Question

Let Q be the image of a point P(1,2) with respect to the line $x+y+1=0$ and R be the image of Q with respect to the line $x-y-1=0$. If M and N are the midpoints of PQ and QR respectively, then MN =

• A. $\sqrt{10}$
• B. 4
• C. $\sqrt{22}$
• D. 5

Hint:

A double reflection across two perpendicular lines is equivalent to a $180^\circ$ rotation about their point of intersection. This means the point of intersection of the lines is the midpoint of the original point and its final image.

Answer 1

The Correct Option is A

We are given points P(1,2), M is the midpoint of PQ, and N is the midpoint of QR.
We need to find the length of the segment MN.
In triangle PQR, M is the midpoint of side PQ and N is the midpoint of side QR.
By the Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
Therefore, $MN = \frac{1}{2} PR$.
We need to find the coordinates of R to calculate the distance PR.
Point Q is the image of P(1,2) in the line $L_1: x+y+1=0$.
Point R is the image of Q in the line $L_2: x-y-1=0$.
Notice that R is the point obtained by two successive reflections of P. The lines $L_1$ and $L_2$ are perpendicular because the product of their slopes is $(-1)(1)=-1$. They intersect at the point which solves $x+y=-1$ and $x-y=1$. Adding them gives $2x=0 \implies x=0$, and so $y=-1$. The intersection point is I(0,-1).
The transformation from P to R is a rotation by $180^\circ$ about the point of intersection I. This is equivalent to saying I is the midpoint of PR.
Let R have coordinates $(x_R, y_R)$.
Using the midpoint formula for PR: $I(0,-1) = \left(\frac{1+x_R}{2}, \frac{2+y_R}{2}\right)$.
$\frac{1+x_R}{2} = 0 \implies 1+x_R = 0 \implies x_R = -1$.
$\frac{2+y_R}{2} = -1 \implies 2+y_R = -2 \implies y_R = -4$.
So, the coordinates of R are $(-1, -4)$.
Now, we find the distance PR using the distance formula:
$PR = \sqrt{(-1-1)^2 + (-4-2)^2} = \sqrt{(-2)^2 + (-6)^2} = \sqrt{4+36} = \sqrt{40}$.
Finally, $MN = \frac{1}{2} PR = \frac{1}{2}\sqrt{40} = \frac{1}{2}\sqrt{4 \times 10} = \frac{1}{2}(2\sqrt{10}) = \sqrt{10}$.