Point-Gradient Form

Take the point $(3, -2)$ and any other point $(x, y)$ on the same line. Write the gradient between them using the gradient formula. Now multiply both sides by the denominator. What have you produced?

Take the known point $(x_1, y_1)$ and a general point $(x, y)$ on the line. The gradient between them must be $m$, so $\dfrac{y - y_1}{x - x_1} = m$. Multiplying up gives the form, and it is the gradient formula with nothing solved yet.

Suppose a line has gradient $m$ and passes through the known point $(x_1, y_1)$. Let $(x, y)$ be any other point on the same line.

Because both points are on the line, the gradient between them must be $m$. The gradient formula from Lesson 2 gives

$\dfrac{y - y_1}{x - x_1} = m$

Multiplying both sides by $x - x_1$ gives $y - y_1 = m(x - x_1)$, which is the form.

Nothing new has been introduced. This is the gradient formula, with one point left general and the fraction cleared. That is worth knowing, because a formula you can rebuild in two lines is one you cannot half-remember with a sign wrong.

Three steps: substitute, expand, tidy.

For gradient $4$ through $(3, -2)$, substituting gives $y - (-2) = 4(x - 3)$, which is $y + 2 = 4(x - 3)$.

Expanding gives $y + 2 = 4x - 12$, and subtracting $2$ gives $y = 4x - 14$.

If general form is wanted instead, collect to one side: $4x - y - 14 = 0$.

The middle step is where the marks go missing. A negative $y_1$ turns into $y - (-2)$, which is $y + 2$, and a negative $x_1$ does the same inside the bracket. Write the double negative down before you simplify it rather than doing both in your head.

The same answer is reachable through $y = mx + c$. Substitute the gradient and the point, solve for $c$, then write the equation out again with $c$ filled in.

For the same example: $-2 = 4(3) + c$, so $-2 = 12 + c$ and $c = -14$, giving $y = 4x - 14$. Identical answer, as it must be.

That route is perfectly valid and is often quicker for very simple numbers. Point–gradient form wins in three situations: when the gradient is a fraction, so that solving for $c$ involves fraction arithmetic; when general form is wanted, since the answer is already unexpanded; and when the point is itself an unknown or an expression, which happens as soon as questions become geometric.

Learn both and choose. A student who only knows the $y = mx + c$ route will still get there; a student who only knows point–gradient form will occasionally take the long way round on an easy question.

What happens if the known point is the y-intercept itself, $(0, c)$?

Substituting gives $y - c = m(x - 0)$, which is $y - c = mx$, and adding $c$ gives $y = mx + c$.

So gradient–intercept form is not a rival to point–gradient form; it is the special case of it where the chosen point happens to sit on the y-axis.

This is worth noticing because it makes the collection of formulas smaller. One idea, a point and a direction, generates both. Formulas that turn out to be the same formula are easier to keep hold of than formulas that must each be remembered separately.

Two special cases need naming before they appear in a question.

A horizontal line through $(x_1, y_1)$ has $m = 0$. Substituting gives $y - y_1 = 0(x - x_1) = 0$, so $y = y_1$. The formula handles it without any special treatment.

A vertical line through $(x_1, y_1)$ has an undefined gradient, so there is nothing to substitute and the formula cannot be used at all. Write the equation down directly instead: $x = x_1$.

The reasoning is the same as in Lesson 4. Every point on a vertical line shares an x-coordinate, and that single fact is the equation. A formula failing here is not a gap in the mathematics; it is the formula correctly declining to describe a line that has no gradient.

Five items. Give each answer in gradient–intercept form unless the item says otherwise.

1. 1 Find the line with gradient $3$ through $(2, 5)$.

   Show answer
   $y - 5 = 3(x - 2)$, then expand.$y = 3x - 1$

2. 2 Find the line with gradient $-1$ through $(4, 0)$.

   Show answer
   $y - 0 = -1(x - 4)$$y = -x + 4$

3. 3 Find the line with gradient $\tfrac{1}{2}$ through $(-6, 1)$.

   Show answer
   $y - 1 = \tfrac{1}{2}(x + 6)$, then expand.$y = \tfrac{1}{2}x + 4$

4. 4 Find the line with gradient $0$ through $(7, -3)$.

   Show answer
   A zero gradient makes the bracket vanish.$y = -3$

5. 5 Find the equation of the vertical line through $(-2, 9)$.

   Show answer
   No gradient exists, so the formula cannot be used. Every point shares the x-coordinate.$x = -2$