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Question: The line AB whose equation is x-y=2 cuts the x axis at A and B is (4,2). The line segment AB is rota...

The line AB whose equation is x-y=2 cuts the x axis at A and B is (4,2). The line segment AB is rotated about A through an angle 45{{45}^{\circ }} in the anticlockwise sense, then the equation AB in the new position is

& A.x=2 \\\ & B.x-1=0 \\\ & C.x-\sqrt{2}y-2=0 \\\ & D.y-2=0 \\\ \end{aligned}$$
Explanation

Solution

For solving this question, we will first find slope of the given line. Then we will rotate the line by 45{{45}^{\circ }} which will give us a value of θ\theta which will give us a new slope of the line. Since coordinate A at x axis will remain same so we will find coordinate of A and then find equation of line using slope and coordinate of A. We will use following formula:
1: For finding a point on the x axis, put y = 0 and find the value of x. Point will be (x,0).
2: Slope of any line is given by tanθ\tan \theta where θ\theta is an angle formed by a line with x axis.
3: Slope of any line is given by y2y1x2x1\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} where (x1,y1) and (x2,y2)\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right) are any two points on lines.
4: Equation of a line having slope tanθ\tan \theta and coordinate of any point as (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) is given by (yy1)=tanθ(xx1)\left( y-{{y}_{1}} \right)=\tan \theta \left( x-{{x}_{1}} \right).

Complete step-by-step answer:
Here we are given the equation of the line AB as xy=2(1)x-y=2\cdots \cdots \cdots \left( 1 \right).
Let us first find the coordinate of the point A.
Since point A cuts the x axis, so y coordinate will be zero.
Putting y = 0 in (1) we get,
x0=2x=2x-0=2\Rightarrow x=2.
Hence A coordinate is (2,0).
Also point B is (4,2).
So two points of the line AB are (2,0) and (4,2)
We know slope m1{{m}_{1}} any line passing through (x1,y1) and (x2,y2)\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right) is given by m1=y2y1x2x1{{m}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} so we get:
m1=2042=22=1{{m}_{1}}=\dfrac{2-0}{4-2}=\dfrac{2}{2}=1.
Hence slope of line x-y=2 is 1.
Also as we know that, the slope of any line is tanθ\tan \theta where θ\theta is the angle that the line makes with the x axis. So we get tanθ=1\tan \theta =1.
As we know tan45=1\tan {{45}^{\circ }}=1 so we get θ=45\theta ={{45}^{\circ }}.
Now line is rotated anticlockwise through an angle of 45{{45}^{\circ }} so now θ\theta becomes 45+45=90{{45}^{\circ }}+{{45}^{\circ }}={{90}^{\circ }}.

Hence θ=90\theta ={{90}^{\circ }} for new lines. So the slope of the line becomes equal to tan90=\tan {{90}^{\circ }}=\infty . \infty can also be written as 10\dfrac{1}{0}.
Now we know that, equation of line having slope m and passing through (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) is given by (yy1)=m(xx1)\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right).
Hence putting in (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) as (2,0) and m as 10\dfrac{1}{0} we get:
y0=10(x2)y=10(x2)y-0=\dfrac{1}{0}\left( x-2 \right)\Rightarrow y=\dfrac{1}{0}\left( x-2 \right).
Cross multiplying we get:
x2=0x=2x-2=0\Rightarrow x=2.
Hence x = 2 is the new equation of the line.

So, the correct answer is “Option A”.

Note: Students should always try to draw diagrams for understanding questions clearly. Take care of signs while forming the equation and while finding the coordinate of the point. Note that, if slope is 10\dfrac{1}{0} then this means that line never touches the y axis.