Point $P$ is on the line $y=5 x+3$. The coordinates of point $Q$ are $(3,-2)$. If $M$ is the midpoint of $P Q$, then $M$ must lie on the line
(A) $y=\frac{5}{2} x-\frac{7}{2}$
(B) $y=5 x+1$
(C) $y=-\frac{1}{5} x-\frac{7}{5}$
(D) $y=\frac{5}{2} x+\frac{1}{2}$
(E) $y=5 x-7$

## Part C: Each correct answer is worth 8.
Point $P$ is on the line $y=5 x+3$. The coordinates of point $Q$ are $(3,-2)$. If $M$ is the midpoint of $P Q$, then $M$ must lie on the line
(A) $y=\frac{5}{2} x-\frac{7}{2}$
(B) $y=5 x+1$
(C) $y=-\frac{1}{5} x-\frac{7}{5}$
(D) $y=\frac{5}{2} x+\frac{1}{2}$
(E) $y=5 x-7$

We start by drawing a diagram and labelling the intercepts.

## Solution 1

Since the point $P$ is on the line $y=5 x+3$, select $P(0,3)$ as a point on this line.

The midpoint of $P Q$ is $M\left(\frac{3+0}{2}, \frac{-2+3}{2}\right)=M\left(\frac{3}{2}, \frac{1}{2}\right)$.

The required line must contain $M$ and be midway between the given point and $y=5 x+3$. The only possible line meeting this requirement is the line containing $M\left(\frac{3}{2}, \frac{1}{2}\right)$ and which has a slope of 5 . The required line will this have as its equation

$$
\begin{aligned}
y-\frac{1}{2} & =5\left(x-\frac{3}{2}\right) \\
\text { or, } \quad y & =5 x-7
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-266.jpg?height=876&width=656&top_left_y=337&top_left_x=1293)

## Solution 2

Let a general point on the line $y=5 x+3$ be represented by $(a, 5 a+3)$. Also, let a point on the required line be $M(x, y)$. Since $M(x, y)$ is the midpoint of $P Q$ then

(1) $x=\frac{a+3}{2}$ and

(2) $y=\frac{(5 a+3)+(-2)}{2}$

$$
y=\frac{5 a+1}{2}
$$

Solving (1) for $a$, we have $a=2 x-3$ and solving (2) for $a$, we have $\frac{2 y-1}{5}=a$.

Equating gives, $2 x-3=\frac{2 y-1}{5}$

$$
\begin{aligned}
10 x-15 & =2 y-1 \\
\text { or, } \quad y & =5 x-7
\end{aligned}
$$

ANSWER: (E)

## Part C