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(B) Given: Focal length $f = -40 \, cm$ (concave mirror),object distance $u = -60 \, cm$,and object height $h_o = 6 \, cm$.Using the mirror formula: $

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$(-120 \, cm, -12 \, cm)$

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(B) Given: Focal length $f = -40 \, cm$ (concave mirror),object distance $u = -60 \, cm$,and object height $h_o = 6 \, cm$.Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$\frac{1}{-40} = \frac{1}{v} + \frac{1}{-60}$$\frac{1}{v} = \frac{1}{60} - \frac{1}{40} = \frac{2 - 3}{120} = -\frac{1}{120}$So,$v = -120 \, cm$.The magnification $m$ is given by $m = -\frac{v}{u} = -\frac{-120}{-60} = -2$.The height of the image $h_i = m 	imes h_o = -2 	imes 6 \, cm = -12 \, cm$.Since the image is formed at $v = -120 \, cm$ and its height is $-12 \, cm$,the coordinates of the image are $(-120 \, cm, -12 \, cm)$.

If the coordinates of the position of an object are $(-60 \, cm, 6 \, cm)$,then the coordinates of the image will be:

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