Okay, scoddyevarau87@gmail.com, let's work through these coordinate geometry problems. Problem 33: Plot (3,4) and (-2,5) on the Cartesian plane. To plot these points: • For point $(3,4)$, start at the origin $(0,0)$, move 3 units to the right along the x-axis, then move 4 units up parallel to the y-axis. Mark this point. • For point $(-2,5)$, start at the origin $(0,0)$, move 2 units to the left along the x-axis, then move 5 units up parallel to the y-axis. Mark this point. Problem 34: Find the distance between (0,0) and (6,8). Step 1: Write the distance formula. The distance $D$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $$ D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} $$ Step 2: Substitute the given coordinates. Let $(x_1, y_1) = (0,0)$ and $(x_2, y_2) = (6,8)$. $$ D = \sqrt{(6-0)^2 + (8-0)^2} $$ Step 3: Calculate the distance. $$ D = \sqrt{6^2 + 8^2} $$ $$ D = \sqrt{36 + 64} $$ $$ D = \sqrt{100} $$ $$ D = \boxed{10} $$ Problem 35: Find the midpoint of the line joining (2,3) and (6,7). Step 1: Write the midpoint formula. The midpoint $M$ of a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $$ M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) $$ Step 2: Substitute the given coordinates. Let $(x_1, y_1) = (2,3)$ and $(x_2, y_2) = (6,7)$. $$ M = \left(\frac{2+6}{2}, \frac{3+7}{2}\right) $$ Step 3: Calculate the midpoint coordinates. $$ M = \left(\frac{8}{2}, \frac{10}{2}\right) $$ $$ M = \boxed{(4,5)} $$ Problem 36: Find the gradient of line AB if A is (2,3) and B is (4,1). Step 1: Write the gradient formula. The gradient (slope) $m$ of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $$ m = \frac{y_2-y_1}{x_2-x_1} $$ Step 2: Substitute the given coordinates. Let $(x_1, y_1) = (2,3)$ and $(x_2, y_2) = (4,1)$. $$ m = \frac{1-3}{4-2} $$ Step 3: Calculate the gradient. $$ m = \frac{-2}{2} $$ $$ m = \boxed{-1} $$ Problem 37: Find the equation of the Line if gradient/slope = 2 and it passes through (2,5). Step 1: Use the point-slope form of a linear equation. The point-slope form is $y - y_1 = m(x - x_1)$, where $m$ is the gradient and $(x_1, y_1)$ is a point on the line. Step 2: Substitute the given gradient and point. Given $m = 2$ and $(x_1, y_1) = (2,5)$. $$ y - 5 = 2(x - 2) $$ Step 3: Simplify the equation to the slope-intercept form ($y = mx + c$). $$ y - 5 = 2x - 4 $$ $$ y = 2x - 4 + 5 $$ $$ y = \boxed{2x + 1} $$ Problem 38: The vertices of a triangle are (1,1), (4,1) and (2,5). Find its area. Step 1: Identify the base and height of the triangle. Notice that two vertices, $(1,1)$ and $(4,1)$, have the same y-coordinate. This means the side connecting these two points is a horizontal base. The length of the base ($b$) is the absolute difference of their x-coordinates: $$ b = |4 - 1| = 3 \text{ units} $$ The height ($h$) of the triangle is the perpendicular distance from the third vertex $(2,5)$ to this base. This is the absolute difference between the y-coordinate of the third vertex and the y-coordinate of the base: $$ h = |5 - 1| = 4 \text{ units} $$ Step 2: Use the formula for the area of a triangle. $$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$ Step 3: Substitute the values and calculate the area. $$ \text{Area} = \frac{1}{2} \times 3 \times 4 $$ $$ \text{Area} = \frac{1}{2} \times 12 $$ $$ \text{Area} = \boxed{6 \text{ square units}} $$ Last free one today — make it count tomorrow, or type /upgrade for unlimited.