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(A) For a convex lens,the lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Using sign convention,$u$ is negative,so let $u = -|u|$ a

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$(2f, 2f)$

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(A) For a convex lens,the lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Using sign convention,$u$ is negative,so let $u = -|u|$ and $v = |v|$.The equation becomes $\frac{1}{|v|} - \frac{1}{-|u|} = \frac{1}{f}$,which simplifies to $\frac{1}{|v|} + \frac{1}{|u|} = \frac{1}{f}$.$A$ straight line passing through the origin making an angle of $45^{\circ}$ with the $x$-axis has the equation $|v| = |u|$.At the intersection point $P$,we substitute $|v| = |u|$ into the lens formula:$\frac{1}{|u|} + \frac{1}{|u|} = \frac{1}{f}$$\frac{2}{|u|} = \frac{1}{f}$$|u| = 2f$.Since $|v| = |u|$,we have $|v| = 2f$.Thus,the coordinates of point $P$ are $(2f, 2f)$.

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