In $y=mx+c$: $m$ is the gradient (steepness), and $c$ is the $y$-intercept — the point $(0,c)$ where the line crosses the $y$-axis (since every point on the $y$-axis has $x=0$). Everything in this topic comes down to finding these two values.

Rearrange to isolate $y$. Example: find $m,c$ for $3x+2y-6=0$.

1. $2y=-3x+6$
2. $y=-\tfrac32x+3$

So $m=-\tfrac32$, $c=3$.

1. Read $c$ directly: where the line crosses the $y$-axis.
2. Pick two clear grid-intersection points and calculate $m=\dfrac{\text{rise}}{\text{run}}$, checking the sign against the direction of slope.
3. Substitute $m$ and $c$ into $y=mx+c$.

If the $y$-intercept isn't visible or isn't an integer, find $m$ from two other clear points first, then substitute one of those points into $y=mx+c$ to solve for $c$ algebraically.

Gradient + one point: substitute the gradient for $m$, then substitute the point's coordinates to solve for $c$.

Two points: first calculate $m=\dfrac{y_2-y_1}{x_2-x_1}$, then proceed exactly as above using either point.

Example: Line through $(-1,6)$ and $(5,-4)$.

1. $m=\dfrac{-4-6}{5-(-1)}=\dfrac{-10}{6}=-\dfrac53$
2. Substitute $(5,-4)$: $-4=-\dfrac53(5)+c \Rightarrow -4=-\dfrac{25}{3}+c \Rightarrow c=\dfrac{13}{3}$

Equation: $y=-\dfrac53x+\dfrac{13}{3}$.

• Horizontal line: $y=c$ (gradient $0$).
• Vertical line: $x=k$ (gradient undefined — a mnemonic: Horizontal → 0 → Y=c; Vertical → Undefined → X=k).
• Parallel lines: same gradient.
• Perpendicular lines: gradients multiply to $-1$ (negative reciprocal).

Worked Example

Points $B(4,5)$ and $D(-2,-1)$ define segment $BD$. Find the equation of its perpendicular bisector, $N$.

1. Gradient of $BD$: $m_{BD}=\dfrac{5-(-1)}{4-(-2)}=\dfrac{6}{6}=1$.
2. Perpendicular gradient: $m_N=-1$.
3. Midpoint of $BD$: $\left(\dfrac{4+(-2)}{2},\dfrac{5+(-1)}{2}\right)=(1,2)$.
4. Substitute into $y=-x+c$: $2=-1+c \Rightarrow c=3$.

Equation of $N$: $y=-x+3$.

Substitute both of your original coordinate points back into your final equation — each one should make the equation balance. This simple habit catches almost every slip in the gradient or the $c$-value before you move on, and takes only a few seconds compared to the marks it protects. It's especially worth doing on multi-step problems (like the perpendicular bisector above), where an error in an early step — the gradient of $BD$, say — would otherwise silently propagate through the rest of the working undetected.