Coordinate Geometry - Parallel and Perpendicular Lines

Find the equation of the line that passes through the point $(4, -2)$ and is parallel to the line $y = 3x + 7$.

$y = 3x - 14$

Since the lines are parallel, they share the same gradient. The gradient of the given line is $m = 3$. Using the point-gradient formula $y - y_1 = m(x - x_1)$ with point $(4, -2)$, we get: $y - (-2) = 3(x - 4) \Rightarrow y + 2 = 3x - 12 \Rightarrow y = 3x - 14$.

Find the equation of the line perpendicular to $2x - 5y = 10$ that passes through the point $(0, 3)$.

$y = -2.5x + 3$

First, find the gradient of the given line by rearranging to $y = mx + c$: $5y = 2x - 10 \Rightarrow y = \frac{2}{5}x - 2$. The gradient $m_1 = \frac{2}{5}$. The perpendicular gradient $m_2$ is the negative reciprocal: $m_2 = -\frac{5}{2} = -2.5$. Since the line passes through $(0, 3)$, the y-intercept $c = 3$. Thus, $y = -2.5x + 3$.

The line $L_1$ passes through $(1, 2)$ and $(3, 5)$. The line $L_2$ is perpendicular to $L_1$ and passes through $(5, 1)$. Determine where $L_2$ crosses the x-axis.

x = 5.75

1. Find gradient of $L_1$: $m_1 = \frac{5-2}{3-1} = \frac{3}{2}$. 2. Find gradient of $L_2$: $m_2 = -\frac{2}{3}$. 3. Equation of $L_2$: $y - 1 = -\frac{2}{3}(x - 5) \Rightarrow y = -\frac{2}{3}x + \frac{10}{3} + 1 \Rightarrow y = -\frac{2}{3}x + \frac{13}{3}$. 4. X-axis crossing (where $y=0$): $0 = -\frac{2}{3}x + \frac{13}{3} \Rightarrow \frac{2}{3}x = \frac{13}{3} \Rightarrow 2x = 13 \Rightarrow x = 6.5$.

1. Calculate the gradient ($m_1$) of line $L_1$ using $A(-2, 1)$ and $B(4, 4)$: $m_1 = \frac{4 - 1}{4 - (-2)} = \frac{3}{6} = 0.5$
2. Since $L_2$ is parallel to $L_1$, its gradient ($m_2$) is the same: $m_2 = 0.5$
3. Use the point-gradient form with $C(0, -2)$: $y - (-2) = 0.5(x - 0)$ $y + 2 = 0.5x$
4. Rearrange into the form $ax + by + c = 0$: $0.5x - y - 2 = 0$ Multiply by 2 to clear decimals: $x - 2y - 4 = 0$

Parallel lines share the same gradient. We first find the gradient of the reference line using two points, then apply that same gradient to the target point to form the new equation.

1. Find the midpoint $M$ of $PQ$: $M = \left( \frac{1 + 5}{2}, \frac{5 + (-3)}{2} \right) = (3, 1)$
2. Find the gradient ($m_{PQ}$) of line segment $PQ$: $m_{PQ} = \frac{-3 - 5}{5 - 1} = \frac{-8}{4} = -2$
3. Determine the perpendicular gradient ($m_{\perp}$): $m_{\perp} = -\frac{1}{-2} = 0.5$
4. Find the equation using $M(3, 1)$ and $m = 0.5$: $y - 1 = 0.5(x - 3)$ $y - 1 = 0.5x - 1.5$ $y = 0.5x - 0.5$ or $x - 2y - 1 = 0$

A perpendicular bisector must pass through the midpoint of the segment and have a gradient that is the negative reciprocal of the segment's gradient.