Coordinate Geometry - Equation of a Straight Line (y = mx + c)

Find the gradient and y-intercept of the line with equation $3x + 2y = 8$.

$2y = -3x + 8 \Rightarrow y = -\frac{3}{2}x + 4$. Therefore, $m = -1.5$ and $c = 4$.

To find $m$ and $c$, rearrange the equation into the form $y = mx + c$ by isolating $y$ on one side.

Find the equation of the line passing through the points $A(2, 5)$ and $B(4, 9)$.

$m = \frac{9 - 5}{4 - 2} = \frac{4}{2} = 2$. Substitute into $y = mx + c$: $5 = 2(2) + c \Rightarrow 5 = 4 + c \Rightarrow c = 1$. Equation: $y = 2x + 1$.

First, calculate the gradient using the two-point formula. Then, substitute one of the points and the gradient into the general equation to solve for $c$.

Find the equation of a line parallel to $y = 3x - 5$ that passes through the point $(1, 7)$.

Parallel lines have the same gradient, so $m = 3$. Using $y - y_1 = m(x - x_1)$: $y - 7 = 3(x - 1) \Rightarrow y - 7 = 3x - 3 \Rightarrow y = 3x + 4$.

Identify that the gradient must be 3 because the lines are parallel. Then use the point-gradient formula or $y=mx+c$ to find the new y-intercept.

$m_1 = \frac{1}{2}$ $m_1 \times m_2 = -1 \implies \frac{1}{2} \times m_2 = -1 \implies m_2 = -2$ $y = mx + c \implies y = -2x + c$ $\text{Substitute } (0, -2): -2 = -2(0) + c \implies c = -2$ $y = -2x - 2$

First, identify the gradient of the given line ($m_1 = \frac{1}{2}$). Since perpendicular lines have gradients that multiply to $-1$, the new gradient is $m_2 = -2$. The point $(0, -2)$ provides the y-intercept ($c = -2$) directly.

$m = \frac{0 - 4}{2 - (-2)} = \frac{-4}{4} = -1$ $y - y_1 = m(x - x_1) \implies y - 0 = -1(x - 2)$ $y = -x + 2$ $\text{At x-axis, } y = 0: 0 = -x + 2 \implies x = 2$ $\text{X-intercept: } (2, 0)$

Calculate the gradient using the two points. Substitute the gradient and one point into the point-gradient form to find the equation. To find the x-intercept, set $y=0$ and solve for $x$.