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Straight lines

Equation of a straight line

Suppose a straight line passes through the points $A(x_1,y_1)$ and $B(x_2,y_2)$ then we can determine the equation of the line through the following process:
  1. Determine the gradient, $m$, by using the formula $$ m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} $$
  2. Use the equation $y - y_\alpha = m(x-x_\alpha)$, (where $(x_\alpha,y_\alpha)$ is any point on the line) to determine the final equation: $$ \begin{align} y - y_\alpha &= m(x-x_\alpha)\\ y - y_1 &= m(x-x_1) \\ y &= mx +(y_1 - mx_1) \end{align} $$
  3. Rewrite the equation in whatever form is required in the question!
Note that when in the form $y = mx + c$, $m$ is the gradient of the line and $c$ is the $y$-intercept.

Equation of straight line 1

Determine the equation of the straight line passing through $A(4,7)$ and $B(7,14)$.

Express your answer in the form $ax + by + c= 0$, where $a,b,c$ are all integers

solution - press button to display
  1. $m = \frac{\Delta y}{\Delta x} = \frac{14-7}{7-4} = \frac{7}{3}$
  2. The equation of the line is therefore $$ y - 7 = \frac{7}{3}(x - 4) $$
  3. This isn't the form required by the question, so multiply both sides by the denominator, 3, to get $$3y - 21 = 7x - 28$$ Moving all terms to one side gives $$3y + (-7)x + 7 =0$$ So the coefficients are $a = 3$, $b = -7$, $c = 7$