1. **State the problem:** Given triangle $\triangle PQR$ with vertices $P(1,3)$, $Q(4,1)$, and $R(6,y)$, and $RQ \perp PQ$, solve the following: 2.1 Find the gradient of $PQ$. 2.2 Interpret the meaning of the gradient from 2.1. 2.3 Find the length of $PQ$, rounded to 2 decimal places. 2.4 Calculate the value of $y$. 2.5 Find coordinates of $S$ such that $\triangle PQR \equiv \triangle SQR$. --- 2.1 **Gradient of $PQ$:** The gradient formula is $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ For $P(1,3)$ and $Q(4,1)$, $$m_{PQ} = \frac{1 - 3}{4 - 1} = \frac{-2}{3} = -\frac{2}{3}$$ 2.2 **Meaning of the gradient:** The gradient $-\frac{2}{3}$ means the line $PQ$ slopes downward, decreasing by 2 units vertically for every 3 units moved horizontally to the right. 2.3 **Length of $PQ$:** Use distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(4 - 1)^2 + (1 - 3)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \approx 3.61$$ 2.4 **Calculate $y$ using perpendicularity:** Since $RQ \perp PQ$, gradients satisfy $$m_{RQ} \times m_{PQ} = -1$$ We know $m_{PQ} = -\frac{2}{3}$, so $$m_{RQ} = \frac{3}{2}$$ Gradient of $RQ$ is $$m_{RQ} = \frac{y - 1}{6 - 4} = \frac{y - 1}{2}$$ Set equal: $$\frac{y - 1}{2} = \frac{3}{2} \implies y - 1 = 3 \implies y = 4$$ 2.5 **Coordinates of $S$ for congruence $\triangle PQR \equiv \triangle SQR$:** Since $S$ replaces $P$ in congruence, $S$ must be the reflection of $P$ across line $QR$. Line $QR$ passes through $Q(4,1)$ and $R(6,4)$. Gradient of $QR$ is $\frac{4-1}{6-4} = \frac{3}{2}$. Equation of $QR$: $$y - 1 = \frac{3}{2}(x - 4) \implies y = \frac{3}{2}x - 5$$ Reflect $P(1,3)$ about $QR$: The formula for reflection of point $(x_0,y_0)$ about line $Ax + By + C = 0$ is: $$x' = x_0 - \frac{2A(Ax_0 + By_0 + C)}{A^2 + B^2}, \quad y' = y_0 - \frac{2B(Ax_0 + By_0 + C)}{A^2 + B^2}$$ Rewrite line $QR$ in standard form: $$y = \frac{3}{2}x - 5 \implies \frac{3}{2}x - y - 5 = 0$$ Multiply by 2: $$3x - 2y - 10 = 0$$ So $A=3$, $B=-2$, $C=-10$. Calculate: $$D = A x_0 + B y_0 + C = 3(1) + (-2)(3) - 10 = 3 - 6 - 10 = -13$$ Denominator: $$A^2 + B^2 = 9 + 4 = 13$$ Coordinates of $S$: $$x' = 1 - \frac{2 \times 3 \times (-13)}{13} = 1 + \frac{78}{13} = 1 + 6 = 7$$ $$y' = 3 - \frac{2 \times (-2) \times (-13)}{13} = 3 - \frac{52}{13} = 3 - 4 = -1$$ Thus, $$S = (7, -1)$$ --- **Final answers:** - Gradient of $PQ$ is $-\frac{2}{3}$. - This means the line slopes downward 2 units for every 3 units right. - Length of $PQ$ is approximately $3.61$. - Value of $y$ is $4$. - Coordinates of $S$ are $(7, -1)$.