1. **Problem statement:** Given points $P(-4,2)$ and $Q(5,-4)$, a line $l$ is drawn through $P$ and perpendicular to the line segment $PQ$. This line $l$ meets the $y$-axis at point $R$. We need to find: a. The equation of line $l$ b. The coordinates of point $R$ c. The area of triangle $PQR$ 2. **Find the slope of line $PQ$:** The slope formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$. For $P(-4,2)$ and $Q(5,-4)$: $$m_{PQ} = \frac{-4 - 2}{5 - (-4)} = \frac{-6}{9} = -\frac{2}{3}$$ 3. **Find the slope of line $l$ (perpendicular to $PQ$):** The slope of a line perpendicular to another with slope $m$ is the negative reciprocal, so: $$m_l = -\frac{1}{m_{PQ}} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2}$$ 4. **Find the equation of line $l$ passing through $P(-4,2)$ with slope $\frac{3}{2}$:** Use point-slope form: $$y - y_1 = m(x - x_1)$$ $$y - 2 = \frac{3}{2}(x + 4)$$ Simplify: $$y - 2 = \frac{3}{2}x + 6$$ $$y = \frac{3}{2}x + 8$$ This is the equation of line $l$. 5. **Find the coordinates of point $R$ where line $l$ meets the $y$-axis:** On the $y$-axis, $x=0$, so substitute $x=0$ into the equation of $l$: $$y = \frac{3}{2}(0) + 8 = 8$$ Thus, $R = (0,8)$. 6. **Find the area of triangle $PQR$:** The vertices are $P(-4,2)$, $Q(5,-4)$, and $R(0,8)$. Use the formula for the area of a triangle given coordinates: $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ Substitute: $$= \frac{1}{2} | -4(-4 - 8) + 5(8 - 2) + 0(2 + 4) |$$ $$= \frac{1}{2} | -4(-12) + 5(6) + 0 |$$ $$= \frac{1}{2} | 48 + 30 | = \frac{1}{2} \times 78 = 39$$ **Final answers:** a. Equation of line $l$: $y = \frac{3}{2}x + 8$ b. Coordinates of $R$: $(0,8)$ c. Area of triangle $PQR$: $39$ square units