Geometry and Measurement - Equations and Graphs of Straight Lines

Find the gradient of the line passing through the points $A(2, 5)$ and $B(4, 11)$.

Using the gradient formula: $m = \frac{11 - 5}{4 - 2}$ $m = \frac{6}{2}$ $m = 3$

Substitute the coordinates $(x_1, y_1) = (2, 5)$ and $(x_2, y_2) = (4, 11)$ into the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ to find the slope.

Identify the gradient and the $y$-intercept for the line given by the equation $2y - 6x = 10$.

First, rearrange the equation into the form $y = mx + c$: $2y = 6x + 10$ Divide every term by $2$: $y = 3x + 5$ Therefore, $m = 3$ and $c = 5$.

To identify $m$ and $c$, the equation must be isolated for $y$. The coefficient of $x$ is the gradient, and the constant term is the $y$-intercept.

Find the equation of a line that has a gradient of $-2$ and passes through the point $(0, 4)$.

The $y$-intercept $c$ is the value of $y$ when $x = 0$. Since the line passes through $(0, 4)$, $c = 4$. Given $m = -2$, substitute these into $y = mx + c$: $y = -2x + 4$

Because the point provided is the $y$-intercept, we can directly plug the gradient and the $y$-coordinate into the slope-intercept equation.

Determine if the lines $y = 4x + 7$ and $4x - y + 10 = 0$ are parallel.

Line 1: $y = 4x + 7$, so $m_1 = 4$. Line 2: Rearrange $4x - y + 10 = 0$ to $y = mx + c$: $-y = -4x - 10$ $y = 4x + 10$ So, $m_2 = 4$. Since $m_1 = m_2$, the lines are parallel.

Parallel lines must have identical gradients. By converting the second equation to slope-intercept form, we can compare its gradient to the first line.

1. Start with the given equation: $3x + 4y = 12$
2. Subtract $3x$ from both sides: $4y = -3x + 12$
3. Divide every term by $4$: $y = -\frac{3}{4}x + 3$
4. The gradient $m = -\frac{3}{4}$ and the $y$-intercept $c = 3$.

To find the gradient and intercept easily, rearrange the standard form $Ax+By=C$ into $y=mx+c$ by isolating $y$.

1. Identify the coordinates of the two points: $P(x_1, y_1) = (-2, -3)$ and $Q(x_2, y_2) = (2, 5)$.
2. Calculate the gradient ($m$): $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - (-3)}{2 - (-2)} = \frac{5 + 3}{2 + 2} = \frac{8}{4} = 2$
3. Use the slope-intercept form $y = mx + c$ and substitute one point, say $Q(2, 5)$, to find $c$: $5 = 2(2) + c$ $5 = 4 + c$ $c = 5 - 4 = 1$
4. Write the final equation: $y = 2x + 1$

To find the equation of a straight line, we first determine the gradient (steepness) by calculating the change in $y$ divided by the change in $x$. Once the gradient is known, we substitute the coordinates of any point on the line into the general linear equation to solve for the $y$-intercept ($c$).