Equation of a line: two-point and intercept forms
Through two points, use the two-point form. If a line cuts the axes at (a, 0) and (0, b), use the intercept form, x/a + y/b = 1.
Where these forms come from
Two points fix a line. Their slope is (y₂ − y₁)/(x₂ − x₁). Put this slope into the point-slope form and rearrange: you get the two-point form. If a line cuts the x-axis at (a, 0) and the y-axis at (0, b), then a and b are its intercepts. Put these two points into the two-point form and you get the intercept form x/a + y/b = 1. A median of a triangle joins a vertex to the midpoint of the opposite side, so it is found with the two-point form.
Example 1: line through two points
Find the equation of the line through A(−1, 3) and B(5, −1).
- Formula: (y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁)
- (y − 3)/(−1 − 3) = (x − (−1))/(5 − (−1))
- (y − 3)/(−4) = (x + 1)/6
- Cross-multiply: 6(y − 3) = −4(x + 1)
- 6y − 18 = −4x − 4
- 4x + 6y − 14 = 0. Divide by 2: 2x + 3y − 7 = 0
- Check with B: 2(5) + 3(−1) − 7 = 10 − 3 − 7 = 0
2x + 3y − 7 = 0
Example 2: line from its intercepts
Find the equation of the line with x-intercept −3 and y-intercept 4.
- Formula: x/a + y/b = 1, with a = −3 and b = 4
- x/(−3) + y/4 = 1
- Multiply by 12 (the LCM of 3 and 4): −4x + 3y = 12
- 4x − 3y + 12 = 0
- Check: (−3, 0) gives −12 + 12 = 0, and (0, 4) gives −12 + 12 = 0
4x − 3y + 12 = 0
Example 3: a canal path
Take the town centre as the origin, the road going east as the x-axis and the road going north as the y-axis (1 unit = 1 km). A straight canal path cuts the east road 6 km from the centre and the north road 4 km from the centre. Find the equation of the path. Does a well at (3, 2) lie on it?
- Formula: x/a + y/b = 1
- Here a = 6 and b = 4
- x/6 + y/4 = 1
- Multiply by 12: 2x + 3y = 12
- For the well: 2(3) + 3(2) = 6 + 6 = 12
- The point fits the equation.
The path is 2x + 3y − 12 = 0, and the well at (3, 2) lies on it.
Town planners draw roads as straight lines on a map, with the town centre as the origin. The intercepts show where a new road will cross the main roads.
2 marks: “Find the equation of the line through (2, 5) and (4, 9).” (y − 5)/(9 − 5) = (x − 2)/(4 − 2), so (y − 5)/4 = (x − 2)/2. Then y − 5 = 2x − 4, giving **2x − y + 1 = 0**.
2 marks: “Find the equation of the line with x-intercept 3 and y-intercept −5.” x/3 + y/(−5) = 1. Multiply by 15: 5x − 3y = 15, so **5x − 3y − 15 = 0**.
5 marks: “Find the equation of the median through A of triangle ABC with A(2, 4), B(−2, 0) and C(4, −2).” Midpoint of BC: D = ((−2 + 4)/2, (0 − 2)/2) = (1, −1). Line through A(2, 4) and D(1, −1): (y − 4)/(−5) = (x − 2)/(−1), so y − 4 = 5(x − 2), giving **5x − y − 6 = 0**.
Two points? Two-point form. Two intercepts? x/a + y/b = 1, with a under x and b under y.