, 2023

Since $Q(5,3)$ is the midpoint
of $P(1,p)$ and $R(r,5)$, then $\dfrac{1+r}{2} = 5$ and $\dfrac{p + 5}{2} = 3$.

Thus, $1+r = 10$ which gives $r = 9$, and $p
+ 5 = 6$which gives$p =
1$.

Therefore, $p=1$ and $r=9$.
Solution 1

The point with coordinates $P(3,6)$ is 6 units above the $x$-axis.

A line with slope 3 moves 2 units to the right as it moves 6 units up.
Therefore, to move from $P(3,6)$ to
the $x$-axis along a line with slope
3 results in a move of 6 units down and $2$ units left. Thus, its $x$-intercept is $3 - 2=1$.

A line with slope $-1$ moves 6 units
to the left as it moves 6 units up. Therefore, to move from $P(3,6)$ to the $x$-axis along a line with slope $-1$ results in a move of 6 units down and
6 units right. Thus, its $x$-intercept is $3+6=9$.

The distance between these $x$-intercepts is $9 - 1 = 8$.

Solution 2

The line with slope $3$ that
passes through $P(3,6)$ has equation
$y - 6 = 3(x - 3)$ or $y = 3x - 3$.

The $x$-intercept of this line has
$y = 0$ and so $0 = 3x - 3$ or $3x = 3$, which gives $x = 1$.

The line with slope $-1$ that passes
through $P(3,6)$ has equation $y - 6 = (-1)(x - 3)$ or $y = -x + 9$.

The $x$-intercept of this line has
$y = 0$ and so $0 = -x + 9$ or $x = 9$.

The distance between these $x$-intercepts is $9 - 1 = 8$.
The line with equation $y = 2x +
7$ has slope 2.

The line with equation $y = tx + t$
has slope $t$.

Since these lines are perpendicular, the product of their slopes is
$-1$ and so $2t = -1$ which gives $t = -\frac{1}{2}$.

We now need to find the point of intersection of the lines with
equations $y = 2x + 7$ and $y = -\frac{1}{2}x - \frac{1}{2}$.

Equating expressions for $y$, we
obtain $2x + 7 = $-\frac{1}{2}x - \frac{1}{2}$$or$$\frac{5}{2}x = -\frac{15}{2}$$, which gives$x =
-3$.

Therefore, $y = 2x + 7 = 2(-3) + 7 =
1, and so the point of intersection of these lines is (-3, 1)$.