The image of point (−1, 3, 4) in the plane $$x − 2y = 0$$​ is:

Correct option is D

Given:
Point $$P = (x_1, y_1, z_1) = (−1, 3, 4)$$​​
Plane: $$x − 2y = 0$$​​
Formula used:
For reflection of a point $$(x_1, y_1, z_1)$$​ in plane $$ax + by + cz + d = 0,$$​​
the image point (x', y', z') is given by
​$$x' = x_1 − \frac{2a(ax_1 + by_1 + cz_1 + d)}{a^{2} + b^{2} + c^{2}}\\y' = y_1 − \frac{2b(ax_1 + by_1 + cz_1 + d)}{a^{2} + b^{2} + c^{2}}\\z' = z_1 − \frac{2c(ax_1 + by_1 + cz_1 + d)}{a^{2} + b^{2} + c^{2}}$$​​
Solution:​
Here a = 1, b = −2, c = 0, d = 0
Compute:​
$$ax_1 + by_1 + cz_1 + d\\= (1)(−1) + (−2)(3) + 0\\= −1 − 6\\= −7\\a^{2} + b^{2} + c^{2}= 1 + 4 + 0= 5$$​​
​$$x' = −1 − \frac{2(1)(−7)}{5}= −1 + \frac{14}{5}= \frac{9}{5}\\[3pt]y' = 3 − \frac{2(−2)(−7)}{5}= 3 − \frac{28}{5}= −\frac{13}{5}\\[3pt]z' = 4$$​​
So the image point is:
​$$\left( \frac{9}{5}, -\frac{13}{5}, 4 \right)$$​​
The correct answer is (d) $$\left( \frac{9}{5}, -\frac{13}{5}, 4 \right).$$​​