, 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 $(-3, 1)$.