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
\[\begin{align}
& (A)\text{ x + y =2} \\
& \text{(B) 8x + y=9} \\
& \text{(C) 7x - y =6} \\
& \text{(D) None of these} \\
\end{align}\]
Answer
Verified583.5k+ views
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
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)
\[\begin{align}
& \Rightarrow \dfrac{2{{x}_{1}}-{{x}_{1}}-1}{2{{x}_{2}}-{{x}_{2}}-1}=-k \\
& \Rightarrow \dfrac{{{x}_{1}}-1}{{{x}_{2}}-1}=-k........(2) \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow \dfrac{{{h}_{1}}-{{x}_{1}}}{2}=\dfrac{{{k}_{1}}-{{x}_{1}}}{-1}=\dfrac{-2(2{{x}_{1}}-{{x}_{1}}-1)}{{{2}^{2}}+{{1}^{2}}} \\
& \Rightarrow \dfrac{{{h}_{1}}-{{x}_{1}}}{2}=\dfrac{{{k}_{1}}-{{x}_{1}}}{-1}=\dfrac{-2({{x}_{1}}-1)}{5}.....(3) \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow {{h}_{1}}-{{x}_{1}}=\dfrac{-4({{x}_{1}}-1)}{5} \\
& \Rightarrow {{h}_{1}}-{{x}_{1}}=\dfrac{-4{{x}_{1}}+4}{5} \\
& \Rightarrow {{h}_{1}}={{x}_{1}}+\left( \dfrac{-4{{x}_{1}}+4}{5} \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow {{h}_{1}}=\dfrac{5{{x}_{1}}-4{{x}_{1}}+4}{5} \\
& \Rightarrow {{h}_{1}}=\dfrac{{{x}_{1}}+4}{5}.......(4) \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow {{k}_{1}}-{{x}_{1}}=\dfrac{2({{x}_{1}}-1)}{5} \\
& \Rightarrow {{k}_{1}}={{x}_{1}}+\dfrac{2({{x}_{1}}-1)}{5} \\
& \Rightarrow {{k}_{1}}=\dfrac{5{{x}_{1}}+2{{x}_{1}}-2}{5} \\
& \Rightarrow {{k}_{1}}=\dfrac{7{{x}_{1}}-2}{5}......(5) \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow \dfrac{{{h}_{2}}-{{x}_{2}}}{2}=\dfrac{{{k}_{2}}-{{x}_{2}}}{-1}=\dfrac{-2(2{{x}_{2}}-{{x}_{2}}-1)}{{{2}^{2}}+{{1}^{2}}} \\
& \Rightarrow \dfrac{{{h}_{2}}-{{x}_{2}}}{2}=\dfrac{{{k}_{2}}-{{x}_{2}}}{-1}=\dfrac{-2({{x}_{2}}-1)}{5}.....(6) \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow {{h}_{2}}-{{x}_{2}}=\dfrac{-4({{x}_{2}}-1)}{5} \\
& \Rightarrow {{h}_{2}}-{{x}_{2}}=\dfrac{-4{{x}_{2}}+4}{5} \\
& \Rightarrow {{h}_{2}}={{x}_{2}}+\left( \dfrac{-4{{x}_{2}}+4}{5} \right) \\
& \Rightarrow {{h}_{2}}=\dfrac{5{{x}_{2}}-4{{x}_{2}}+4}{5} \\
& \Rightarrow {{h}_{2}}=\dfrac{{{x}_{2}}+4}{5}.......(7) \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow {{k}_{2}}-{{x}_{2}}=\dfrac{2({{x}_{2}}-1)}{5} \\
& \Rightarrow {{k}_{2}}={{x}_{2}}+\dfrac{2({{x}_{2}}-1)}{5} \\
& \Rightarrow {{k}_{2}}=\dfrac{5{{x}_{2}}+2{{x}_{2}}-2}{5} \\
& \Rightarrow {{k}_{2}}=\dfrac{7{{x}_{2}}-2}{5}......(8) \\
\end{align}\]
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
\[\begin{align}
& y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=\left( \dfrac{\left( \dfrac{7{{x}_{2}}-2}{5} \right)-\left( \dfrac{7{{x}_{1}}-2}{5} \right)}{\left( \dfrac{{{x}_{2}}+4}{5} \right)-\left( \dfrac{{{x}_{1}}+4}{5} \right)} \right)\left( x-\left( \dfrac{{{x}_{1}}+4}{5} \right) \right) \\
& \Rightarrow y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=\left( \dfrac{\dfrac{(7{{x}_{2}}-2)-(7{{x}_{1}}-2)}{5}}{\dfrac{({{x}_{2}}+4)-({{x}_{1}}+4)}{5}} \right)\left( x-\left( \dfrac{{{x}_{1}}+4}{5} \right) \right) \\
& \Rightarrow y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=\left( \dfrac{(7{{x}_{2}}-2)-(7{{x}_{1}}-2)}{({{x}_{2}}+4)-({{x}_{1}}+4)} \right)\left( x-\left( \dfrac{{{x}_{1}}+4}{5} \right) \right) \\
& \Rightarrow y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=\left( \dfrac{7({{x}_{2}}-{{x}_{1}})}{({{x}_{2}}-{{x}_{1}})} \right)\left( x-\left( \dfrac{{{x}_{1}}+4}{5} \right) \right) \\
& \Rightarrow y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=7\left( x-\left( \dfrac{{{x}_{1}}+4}{5} \right) \right) \\
& \Rightarrow y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=7x-7\left( \dfrac{{{x}_{1}}+4}{5} \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow 7x-y=7\left( \dfrac{{{x}_{1}}+4}{5} \right)-\left( \dfrac{7{{x}_{1}}-2}{5} \right) \\
& \Rightarrow 7x-y=\dfrac{7({{x}_{1}}+4)-(7{{x}_{1}}-2)}{5} \\
\end{align}\]
\[\begin{align}
& \Rightarrow 7x-y=\dfrac{7{{x}_{1}}+28-7{{x}_{1}}+2}{5} \\
& \Rightarrow 7x-y=6 \\
\end{align}\]
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
\[\begin{align}
& y=x......(1) \\
& 2x-y=1.....(2) \\
\end{align}\]
Now, add equation (1) with equation (2)
\[\begin{align}
& y+(2x-y)=x+1 \\
& \Rightarrow y+2x-y=x+1 \\
& \Rightarrow 2x=x+1 \\
& \Rightarrow x=1....(3) \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow \dfrac{-1}{3}=\dfrac{a+2b}{-2a+b} \\
& \Rightarrow -(-2a+b)=3(a+2b) \\
& \Rightarrow 2a-b=3a+6b \\
& \Rightarrow a=-7b......(8) \\
& \\
\end{align}\]
Now we will substitute equation (8) in equation (6)
\[\begin{align}
& -7b+b+c=0 \\
& \Rightarrow -6b+c=0 \\
& \Rightarrow c=6b.....(9) \\
\end{align}\]
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.
By comparing 2x-y=1 with \[L=ax+by+c\], we get a=2, b=-1 and c=-1.
From (1)
\[\begin{align}
& \Rightarrow \dfrac{2{{x}_{1}}-{{x}_{1}}-1}{2{{x}_{2}}-{{x}_{2}}-1}=-k \\
& \Rightarrow \dfrac{{{x}_{1}}-1}{{{x}_{2}}-1}=-k........(2) \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow \dfrac{{{h}_{1}}-{{x}_{1}}}{2}=\dfrac{{{k}_{1}}-{{x}_{1}}}{-1}=\dfrac{-2(2{{x}_{1}}-{{x}_{1}}-1)}{{{2}^{2}}+{{1}^{2}}} \\
& \Rightarrow \dfrac{{{h}_{1}}-{{x}_{1}}}{2}=\dfrac{{{k}_{1}}-{{x}_{1}}}{-1}=\dfrac{-2({{x}_{1}}-1)}{5}.....(3) \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow {{h}_{1}}-{{x}_{1}}=\dfrac{-4({{x}_{1}}-1)}{5} \\
& \Rightarrow {{h}_{1}}-{{x}_{1}}=\dfrac{-4{{x}_{1}}+4}{5} \\
& \Rightarrow {{h}_{1}}={{x}_{1}}+\left( \dfrac{-4{{x}_{1}}+4}{5} \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow {{h}_{1}}=\dfrac{5{{x}_{1}}-4{{x}_{1}}+4}{5} \\
& \Rightarrow {{h}_{1}}=\dfrac{{{x}_{1}}+4}{5}.......(4) \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow {{k}_{1}}-{{x}_{1}}=\dfrac{2({{x}_{1}}-1)}{5} \\
& \Rightarrow {{k}_{1}}={{x}_{1}}+\dfrac{2({{x}_{1}}-1)}{5} \\
& \Rightarrow {{k}_{1}}=\dfrac{5{{x}_{1}}+2{{x}_{1}}-2}{5} \\
& \Rightarrow {{k}_{1}}=\dfrac{7{{x}_{1}}-2}{5}......(5) \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow \dfrac{{{h}_{2}}-{{x}_{2}}}{2}=\dfrac{{{k}_{2}}-{{x}_{2}}}{-1}=\dfrac{-2(2{{x}_{2}}-{{x}_{2}}-1)}{{{2}^{2}}+{{1}^{2}}} \\
& \Rightarrow \dfrac{{{h}_{2}}-{{x}_{2}}}{2}=\dfrac{{{k}_{2}}-{{x}_{2}}}{-1}=\dfrac{-2({{x}_{2}}-1)}{5}.....(6) \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow {{h}_{2}}-{{x}_{2}}=\dfrac{-4({{x}_{2}}-1)}{5} \\
& \Rightarrow {{h}_{2}}-{{x}_{2}}=\dfrac{-4{{x}_{2}}+4}{5} \\
& \Rightarrow {{h}_{2}}={{x}_{2}}+\left( \dfrac{-4{{x}_{2}}+4}{5} \right) \\
& \Rightarrow {{h}_{2}}=\dfrac{5{{x}_{2}}-4{{x}_{2}}+4}{5} \\
& \Rightarrow {{h}_{2}}=\dfrac{{{x}_{2}}+4}{5}.......(7) \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow {{k}_{2}}-{{x}_{2}}=\dfrac{2({{x}_{2}}-1)}{5} \\
& \Rightarrow {{k}_{2}}={{x}_{2}}+\dfrac{2({{x}_{2}}-1)}{5} \\
& \Rightarrow {{k}_{2}}=\dfrac{5{{x}_{2}}+2{{x}_{2}}-2}{5} \\
& \Rightarrow {{k}_{2}}=\dfrac{7{{x}_{2}}-2}{5}......(8) \\
\end{align}\]
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
\[\begin{align}
& y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=\left( \dfrac{\left( \dfrac{7{{x}_{2}}-2}{5} \right)-\left( \dfrac{7{{x}_{1}}-2}{5} \right)}{\left( \dfrac{{{x}_{2}}+4}{5} \right)-\left( \dfrac{{{x}_{1}}+4}{5} \right)} \right)\left( x-\left( \dfrac{{{x}_{1}}+4}{5} \right) \right) \\
& \Rightarrow y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=\left( \dfrac{\dfrac{(7{{x}_{2}}-2)-(7{{x}_{1}}-2)}{5}}{\dfrac{({{x}_{2}}+4)-({{x}_{1}}+4)}{5}} \right)\left( x-\left( \dfrac{{{x}_{1}}+4}{5} \right) \right) \\
& \Rightarrow y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=\left( \dfrac{(7{{x}_{2}}-2)-(7{{x}_{1}}-2)}{({{x}_{2}}+4)-({{x}_{1}}+4)} \right)\left( x-\left( \dfrac{{{x}_{1}}+4}{5} \right) \right) \\
& \Rightarrow y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=\left( \dfrac{7({{x}_{2}}-{{x}_{1}})}{({{x}_{2}}-{{x}_{1}})} \right)\left( x-\left( \dfrac{{{x}_{1}}+4}{5} \right) \right) \\
& \Rightarrow y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=7\left( x-\left( \dfrac{{{x}_{1}}+4}{5} \right) \right) \\
& \Rightarrow y-\left( \dfrac{7{{x}_{1}}-2}{5} \right)=7x-7\left( \dfrac{{{x}_{1}}+4}{5} \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow 7x-y=7\left( \dfrac{{{x}_{1}}+4}{5} \right)-\left( \dfrac{7{{x}_{1}}-2}{5} \right) \\
& \Rightarrow 7x-y=\dfrac{7({{x}_{1}}+4)-(7{{x}_{1}}-2)}{5} \\
\end{align}\]
\[\begin{align}
& \Rightarrow 7x-y=\dfrac{7{{x}_{1}}+28-7{{x}_{1}}+2}{5} \\
& \Rightarrow 7x-y=6 \\
\end{align}\]
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
\[\begin{align}
& y=x......(1) \\
& 2x-y=1.....(2) \\
\end{align}\]
Now, add equation (1) with equation (2)
\[\begin{align}
& y+(2x-y)=x+1 \\
& \Rightarrow y+2x-y=x+1 \\
& \Rightarrow 2x=x+1 \\
& \Rightarrow x=1....(3) \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow \dfrac{-1}{3}=\dfrac{a+2b}{-2a+b} \\
& \Rightarrow -(-2a+b)=3(a+2b) \\
& \Rightarrow 2a-b=3a+6b \\
& \Rightarrow a=-7b......(8) \\
& \\
\end{align}\]
Now we will substitute equation (8) in equation (6)
\[\begin{align}
& -7b+b+c=0 \\
& \Rightarrow -6b+c=0 \\
& \Rightarrow c=6b.....(9) \\
\end{align}\]
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.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success
Master Class 12 Economics: Engaging Questions & Answers for Success
Master Class 12 English: Engaging Questions & Answers for Success
Master Class 12 Maths: Engaging Questions & Answers for Success
Master Class 12 Social Science: Engaging Questions & Answers for Success
Master Class 12 Chemistry: Engaging Questions & Answers for Success
Trending doubts
What are the major means of transport Explain each class 12 social science CBSE
Which are the Top 10 Largest Countries of the World?
Draw a labelled sketch of the human eye class 12 physics CBSE
How much time does it take to bleed after eating p class 12 biology CBSE
Explain sex determination in humans with line diag class 12 biology CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE