The image of the point 0-0 about the line x - y -1-0 is -A- 1-1 B- -1-1C- 2-2

The correct option is B ( 1, 1)If the image of (X_1 , Y_1) w.r.t ax+by+c = 0 is (x,y), then x X_1/a = y Y_1/b = 2(aX_1 + bY_1 + c)/a^2 + b^2 Here, x 0/1 ...

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Standard X

Mathematics

Image of a Point with Respect to X = a

The image of ...

Question

The image of the point (0,0) about the line $x + y + 1 = 0$ is .

(1,1)

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(-1,-1)

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(2,2)

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Solution

The correct option is B (-1,-1)
If the image of $(X_{1}, Y_{1})$ w.r.t $a x + b y + c = 0$ is (x,y), then

$\frac{x - X_{1}}{a} = \frac{y - Y_{1}}{b} = - \frac{2 (a X_{1} + b Y_{1} + c)}{a^{2} + b^{2}}$

Here, $\frac{x - 0}{1} = \frac{y - 0}{1} = \frac{- 2 (1)}{2}$ $\Rightarrow (x, y) = (- 1, - 1)$

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The image of the point (0,0) about the line x+y+1=0 is .

The correct option is B (-1,-1)If the image of (X1,Y1) w.r.t ax+by+c=0 is (x,y), then x−X1a=y−Y1b=−2(aX1+bY1+c)a2+b2 Here, x−01=y−01=−2(1)2 ⇒(x,y)=(−1,−1)

The image of the point (0,0) about the line $x + y + 1 = 0$ is .

The correct option is B (-1,-1)
If the image of $(X_{1}, Y_{1})$ w.r.t $a x + b y + c = 0$ is (x,y), then

$\frac{x - X_{1}}{a} = \frac{y - Y_{1}}{b} = - \frac{2 (a X_{1} + b Y_{1} + c)}{a^{2} + b^{2}}$

Here, $\frac{x - 0}{1} = \frac{y - 0}{1} = \frac{- 2 (1)}{2}$ $\Rightarrow (x, y) = (- 1, - 1)$