Two Points Are All You Need to Write the Equation of Any Straight Line

A straight line is fixed completely by one point and its slope, or by two points. From either, you can write an equation that every point on the line satisfies:

$$y-y_1=m(x-x_1)$$

This part covers the point-slope and two-point forms, parallel and perpendicular lines through a point, lines in a triangle, and intercepts and intersections.

**With a point $(x_1,y_1)$ and slope $m$, write $y-y_1=m(x-x_1)$; with two points, first find $m=\frac{y_2-y_1}{x_2-x_1}$ and then use either point.

Worked example 1 — point and slope.** Through $(2,-3)$ with slope $4$:

$$y+3=4(x-2)y=4x-114x-y-11=0$$

Worked example 2 — two points. Through $(-2,5)$ and $(3,-5)$:

$$m=\frac{-5-5}{3+2}=-2y-5=-2(x+2)2x+y-1=0$$

Check with $(3,-5)$: $6-5-1=0$.

Worked example 3 — inclination. Through $(0,4)$ making $135^{\circ }$ with the x-axis: $m=\tan 135^{\circ }=-1$, so $y=-x+4$.

An everyday example. **Suppose a gas cylinder has $12$ kg left after $2$ days and $8$ kg after $6$ days.** The slope is $\frac{8-12}{6-2}=-1$ kg per day, so $y-12=-1(x-2)$, giving $y=-x+14$: it started with $14$ kg.

The substance. Substitute the second point to check — it must satisfy the final equation.

Find the slope of the given line; use the same slope for a parallel line or the negative reciprocal for a perpendicular line, then apply the point-slope form.

Worked example 1 — parallel. Through $(4,1)$, parallel to $3x-2y+5=0$, which has slope $\frac{3}{2}$:

$$y-1=\frac{3}{2}(x-4)2y-2=3x-123x-2y-10=0$$

Worked example 2 — perpendicular. Through $(4,1)$, perpendicular to the same line, so slope $-\frac{2}{3}$:

$$y-1=-\frac{2}{3}(x-4)3y-3=-2x+82x+3y-11=0$$

Check: $2(4)+3(1)-11=0$.

Shortcut. A line parallel to $ax+by+c=0$ has the form $ax+by=k$; a perpendicular line has the form $bx-ay=k$. Substitute the point to find $k$.

An everyday example. A new service road built alongside a straight highway through a village follows a line parallel to the highway, passing through one fixed point.

The substance. Parallel lines differ only in the constant term, which is why the shortcut works.

A median joins a vertex to the mid-point of the opposite side, an altitude passes through a vertex perpendicular to the opposite side, and a perpendicular bisector passes through the mid-point of a side perpendicular to it.

Take the triangle $A(2,5)$, $B(-2,1)$, $C(4,-1)$.

**Median from $A$.** Mid-point of $BC$ is $M(1,0)$.

$$m_{AM}=\frac{5-0}{2-1}=5y=5(x-1)y=5x-5$$

**Altitude from $A$.** Slope of $BC=\frac{-1-1}{4+2}=-\frac{1}{3}$, so the altitude has slope $3$.

$$y-5=3(x-2)y=3x-1$$

**Perpendicular bisector of $BC$.** Through $M(1,0)$ with slope $3$:

$$y=3(x-1)y=3x-3$$

An everyday example. A street light meant to be equally far from two houses must stand on the perpendicular bisector of the line joining them.

The substance. Here the altitude and the perpendicular bisector are parallel but different; they coincide only when $AB=AC$.

**Put $y=0$ to get the x-intercept and $x=0$ to get the y-intercept; to find where two lines meet, solve their equations simultaneously.

Worked example 1 — intercepts.** For $3x+4y=12$:

$$y=0:x=4x=0:y=3$$

It cuts the axes at $(4,0)$ and $(0,3)$, enclosing a triangle of area $\frac{1}{2}\times 4\times 3=6$ square units.

Worked example 2 — intercepts. For $2x-5y+10=0$: x-intercept $-5$, y-intercept $2$.

Worked example 3 — intersection. Solve $2x+3y=13$ and $3x-y=3$.

$$y=3x-32x+9x-9=1311x=22x=2,y=3$$

The lines meet at $(2,3)$.

An everyday example. Two straight bus routes drawn on a city map cross at the junction given by solving their equations together.

The substance. **Dividing $3x+4y=12$ by $12$ gives $\frac{x}{4}+\frac{y}{3}=1$**, where the intercepts can be read directly.