The image of the point P(3, 8) with respect t | vedclass

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(D) Let the image be $P'(h, k)$.Since $PP'$ is perpendicular to the given line $x + 3y = 7$,the slope of $PP'$ is $m_1 = \frac{k - 8}{h - 3}$.The slop

Vedclass · Accepted Answer

$(-1, -4)$

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(D) Let the image be $P'(h, k)$.Since $PP'$ is perpendicular to the given line $x + 3y = 7$,the slope of $PP'$ is $m_1 = \frac{k - 8}{h - 3}$.The slope of the line $x + 3y = 7$ is $m_2 = -\frac{1}{3}$.Since $PP' \perp 	ext{line}$,$m_1 	imes m_2 = -1$ $\Rightarrow \frac{k - 8}{h - 3} 	imes (-\frac{1}{3}) = -1$ $\Rightarrow k - 8 = 3(h - 3)$ $\Rightarrow 3h - k = 1 \quad \dots(i)$.The midpoint of $PP'$ is $(\frac{h + 3}{2}, \frac{k + 8}{2})$,which lies on the line $x + 3y = 7$.So,$\frac{h + 3}{2} + 3(\frac{k + 8}{2}) = 7$ $\Rightarrow h + 3 + 3k + 24 = 14$ $\Rightarrow h + 3k = -13 \quad \dots(ii)$.Solving equations $(i)$ and $(ii)$:From $(i)$,$k = 3h - 1$.Substituting into $(ii)$: $h + 3(3h - 1) = -13$ $\Rightarrow h + 9h - 3 = -13$ $\Rightarrow 10h = -10$ $\Rightarrow h = -1$.Then $k = 3(-1) - 1 = -4$.Thus,the image is $(-1, -4)$.

The image of the point $P(3, 8)$ with respect to the line $x + 3y = 7$,assuming the line to be a plane mirror,is equal to .........

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