Question

Assuming that the straight line works as a plane mirror for a point, the image of the point \((1,2)\) in the line \[ x-3y+4=0 \] is

• A. \(\left(\frac{5}{6},\frac{7}{5}\right)\)
• B. \(\left(\frac{6}{5},\frac{5}{7}\right)\)
• C. \(\left(\frac{5}{6},\frac{5}{7}\right)\)
• D. \(\left(\frac{6}{5},\frac{7}{5}\right)\)

Hint:

Reflection of a point \((x_1,y_1)\) in the line \[ ax+by+c=0 \] can be found directly using the image formula. This avoids finding the foot of the perpendicular separately.

Answer 1

The Correct Option is D

Concept: The image of a point \((x_1,y_1)\) in the line \[ ax+by+c=0 \] is \[ \left( x_1-\frac{2a(ax_1+by_1+c)}{a^2+b^2}, \; y_1-\frac{2b(ax_1+by_1+c)}{a^2+b^2} \right) \]

Step 1: Identify the values of \(a,b,c\). \[ a=1,\qquad b=-3,\qquad c=4 \] and \[ (x_1,y_1)=(1,2) \]

Step 2: Compute \(ax_1+by_1+c\). \[\begin{aligned} 1(1)+(-3)(2)+4 &=1-6+4\\ &=-1 \end{aligned}\] Also, \[ a^2+b^2=1+9=10 \]

Step 3: Find the image coordinates. \[\begin{aligned} x' &= 1-\frac{2(1)(-1)}{10}\\ &= 1+\frac15\\ &=\frac65 \end{aligned}\] \[\begin{aligned} y' &= 2-\frac{2(-3)(-1)}{10}\\ &= 2-\frac35\\ &=\frac75 \end{aligned}\] Hence the image point is \[ \left(\frac65,\frac75\right) \] \[\begin{aligned} \boxed{\left(\frac65,\frac75\right)} \end{aligned}\] Hence, option \(\mathbf{(D)}\) is correct.