The line from $G$ through the midpoint $M$ of $O H$ intersects the $y$-axis at $P(0,-4)$. What are the coordinates of $G$ ?
(A) $(12,3)$
(B) $(12,7)$
(C) $(12,5)$
(D) $(12,6)$
(E) $(12,4)$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-093.jpg?height=401&width=537&top_left_y=1144&top_left_x=1222)
Solution 1

Since $\angle O M P$ and $\angle G M H$ are opposite angles, then $\angle O M P=\angle G M H$.

The $x$-axis and the $y$-axis are perpendicular, so $\angle P O M=90^{\circ}$, so $\angle P O M=\angle G H M$.

Since $M$ is the midpoint of $O H$, then $O M=H M$.

Therefore, $\triangle P O M$ and $\triangle G H M$ are congruent by Angle-Side-Angle.

Thus, $G H=O P=4$.

Since $G H$ is perpendicular to the $x$-axis, then the $x$-coordinate of $G$ is 12 , so $G$ has coordinates $(12,4)$.

## Solution 2

Since $G H$ is perpendicular to the $x$-axis, then the $x$-coordinate of $G$ is 12 , so $G$ has coordinates $(12, g)$, for some value of $g$.

Since $O H=12$ and $M$ is the midpoint of $O H$, then $O M=\frac{1}{2}(12)=6$, so $M$ has coordinates $(6,0)$.

To get from $P$ to $M$, we move 6 units right and 4 units up.

To get from $M$ to $G$, we also move 6 units right. Since $P, M$ and $G$ lie on a line, then we must also move 4 units up to get from $M$ to $G$.

Therefore, the coordinates of $G$ are $(12,4)$.

ANswer: (E)