Geometry and Trigonometry - Coordinate geometry: distance, midpoint, and gradient of a line

Given two points $A(-2, 3)$ and $B(4, 11)$, calculate the gradient, the midpoint, and the distance between them.

1. Gradient ($m$): $m = \frac{11 - 3}{4 - (-2)} = \frac{8}{6} = \frac{4}{3}$.
2. Midpoint ($M$): $M = \left( \frac{-2 + 4}{2}, \frac{3 + 11}{2} \right) = \left( \frac{2}{2}, \frac{14}{2} \right) = (1, 7)$.
3. Distance ($d$): $d = \sqrt{(4 - (-2))^{2} + (11 - 3)^{2}} = \sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10$.

To solve this, we identify $x_1 = -2, y_1 = 3, x_2 = 4, y_2 = 11$ and substitute them into the standard formulas for coordinate geometry.

The gradient of a line connecting $P(1, 4)$ and $Q(5, k)$ is $2$. Find the value of $k$.

1. Use the gradient formula: $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$.
2. Substitute known values: $2 = \frac{k - 4}{5 - 1}$.
3. Simplify the denominator: $2 = \frac{k - 4}{4}$.
4. Multiply both sides by 4: $8 = k - 4$.
5. Solve for $k$: $k = 12$.

This problem requires rearranging the gradient formula to solve for an unknown coordinate component when the slope is already known.

$x_{1} = 2, y_{1} = -1, x_{2} = -4, y_{2} = 7$ $d = \sqrt{(-4 - 2)^{2} + (7 - (-1))^{2}}$ $d = \sqrt{(-6)^{2} + (8)^{2}}$ $d = \sqrt{36 + 64}$ $d = \sqrt{100}$ $d = 10$ units

Substitute the coordinates into the distance formula. Be careful with negative signs when calculating the differences; subtracting a negative results in addition.

Let $S = (x, y)$. $\frac{3 + x}{2} = 5 \implies 3 + x = 10 \implies x = 7$ $\frac{5 + y}{2} = 2 \implies 5 + y = 4 \implies y = -1$ $S = (7, -1)$

Use the midpoint formula in reverse. Set the average of the endpoints equal to the known midpoint coordinates and solve for the unknown variables $x$ and $y$.