligned} & y+(2x-y)=x+1 \\\ & \Rightarrow y+2x-y=x+1 \\\ & \Rightarrow 2x=x+1 \\\ & \Rightarrow x=1....(3) \\\ \end{aligned}$$ Now substitute equation (3) in equation (1) $$\Rightarrow y=1.....(4)$$ So, the intersection point of mirror line and incident ray is (1,1). Now few needed to find the angle between y=x and 2x-y=1. We know that if $${{m}_{1}}$$and $${{m}_{2}}$$are slopes of two lines then the angle between two lines is $$\theta =Ta{{n}^{-1}}\left( \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right)$$. $${{m}_{1}}$$= Slope of line x-y=0 is 1. $${{m}_{2}}$$= Slope of line 2x-y-1=0 is 2. $$\theta =Ta{{n}^{-1}}\left( \dfrac{1-2}{1+(1)(2)} \right)=Ta{{n}^{-1}}\left( \dfrac{-1}{3} \right).....(5)$$ So, $$\theta $$ is also the angle between the mirror line and reflected line. Let us assume the reflected line as a x + b y + c=0. $$L=ax+by+c$$ also passes through (1,1). $$\Rightarrow a+b+c=0.....(6)$$ Slope of$$L=ax+by+c$$ is $$\dfrac{-a}{b}$$. Angle between $$L=ax+by+c$$ and 2x-y-1=0 is $$\theta =Ta{{n}^{-1}}\left( \dfrac{2-\left( \dfrac{-a}{b} \right)}{1+2\left( \dfrac{-a}{b} \right)} \right)=Ta{{n}^{-1}}\left( \dfrac{2+\dfrac{a}{b}}{1-\dfrac{2a}{b}} \right)=Ta{{n}^{-1}}\left( \dfrac{\dfrac{2b+a}{b}}{\dfrac{b-2a}{b}} \right)=Ta{{n}^{-1}}\left( \dfrac{a+2b}{-2a+b} \right)....(7)$$ From equation (5) and equation (7), $$Ta{{n}^{-1}}\left( \dfrac{-1}{3} \right)=$$$$Ta{{n}^{-1}}\left( \dfrac{a+2b}{-2a+b} \right)$$ Now we will apply Tan on both sides $$Tan\left( Ta{{n}^{-1}}\left( \dfrac{-1}{3} \right) \right)=Tan\left( Ta{{n}^{-1}}\left( \dfrac{a+2b}{-2a+b} \right) \right)$$ $$\Rightarrow \dfrac{-1}{3}=\dfrac{a+2b}{-2a+b}$$ By using cross multiplication, we get $$\begin{aligned} & \Rightarrow \dfrac{-1}{3}=\dfrac{a+2b}{-2a+b} \\\ & \Rightarrow -(-2a+b)=3(a+2b) \\\ & \Rightarrow 2a-b=3a+6b \\\ & \Rightarrow a=-7b......(8) \\\ & \\\ \end{aligned}$$ Now we will substitute equation (8) in equation (6) $$\begin{aligned} & -7b+b+c=0 \\\ & \Rightarrow -6b+c=0 \\\ & \Rightarrow c=6b.....(9) \\\ \end{aligned}$$ From equation (8) and equation (9), We get that the reflected ray is $$(-7b)x+by+(6b)=0$$. Now we will divide the equation by (-b) on both sides. $$7x-y=6.....(10)$$ From equation (10), it is clear that the reflected ray is $$7x-y=6$$. The reflected line equation is a line equation which passes through the image of incident line equation. Hence, the image of AB is $$7x-y=6$$. Therefore, option (c) is correct.