Can we reflect this point in this line?

Can we reflect this point in this line?

Question

What is the reflection of the point $(3,4)$ in the line $3x + 4y = 50$?

1. $(9,12)$,

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1. $(6,8)$,

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1. $(12,16)$,

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1. $(16,12)$.

Suggestion

It might be helpful to sketch the situation. How are the point $(3,4)$ and its reflection related?

Solution

Let $L$ be the line with equation $3x + 4y = 50$.

The point $P(3,4)$ and its reflection $P'$ in the line $L$ are related in two ways: the line $PP'$ is perpendicular to the line $L$, and $P$ and $P'$ are equidistant (equal distances) from $L$. (So in terms of vectors, $\overrightarrow{PM}=\overrightarrow{MP'}$, where $M$ is the point of intersection of $PP'$ and $L$.)

Now $L$ has gradient $-\frac{3}{4}$, so any line perpendicular to $L$ will have gradient $\frac{4}{3}$.

The perpendicular line $PP'$ passing through $P(3,4)$ therefore has equation $y - 4 = \frac{4}{3}(x - 3),$ that is, $y = \frac{4}{3}x.$

From substituting this into the equation for $L$, we see that $L$ and $PP'$ intersect at $M$, where $3x + 4 \times \frac{4}{3}x = 50,$ and rearranging this gives $\frac{25}{3} x = 50,$ that is, $x = 6.$ Then substituting this into either equation gives $y = 8$, so $M$ lies at $(6,8)$.

We must now use this information to find the reflected point. We can see that the difference in $x$-coordinates between $(3,4)$ and $(6,8)$ is $3$, and the difference in $y$-coordinates is $4$. So to get the reflected point $P'$, we must add $3$ to the $x$-coordinate and $4$ to the $y$-coordinate of $M(6,8)$, as illustrated below:

This tells us that the coordinate of the reflected point is $(9,12)$.

The answer is (a).