Functions - Lines (Linear functions)

Find the equation of the line passing through the points $A(1, 4)$ and $B(3, 10)$ in the form $y = mx + c$.

First, find the gradient $m$: $m = \frac{10 - 4}{3 - 1} = \frac{6}{2} = 3$ Now, use the point-gradient form with point $A(1, 4)$: $y - 4 = 3(x - 1)$ $y - 4 = 3x - 3$ $y = 3x + 1$

To find the linear equation, we first determine the slope using the two-point formula and then substitute one point into the point-gradient equation to solve for $y$.

Line $L_1$ has the equation $y = 2x + 5$. Line $L_2$ is perpendicular to $L_1$ and passes through the point $(4, 1)$. Find the equation of $L_2$.

The gradient of $L_1$ is $m_1 = 2$. Since $L_2 \perp L_1$, the gradient of $L_2$ is: $m_2 = -\frac{1}{m_1} = -\frac{1}{2}$ Using the point $(4, 1)$ in the point-gradient form: $y - 1 = -\frac{1}{2}(x - 4)$ $y - 1 = -\frac{1}{2}x + 2$ $y = -\frac{1}{2}x + 3$

Perpendicular lines have gradients that multiply to $-1$. Once the new gradient is found, we use the given point to construct the specific linear equation.

Find the distance between the points $P(-1, 2)$ and $Q(3, 5)$.

Using the distance formula: $d = \sqrt{(3 - (-1))^2 + (5 - 2)^2}$ $d = \sqrt{(4)^2 + (3)^2}$ $d = \sqrt{16 + 9}$ $d = \sqrt{25} = 5$

The distance formula applies the Pythagorean theorem to the horizontal and vertical differences between two points.

1. Identify coordinates: $(x_1, y_1) = (-2, -1)$ and $(x_2, y_2) = (4, 3)$.
2. Use the midpoint formula: $M = \left( \frac{-2 + 4}{2}, \frac{-1 + 3}{2} \right)$
3. Calculate the values: $M = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1)$ The midpoint is $M(1, 1)$.

The midpoint is found by averaging the x-coordinates and y-coordinates of the endpoints.

1. Start with the gradient-intercept form $y = mx + c$. Since the line passes through $(0, -2)$, the $y$-intercept $c = -2$.
2. The equation is: $y = \frac{1}{2}x - 2$
3. To find the $x$-intercept, set $y = 0$: $0 = \frac{1}{2}x - 2$
4. Solve for $x$: $2 = \frac{1}{2}x$ $x = 4$ The $x$-intercept is $(4, 0)$.

By substituting the gradient and y-intercept into the slope-intercept form, we can then solve for the horizontal root by setting the vertical value to zero.