Straight Lines

In this chapter: distance, section and area formulas (recap), inclination and slope, parallel and perpendicular lines, angle between two lines, collinearity, the standard forms of the equation of a line, the general equation, distance of a point from a line and distance between parallel lines.

Quick recap from Class 10

If the area comes out as zero, the three points are collinear.

Slope of a line

The inclination θ of a line is the angle it makes with the positive direction of the x-axis, measured anticlockwise, with 0° ≤ θ < 180°. The slope (or gradient) is

The x-axis and all horizontal lines have slope 0. The y-axis and all vertical lines have undefined slope. A line rising from left to right has positive slope (acute θ); a falling line has negative slope (obtuse θ).

Parallel and perpendicular lines

• Two non-vertical lines are parallel if and only if m₁ = m₂.
• Two non-vertical lines are perpendicular if and only if m₁m₂ = −1, that is, each slope is the negative reciprocal of the other.

Three points A, B, C are collinear if slope of AB = slope of BC. This is often faster than the area formula.

Angle between two lines

If two non-vertical lines have slopes m₁ and m₂, with 1 + m₁m₂ ≠ 0, the acute angle θ between them satisfies

The obtuse angle between the lines is 180° − θ. If 1 + m₁m₂ = 0 the lines are perpendicular.

Worked example: Find the acute angle between two lines with slopes 2 and −3.
Solution: tan θ = |(−3 − 2)/(1 + 2(−3))| = |−5/−5| = 1, so θ = 45°. The obtuse angle between them is 135°.

Forms of the equation of a line

Every one of these can be rearranged into the general form Ax + By + C = 0, with A and B not both zero. From the general form, when B ≠ 0 the slope is −A/B and the y-intercept is −C/B; when A ≠ 0 the x-intercept is −C/A.

Worked example: (a) Find the equation of the line through (−2, 3) with slope −4. (b) Find the line whose x- and y-intercepts are 3 and −2.
Solution: (a) Point-slope form: y − 3 = −4(x + 2), so y = −4x − 5, or 4x + y + 5 = 0. (b) Intercept form: x/3 + y/(−2) = 1. Multiplying by 6: 2x − 3y = 6.

A useful shortcut: a line parallel to Ax + By + C = 0 has the form Ax + By + k = 0, and a line perpendicular to it has the form Bx − Ay + k = 0. Find k from the given point.

Distance of a point from a line

The perpendicular distance of P(x₁, y₁) from the line Ax + By + C = 0 is

For two parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0, the distance between them is

Before using the second formula, make the coefficients of x and y identical in both equations. For 3x − 4y + 7 = 0 and 6x − 8y + 5 = 0, first rewrite the second as 3x − 4y + 5/2 = 0.

Worked example: Find the distance of the point (3, −5) from the line 3x − 4y − 26 = 0.
Solution: d = |3(3) − 4(−5) − 26| / √(9 + 16) = |9 + 20 − 26| / 5 = 3/5.

Common mistakes: (1) Writing the slope as (x₂ − x₁)/(y₂ − y₁). (2) Reading the slope of 2x − 3y + 5 = 0 as 2 or −2/3; it is −A/B = 2/3. (3) Dropping the modulus in the angle formula and reporting an obtuse angle as the acute one. (4) Using the parallel-lines distance formula when the x and y coefficients are not the same in both equations. (5) Forgetting that vertical lines have no slope, so m₁m₂ = −1 cannot be used for a vertical and a horizontal line (they are perpendicular anyway). (6) Using the intercept form for a line through the origin, where both intercepts are zero.

Exam focus

• Slope from two points or from the general equation; conditions for parallel and perpendicular lines.
• Acute angle between two lines, and finding a slope when the angle is given.
• Choosing the right form of the equation from the given data.
• Distance of a point from a line and between parallel lines; foot of the perpendicular.
• Collinearity, area of a triangle, and section formula problems.

Practice questions