The equation of the image of the pair of line | vedclass

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(C) The equation $y = |x - 1|$ represents two rays: $y = x - 1$ for $x \ge 1$ and $y = 1 - x$ for $x < 1$.To find the image of these lines in the $y$-

Vedclass · Accepted Answer

$x^2 - y^2 + 2x + 1 = 0$

Vedclass · Answer

(C) The equation $y = |x - 1|$ represents two rays: $y = x - 1$ for $x \ge 1$ and $y = 1 - x$ for $x To find the image of these lines in the $y$-axis,we replace $x$ with $-x$.For the line $y = x - 1$,the image is $y = -x - 1$,which can be written as $x + y + 1 = 0$.For the line $y = 1 - x$,the image is $y = 1 - (-x)$,which is $y = x + 1$,or $x - y + 1 = 0$.The joint equation of these two lines is $(x + y + 1)(x - y + 1) = 0$.This simplifies to $((x + 1) + y)((x + 1) - y) = 0$.Using the identity $(a + b)(a - b) = a^2 - b^2$,we get $(x + 1)^2 - y^2 = 0$.Expanding this,we have $x^2 + 2x + 1 - y^2 = 0$,or $x^2 - y^2 + 2x + 1 = 0$.

The equation of the image of the pair of lines $y = |x - 1|$ in the $y$-axis is:

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