# The image of the point (–1, 3, 4) in the 3 plane $\dpi{100} x-2y=0$  is Option 1) $\left ( -\frac{17}{3},-\frac{19}{3},4 \right )\;$ Option 2) $(15,11,4)$ Option 3) $\left ( -\frac{17}{3},-\frac{19}{3},1 \right )\; \;$ Option 4) $\left ( \frac{9}{5},-\frac{13}{5},4 \right )\; \;$

P Plabita

As we learnt in

Image of a point -

Let $L'$ be the image of point $P\left ( \alpha ,\beta ,\gamma \right )$ in the plane $ax+by+cz+d=0$

$L'$ will be given by the formula

$\frac{x-\alpha }{a}=\frac{y-\beta }{b}=\frac{z-\gamma }{c}= \frac{-2\left ( a\alpha +b\beta +c\gamma +d \right )}{a^{2}+b^{2}+c^{2}}$

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To find image we take

$\frac{x+1}{1}=\frac{y-3}{-2}=\frac{z-4}{0}=\frac{-2\left ( -1-6 \right )}{5}$

$x=\frac{9}{5} ;\, \, y=\frac{-13}{5}; \, \, z=4$

Option 1)

$\left ( -\frac{17}{3},-\frac{19}{3},4 \right )\;$

Incorrect Option

Option 2)

$(15,11,4)$

Incorrect Option

Option 3)

$\left ( -\frac{17}{3},-\frac{19}{3},1 \right )\; \;$

Incorrect Option

Option 4)

$\left ( \frac{9}{5},-\frac{13}{5},4 \right )\; \;$

Correct Option

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