Various Forms of the Equation of a Line

What "Equation of a Line" Means

The equation of a line $L$ is a condition on $(x, y)$ that is true exactly when the point lies on $L$. Every form below is the same idea — slope read between a fixed point and the moving point $(x, y)$.

Horizontal and vertical lines come first: a horizontal line at distance $a$ from the $x$-axis is $y = a$ or $y = -a$; a vertical line at distance $b$ from the $y$-axis is $x = b$ or $x = -b$. Through $(-2, 3)$: the horizontal line is $y = 3$ and the vertical line is $x = -2$.

Key Point (Point-slope form): The line through $(x_0, y_0)$ with slope $m$ is

$$y - y_0 = m(x - x_0)$$

Key Point (Two-point form): Through $(x_1, y_1)$ and $(x_2, y_2)$:

$$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$

For example: through $(-2, 3)$ with $m = -4$: $y - 3 = -4(x + 2)$, i.e. $4x + y + 5 = 0$; and through $(1, -1)$ and $(3, 5)$: $-3x + y + 4 = 0$.

Intercept Forms and the General Equation

Key Point (Slope-intercept forms): With slope $m$ and $y$-intercept $c$: $y = mx + c$. With slope $m$ and $x$-intercept $d$: $y = m(x - d)$. The sign of $c$ records whether the line meets the $y$-axis above or below the origin.

Key Point (Intercept form): Making intercepts $a$ and $b$ on the axes:

$$\frac{x}{a} + \frac{y}{b} = 1$$

For example: intercepts $-3$ and $2$ give $\frac{x}{-3} + \frac{y}{2} = 1$, i.e. $2x - 3y + 6 = 0$.

Every line — including vertical ones — fits the general equation $Ax + By + C = 0$ ($A, B$ not both zero), with slope $-\frac{A}{B}$ when $B \ne 0$.

Lines as models: "$L$ is a linear function of $C$" means a line through two data points — find it by two-point form, then evaluate.

[Board Tip] Present final answers in the form the question requests — "in intercept form" or "in general form." Converting at the end costs seconds; skipping the conversion costs the mark.

Solved Examples

Example 1: Point-slope

Find the equation of the line through $(-2, 3)$ with slope $-4$.

Solution:

Step 1 — Substitute into point-slope form. $y - 3 = -4(x - (-2)) = -4(x + 2)$.

Step 2 — Expand. $y - 3 = -4x - 8$, so $4x + y + 5 = 0$.

Step 3 — Check. At $(-2, 3)$: $-8 + 3 + 5 = 0$ ✓.

Takeaway: Substituting the given point back is a two-second full check on any line equation.

Example 2: Two-point

Write the equation of the line through $(1, -1)$ and $(3, 5)$.

Solution:

Step 1 — Slope. $m = \frac{5 - (-1)}{3 - 1} = \frac{6}{2} = 3$.

Step 2 — Point-slope from either point. $y + 1 = 3(x - 1)$.

Step 3 — Simplify. $y = 3x - 4$, i.e. $3x - y - 4 = 0$. Check with $(3, 5)$: $9 - 5 - 4 = 0$ ✓.

Takeaway: Two-point form is point-slope with the slope computed first — use whichever point makes the arithmetic lighter.

Example 3: Both intercept flavours

Write the equations of the lines with $\tan\theta = \frac{1}{2}$ and (i) $y$-intercept $-\frac{3}{2}$ (ii) $x$-intercept 4.

Solution:

Step 1 — (i) Slope-intercept. $y = \frac{1}{2}x - \frac{3}{2}$; multiply by 2: $2y - x + 3 = 0$.

Step 2 — (ii) x-intercept form. $y = \frac{1}{2}(x - 4)$; multiply by 2: $2y - x + 4 = 0$.

Takeaway: $c$ is where the line crosses the $y$-axis; $d$ is where it crosses the $x$-axis — same slope, different anchors.

Example 4: Intercept form

Find the equation of the line making intercepts $-3$ and $2$ on the axes.

Solution:

Step 1 — Substitute into intercept form. $\frac{x}{-3} + \frac{y}{2} = 1$.

Step 2 — Clear denominators. Multiply by 6: $-2x + 3y = 6$, i.e. $2x - 3y + 6 = 0$.

Step 3 — Check the intercepts. $y = 0$: $x = -3$ ✓; $x = 0$: $y = 2$ ✓.

Takeaway: Negative intercepts go straight into the form — the signs sort themselves out on clearing.

Example 5: From an inclination

Find the equation of the line through $\left(2, 2\sqrt{3}\right)$ inclined at $75°$ to the $x$-axis.

Solution:

Step 1 — Convert the angle to a slope. $m = \tan 75° = \tan(45° + 30°) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = 2 + \sqrt{3}$.

Step 2 — Point-slope. $y - 2\sqrt{3} = (2 + \sqrt{3})(x - 2)$.

Takeaway: Exact-angle slopes come from the tangent addition formula — memorise $\tan 75° = 2 + \sqrt{3}$ and $\tan 15° = 2 - \sqrt{3}$.

Example 6: Intercept data

(i) Line with $x$-intercept $-3$ and slope $-2$. (ii) Line meeting the $y$-axis 2 units above the origin at $30°$ to the positive $x$-axis.

Solution:

Step 1 — (i). Through $(-3, 0)$ with $m = -2$: $y = -2(x + 3)$, i.e. $2x + y + 6 = 0$.

Step 2 — (ii). $m = \tan 30° = \frac{1}{\sqrt{3}}$, $c = 2$: $y = \frac{x}{\sqrt{3}} + 2$; multiply by $\sqrt{3}$: $x - \sqrt{3}y + 2\sqrt{3} = 0$.

Takeaway: "2 units above the origin on the $y$-axis" is just $c = 2$ — translate words to parameters before writing anything.

Example 7: A median

The vertices of $\triangle PQR$ are $P(2, 1)$, $Q(-2, 3)$, $R(4, 5)$. Find the equation of the median through $R$.

Solution:

Step 1 — Find the opposite side's midpoint. Midpoint of $PQ$: $\left(\frac{2-2}{2}, \frac{1+3}{2}\right) = (0, 2)$.

Step 2 — Slope of the median. From $(0,2)$ to $R(4,5)$: $m = \frac{5-2}{4-0} = \frac{3}{4}$.

Step 3 — Write the line. $y - 2 = \frac{3}{4}x$, i.e. $3x - 4y + 8 = 0$. Check at $R$: $12 - 20 + 8 = 0$ ✓.

Takeaway: A median is just "line through two points" once the midpoint is computed.

Example 8: A perpendicular through a point

Find the line through $(-3, 5)$ perpendicular to the line through $(2, 5)$ and $(-3, 6)$.

Solution:

Step 1 — Slope of the given line. $\frac{6 - 5}{-3 - 2} = -\frac{1}{5}$.

Step 2 — Perpendicular slope. Negative reciprocal: $5$.

Step 3 — Point-slope. $y - 5 = 5(x + 3)$, i.e. $5x - y + 20 = 0$.

Takeaway: Flip and negate the slope, then it is one more point-slope application.

Example 9: Equal intercepts

Find the line with equal intercepts through $(2, 3)$.

Solution:

Step 1 — Impose $a = b$. $\frac{x}{a} + \frac{y}{a} = 1$, i.e. $x + y = a$.

Step 2 — Push through the point. $2 + 3 = a = 5$.

Step 3 — Write the line. $x + y = 5$.

Takeaway: Equal intercepts collapse the intercept form to $x + y = a$ — the point then reads off $a$ directly.

Example 10: Intercepts with a given sum

Find the line through $(2, 2)$ whose intercepts sum to 9.

Solution:

Step 1 — Parametrise. Intercepts $a$ and $9 - a$: $\frac{x}{a} + \frac{y}{9-a} = 1$.

Step 2 — Push through $(2, 2)$. $\frac{2}{a} + \frac{2}{9-a} = 1$; multiply by $a(9-a)$: $2(9-a) + 2a = a(9-a)$.

Step 3 — Solve the quadratic. $18 = 9a - a^2$, so $a^2 - 9a + 18 = 0$: $a = 3$ or $6$.

Step 4 — Write both lines. $a = 3$: $\frac{x}{3} + \frac{y}{6} = 1$, i.e. $2x + y = 6$; $a = 6$: $x + 2y = 6$.

Takeaway: A sum condition plus a point gives a quadratic in $a$ — expect (and report) two lines.

Example 11: Line from its perpendicular's foot

The perpendicular from the origin to a line meets it at $(-2, 9)$. Find the equation of the line.

Solution:

Step 1 — Slope of the perpendicular. From the origin to $(-2, 9)$: $-\frac{9}{2}$.

Step 2 — Slope of the line. Negative reciprocal: $\frac{2}{9}$.

Step 3 — Point-slope at the foot. $y - 9 = \frac{2}{9}(x + 2)$; multiply by 9: $2x - 9y + 85 = 0$.

Takeaway: The foot of the perpendicular is ON the line — it supplies both the point and (via the origin) the slope.

Example 12: The milk model

A store sells 980 litres weekly at Rs 14/litre and 1220 litres at Rs 16/litre. Assuming linearity, how much at Rs 17/litre?

Solution:

Step 1 — Slope from the two data points. $\frac{1220 - 980}{16 - 14} = \frac{240}{2} = 120$ litres per rupee.

Step 2 — Point-slope model. $D = 980 + 120(P - 14)$.

Step 3 — Evaluate. At $P = 17$: $D = 980 + 120 \times 3 = 1340$ litres.

Takeaway: "Linear relationship" = two-point form; the slope's units (litres/rupee) tell the story.

Example 13: Midpoints and ratios on a segment between axes

(i) $P(a, b)$ is the midpoint of a segment between the axes: show the line is $\frac{x}{a} + \frac{y}{b} = 2$. (ii) $R(h, k)$ divides such a segment in ratio $1 : 2$: find the line.

Solution:

Step 1 — (i) Name the endpoints. $(p, 0)$ and $(0, q)$: the midpoint is $\left(\frac{p}{2}, \frac{q}{2}\right) = (a, b)$, so $p = 2a$, $q = 2b$.

Step 2 — (i) Write the intercept form. $\frac{x}{2a} + \frac{y}{2b} = 1$, i.e. $\frac{x}{a} + \frac{y}{b} = 2$. ∎

Step 3 — (ii) Section formula. $R$ divides $(p,0)$ to $(0,q)$ in $1:2$: $h = \frac{2p}{3}$, $k = \frac{q}{3}$, so $p = \frac{3h}{2}$, $q = 3k$.

Step 4 — (ii) Intercept form. $\frac{x}{3h/2} + \frac{y}{3k} = 1$, i.e. $\frac{2x}{3h} + \frac{y}{3k} = 1$, or $2kx + hy = 3hk$.

Takeaway: Segments between the axes are begging for intercept form — name the endpoints $(p, 0)$, $(0, q)$ and use the section formula.