The equation of a straight line with slope m through (x₁, y₁) is y − y₁ = m(x − x₁), or y = mx + c with y-intercept c. HKDSE answers usually use the general form Ax + By + C = 0, whose slope is −A/B. The slope through two points is m = (y₂ − y₁) / (x₂ − x₁).
m = (y₂ − y₁) / (x₂ − x₁)y − y₁ = m(x − x₁)(y − y₁) / (x − x₁) = (y₂ − y₁) / (x₂ − x₁)y = mx + cx/a + y/b = 1 (x-intercept a, y-intercept b)Ax + By + C = 0: slope −A/B, y-intercept −C/B, x-intercept −C/Am₁ = m₂ / m₁ × m₂ = −1|Ax₀ + By₀ + C| / √(A² + B²)y − y₁ = m(x − x₁).m = (y₂ − y₁) / (x₂ − x₁), then use point-slope form.y = mx + c.x/a + y/b = 1.Ax + By + C = 0 with integer A, B, C.Rearrange Ax + By + C = 0 to y = −(A/B)x − C/B. So the slope is −A/B, the y-intercept is −C/B and the x-intercept is −C/A. For example, 2x − 3y + 12 = 0 has slope 2/3, y-intercept 4 and x-intercept −6 (Example 2 below). A vertical line x = h has no slope; a horizontal line y = k has slope 0.
Scan this compact reference table to review every essential concept, step, and shortcut in under 10 seconds.
| Topic | Key Concept & Steps | Exam Shortcuts & Tips |
|---|---|---|
| 1. Slope (Gradient) | m = (y₂ − y₁) / (x₂ − x₁) = tan θ | • m > 0: Uphill (0° < θ < 90°) • m < 0: Downhill (90° < θ < 180°) • m = 0: Horizontal line Parallel: m₁ = m₂Perpendicular: m₁ × m₂ = −1 |
| 2. Slope-Intercept Form | y = mx + c• m = slope • c = y-intercept (where x = 0) | Use this form to instantly sketch a line. The line passes through (0, c) with steepness m. |
| 3. Point-Slope Form | y − y₁ = m(x − x₁)Passes through (x₁, y₁) with slope m. | Your Go-To Formula: If given two points, find m first, then plug either point into this formula. |
| 4. General Form | Ax + By + C = 0(A, B, C ∈ ℤ and A > 0) | Fast Coefficients Read (Paper 2): • Slope m = −A/B• y-intercept = −C/B• x-intercept = −C/A |
| 5. Line Intersections | Solve L₁: y = m₁x + c₁ and L₂: y = m₂x + c₂ simultaneously.• 1 solution: Intersecting ( m₁ ≠ m₂)• 0 solutions: Parallel ( m₁ = m₂, c₁ ≠ c₂)• ∞ solutions: Coincident ( m₁ = m₂, c₁ = c₂) | Use substitution or elimination. Plug solved x back into either equation to find y. |
Find the equation of the straight line passing through A(2, −1) and perpendicular to 3x − 4y + 5 = 0. Give the answer in general form.
= −A/B = −3/(−4) = 3/4.= −1 ÷ (3/4) = −4/3.y + 1 = −(4/3)(x − 2).3y + 3 = −4x + 8, so 4x + 3y − 5 = 0. Check: 4(2) + 3(−1) − 5 = 0 ✓Answer: 4x + 3y − 5 = 0
Find the slope, x-intercept and y-intercept of the line 2x − 3y + 12 = 0.
A = 2, B = −3, C = 12.= −A/B = −2/(−3) = 2/3.= −C/B = −12/(−3) = 4; x-intercept = −C/A = −12/2 = −6.(0, 4) and (−6, 0) both satisfy the equation ✓Answer: Slope 2/3, x-intercept −6, y-intercept 4.
m = −1 ⟹ θ = tan⁻¹(−1) = −45°). Rule: The inclination must be measured from the positive x-axis (0° ≤ θ < 180°). You must add 180° to get the correct obtuse angle: θ = −45° + 180° = 135°.3x + 4y − 12 = 0 (slope m₁ = −3/4), any line perpendicular to it must have the negative reciprocal slope: m₂ = −1/m₁ = +4/3. Use this to write perpendicular slopes instantly without working steps.m = 0 (equation: y = k). Vertical lines have an undefined slope (equation: x = h). Never write m = ∞ in your Paper 1 steps; simply state: "Since the line is vertical, its equation is x = h."(x₁, y₁) lies on a line Ax + By + C = 0, substitute x₁ and y₁ into the LHS. If the resulting value is exactly 0, the point lies on the line. If it is not 0, the point does not lie on the line.How to study: Click any question below to instantly load it on the coordinate grid and check your steps.
(1, 2) and B(4, 8) in General Form.2x + 5y − 10 = 0.y = 2x + 1 and x + y = 4.y = x + k does not cut the circle x² + y² − 4x = 0.DSE-style practice sets from the SketchMath question bank, with solutions for every MC and long question.