Question

The image of a point A(3,8) in the line x+3y-7=0, is

• A. (-1, -4)
• B. (-3, -8)
• C. (1, -4)
• D. (3, 8)

Answer 1

The correct option is A (-1, -4)

Equation of the line passing through (3, 8) and perpendicular to x+3y-7=0 is 3x-y-1=0. The intersection point of both the lines is (1, 2).
Now let the image of $A(3,8)$ be $A^{\prime }(x_1,y_1)$, then point (1, 2) will be the mid point of AA' .
$\Rightarrow \frac{x_1+3}{2}=1\Rightarrow x_1=-1and\frac{y_1+8}{2}=2\Rightarrow y_1=-4$.
Hence the image is (-1, -4).