Gradient of a Straight Line
Formula · Interpretation · Special cases · §6.1
The gradient formula
The gradient m measures the steepness and direction of a line — how much y changes for each unit increase in x.
\(m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{\text{rise}}{\text{run}}\)
Choose any two points \((x_1,y_1)\) and \((x_2,y_2)\) on the line. The order doesn't matter — just stay consistent (top/bottom match left/right).
Watch signs: if the line falls from left to right, \(y_2 - y_1\) is negative → gradient is negative. Always write the subtraction in the same order for numerator and denominator.
Interpreting the gradient
m > 0Positive — line rises left to right
m < 0Negative — line falls left to right
m = 0Zero — horizontal line (y = constant)
m = ∞Undefined — vertical line (x = constant)
Find gradient of line through (−1, 3) and (4, −2)
Formula
\(m = \dfrac{-2-3}{4-(-1)} = \dfrac{-5}{5}\)
✓
\(m = -1\)
Interactive — Gradient Explorer
Drag the sliders to place two points and see the gradient calculated
Point A (xâ‚, yâ‚)
Point B (xâ‚‚, yâ‚‚)
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Equations of a Straight Line
Three forms · Finding the equation · Special lines · §6.2
Slope-intercept form
\(y = mx + c\)
m = gradient, c = y-intercept. Read both values directly off the equation. Most common form for sketching.
\(y = 3x - 2\): m = 3, c = −2
Point-gradient form
\(y - y_1 = m(x - x_1)\)
Use when you know a point \((x_1,y_1)\) and gradient m. Expand to get y = mx + c form.
m = 2, passes through (3,1):
\(y-1=2(x-3)\)
\(y-1=2(x-3)\)
General form
\(ax + by + c = 0\)
All terms on one side, often with integer coefficients. Common in exam answers. Rearrange to slope-intercept to read m and c.
\(2x - 3y + 6 = 0\)
→ \(y = \tfrac{2}{3}x + 2\)
→ \(y = \tfrac{2}{3}x + 2\)
Find equation of line through (2, 5) with gradient 3
Point-grad
\(y - 5 = 3(x - 2)\)
Expand
\(y - 5 = 3x - 6\)
✓
\(y = 3x - 1\)
Find equation of line through (−1, 4) and (3, −4)
Gradient
\(m = \dfrac{-4-4}{3-(-1)} = \dfrac{-8}{4} = -2\)
Point-grad
\(y - 4 = -2(x - (-1))\)
Expand
\(y - 4 = -2x - 2\)
✓
\(y = -2x + 2\)
Special lines
Horizontal line
\(y = k\) — passes through all points with y-coordinate k. Gradient = 0.
Vertical line
\(x = k\) — passes through all points with x-coordinate k. Gradient is undefined.
Line through origin
\(y = mx\) — y-intercept c = 0. The ratio y:x = m at every point.
Reading m and c: always rearrange to \(y = mx + c\) first. In \(2y = 6x + 4\), divide by 2: m = 3, c = 2 — not m = 6, c = 4.
Interactive — Line Equation Builder
Adjust m and c and see all three equation forms update live
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Parallel & Perpendicular Lines
Gradient conditions · Finding equations · Perpendicular bisector · §6.3
Parallel Lines
\(m_1 = m_2\)
Parallel lines have equal gradients and different y-intercepts. They never intersect.
Line \(y = 3x + 1\) is parallel to \(y = 3x - 5\). Both have m = 3.
Perpendicular Lines
\(m_1 \times m_2 = -1\)
Perpendicular lines have gradients that are negative reciprocals. They meet at a right angle.
If \(m_1 = 3\), then \(m_2 = -\tfrac{1}{3}\). Product: \(3 \times (-\tfrac{1}{3}) = -1\) ✓
Find equation of line parallel to \(y = 2x + 3\) through (1, 5)
Gradient
Parallel → same gradient: \(m = 2\)
Point-grad
\(y - 5 = 2(x - 1)\)
✓
\(y = 2x + 3\) — wait! Check: different y-intercept means it's a distinct line. \(y = 2x + 3\) passes through (1,5) since 2(1)+3=5. But that's the original line. So actually c = 5−2 = 3, giving y=2x+3 — same line! Try (1,7): y−7=2(x−1) → y=2x+5.
Find equation of line perpendicular to \(3x + y = 5\) through (3, 2)
Rearrange
\(y = -3x + 5\), so \(m_1 = -3\)
Perp grad
\(m_2 = -\dfrac{1}{m_1} = -\dfrac{1}{-3} = \dfrac{1}{3}\)
Point-grad
\(y - 2 = \tfrac{1}{3}(x - 3)\)
✓
\(y = \tfrac{1}{3}x + 1\)
Perpendicular bisector — method
The perpendicular bisector of a line segment AB is the line that:
- Passes through the midpoint M of AB
- Is perpendicular to AB (gradient = \(-1/m_{AB}\))
Key use: any point on the perpendicular bisector is equidistant from A and B. Used in locus problems.
Perp bisector of A(1, 2) and B(5, 8)
Midpoint
\(M = \left(\dfrac{1+5}{2}, \dfrac{2+8}{2}\right) = (3, 5)\)
Gradient AB
\(m_{AB} = \dfrac{8-2}{5-1} = \dfrac{6}{4} = \dfrac{3}{2}\)
Perp grad
\(m_\perp = -\dfrac{2}{3}\)
Equation
\(y - 5 = -\tfrac{2}{3}(x - 3)\)
✓
\(y = -\tfrac{2}{3}x + 7\)
Distance & Midpoint
Between two points · Area of shapes · §6.4
Midpoint
\(M = \left(\dfrac{x_1+x_2}{2},\, \dfrac{y_1+y_2}{2}\right)\)
Average the x-coordinates, average the y-coordinates. Works in all cases including negative coordinates.
Distance (Length)
\(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
Pythagoras in the coordinate plane. The horizontal and vertical separations form the two legs of a right triangle.
Area of Triangle
\(A = \tfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\)
Shoelace formula — given three vertices \((x_1,y_1)\), \((x_2,y_2)\), \((x_3,y_3)\). The absolute value ensures a positive area.
Find midpoint and distance between A(−2, 1) and B(4, 9)
Midpoint
\(M = \left(\dfrac{-2+4}{2}, \dfrac{1+9}{2}\right) = (1, 5)\)
Distance
\(|AB| = \sqrt{(4-(-2))^2 + (9-1)^2} = \sqrt{36 + 64} = \sqrt{100}\)
✓
\(M = (1, 5),\quad |AB| = 10\)
Area of triangle with vertices O(0,0), A(6,0), B(2,5)
Shoelace
\(A = \tfrac{1}{2}|0(0-5) + 6(5-0) + 2(0-0)|\)
Simplify
\(= \tfrac{1}{2}|0 + 30 + 0|\)
✓
\(= 15 \text{ units}^2\)
Interactive — Distance & Midpoint
A: xâ‚
-2
A: yâ‚
1
B: xâ‚‚
4
B: yâ‚‚
5
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Visual Guide — The Shoelace Method, Step by Step
Area of triangle with vertices A, B, C — click each stage to follow the diagonal products
Loading…
The formula
\[A = \tfrac{1}{2}\,|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\]
Equivalent "list" form — multiply coordinates along diagonals
Worked example — A(1,6), B(−2,1), C(4,0)
List the vertices and repeat the first at the bottom:
A: (1, 6)
B: (−2, 1)
C: (4, 0)
A: (1, 6) ↠repeat
A: (1, 6)
B: (−2, 1)
C: (4, 0)
A: (1, 6) ↠repeat
Intersections of Lines
Simultaneous equations · Special cases · Lines & curves · §6.5
Finding the intersection point — general method
Two lines intersect where they have the same (x, y). Set the equations equal and solve the simultaneous system.
- Write both equations in \(y = \ldots\) form (or use elimination)
- Set right-hand sides equal: \(m_1 x + c_1 = m_2 x + c_2\)
- Solve for x, then substitute back to find y
- Write answer as a coordinate pair
Special cases
If you get 0 = k (k ≠0): parallel lines — no intersection.
If you get 0 = 0: identical lines — infinitely many intersections.
If you get 0 = 0: identical lines — infinitely many intersections.
Find intersection of \(y = 2x - 1\) and \(3x + y = 14\)
Substitute
Sub \(y = 2x-1\) into \(3x + y = 14\):
\(3x + (2x-1) = 14\)
\(3x + (2x-1) = 14\)
Solve x
\(5x = 15 \Rightarrow x = 3\)
Solve y
\(y = 2(3) - 1 = 5\)
✓
Intersection: \((3,\ 5)\)
Line meets a curve
Substitute the line equation into the curve equation. The resulting equation's solutions give the x-coordinates of intersection points.
Find where \(y = x + 3\) meets \(y = x^2 - x + 1\)
Equate
\(x + 3 = x^2 - x + 1\)
Rearrange
\(x^2 - 2x - 2 = 0\)
Formula
\(x = \dfrac{2 \pm \sqrt{4+8}}{2} = 1 \pm \sqrt{3}\)
✓
\(x = 1+\sqrt{3}\) or \(x = 1-\sqrt{3}\)
Number of intersections — discriminant
When substituting a line into a quadratic gives \(ax^2+bx+c=0\), the discriminant \(\Delta = b^2 - 4ac\) tells you how many intersections exist:
Δ > 0Two distinct intersection points
Δ = 0One intersection — line is tangent to the curve
Δ < 0No real intersections — line misses the curve
Tangent condition: if asked "find k such that line is tangent", set Δ = 0 and solve for k.
Linear Law — Reducing to Straight Line Form
Transforming non-linear relationships · Reading constants off graphs · §6.6
Why we do this
Many real-world relationships are not linear — plotting y against x gives a curve, not a straight line. By substituting variables (often taking logs), we convert to a linear form \(Y = mX + c\), so we can find constants from a straight-line graph.
\(Y = mX + c\)
where X and Y are expressions in x and y
where X and Y are expressions in x and y
From the straight-line graph: read the gradient (= m) and y-intercept (= c), then reverse the substitution to find the original constants.
Strategy for any non-linear equation
- Identify the form of the equation (power law, exponential, product, etc.)
- Apply appropriate transformation (take logs, reciprocal, square both sides, etc.)
- Write in \(Y = mX + c\) form — label what X, Y, m, c represent
- From graph data: read gradient and intercept, back-substitute to find original constants
Common transformations — at a glance
| Original equation | Plot Y vs X | Gradient | Y-intercept |
|---|---|---|---|
| \(y = ax^n\) | \(\lg y\) vs \(\lg x\) | n | \(\lg a\) |
| \(y = Ab^x\) | \(\lg y\) vs \(x\) | \(\lg b\) | \(\lg A\) |
| \(y = Ae^{bx}\) | \(\ln y\) vs \(x\) | b | \(\ln A\) |
| \(y = a + bx^2\) | \(y\) vs \(x^2\) | b | a |
| \(xy = a + bx\) | \(xy\) vs \(x\) | b | a |
| \(\dfrac{1}{y} = a + \dfrac{b}{x}\) | \(\dfrac{1}{y}\) vs \(\dfrac{1}{x}\) | b | a |
The graph of \(\lg y\) against \(x\) passes through (0, 2) and (3, 5). Find A and b in \(y = Ab^x\).
Linear form
\(\lg y = x\lg b + \lg A\) → this is \(Y = mx + c\)
Y-intercept
At \(x=0\): \(\lg A = 2 \Rightarrow A = 10^2 = 100\)
Gradient
\(\lg b = \dfrac{5-2}{3-0} = 1 \Rightarrow b = 10^1 = 10\)
✓
\(A = 100,\quad b = 10\) so \(y = 100 \cdot 10^x\)
The relationship \(y = ax^n\). Graph of \(\lg y\) against \(\lg x\) has gradient \(2.5\) and passes through \((0, -0.5)\). Find a and n.
Identify
Gradient = n = 2.5
Intercept
\(\lg a = -0.5 \Rightarrow a = 10^{-0.5} = \dfrac{1}{\sqrt{10}} \approx 0.316\)
✓
\(n = 2.5,\quad a = 10^{-0.5} \approx 0.316\)
Practice Questions
3 Easy · 3 Medium · 3 Hard — Straight Lines & Linear Law
Easy 1
[2] Find the gradient and y-intercept of \(3x-2y+8=0\).
SOLUTION
Rearrange
\(y=\frac{3}{2}x+4\). Gradient \(\boldsymbol{=\frac32}\), y-intercept \(\boldsymbol{=4}\)
Easy 2
[3] Find the equation of the line through \((2,5)\) perpendicular to \(y=4x-1\).
SOLUTION
Perp. gradient
Original gradient \(=4\). Perpendicular gradient \(=-\tfrac14\)
Equation
\(y-5=-\tfrac14(x-2) \Rightarrow \boldsymbol{y=-\tfrac14x+\tfrac{11}{2}}\)
Easy 3
[2] Find the distance between \((1,-2)\) and \((4,2)\).
SOLUTION
Formula
\(d=\sqrt{(4-1)^2+(2-(-2))^2}=\sqrt{9+16}=\boldsymbol{5}\)
Medium 1
[4] The points \(A(1,3)\), \(B(5,1)\), \(C(7,5)\) form a triangle. Find the equation of the perpendicular bisector of \(AB\).
SOLUTION
Midpoint AB
\(M=\left(\frac{1+5}{2},\frac{3+1}{2}\right)=(3,2)\)
Gradient AB
\(m_{AB}=\frac{1-3}{5-1}=-\tfrac12\). Perp. gradient \(=2\)
Equation
\(y-2=2(x-3) \Rightarrow \boldsymbol{y=2x-4}\)
Medium 2
[4] Variables \(x\) and \(y\) are such that \(y=ax^n\). A graph of \(\ln y\) against \(\ln x\) is a straight line through \((0,\,2)\) with gradient 3. Find \(a\) and \(n\).
SOLUTION — linear law
Linearise
\(\ln y=n\ln x+\ln a\)
Read off
Gradient \(=n=\boldsymbol{3}\). y-intercept \(=\ln a=2\Rightarrow a=e^2\approx\boldsymbol{7.39}\)
Medium 3
[4] Find the area of the triangle formed by the line \(3x+4y=24\) and the coordinate axes.
SOLUTION
Intercepts
x-intercept (\(y=0\)): \(x=8\). y-intercept (\(x=0\)): \(y=6\).
Area
\(A=\tfrac12\times8\times6=\boldsymbol{24}\) units²
Hard 1
[5] \(A=(2,1)\), \(B=(6,9)\). \(P\) is a point on \(AB\) such that \(AP:PB=3:1\). Find the coordinates of \(P\).
SOLUTION — section formula
Section formula
\(P=\left(\frac{3(6)+1(2)}{3+1},\frac{3(9)+1(1)}{3+1}\right)=\left(\frac{20}{4},\frac{28}{4}\right)=\boldsymbol{(5,7)}\)
Hard 2
[5] The line \(l_1:y=2x+1\) and line \(l_2\) are perpendicular and intersect at \(P(1,3)\). \(l_2\) meets the x-axis at \(Q\). Find the area of triangle \(OPQ\) where \(O\) is the origin.
SOLUTION
\(l_2\) equation
Gradient of \(l_2=-\tfrac12\). Through \((1,3)\): \(y=-\tfrac12x+\tfrac72\)
Point Q
At \(y=0\): \(x=7\). So \(Q=(7,0)\).
Area OPQ
Using \(\tfrac12|x_O(y_P-y_Q)+x_P(y_Q-y_O)+x_Q(y_O-y_P)|\)
\(=\tfrac12|0(3-0)+1(0-0)+7(0-3)|=\tfrac12\cdot21=\boldsymbol{\tfrac{21}{2}}\)
\(=\tfrac12|0(3-0)+1(0-0)+7(0-3)|=\tfrac12\cdot21=\boldsymbol{\tfrac{21}{2}}\)
Hard 3
[6] \(y=px^2+q\) where \(p,q\) are constants. A straight line graph is drawn plotting \(y\) against \(x^2\). Given it passes through \((1,7)\) and \((4,19)\) in \((x^2,y)\) coordinates, find \(p\), \(q\), and the value of \(y\) when \(x=3\).
SOLUTION — linear law
Gradient
\(p=\frac{19-7}{4-1}=\frac{12}{3}=\boldsymbol{4}\)
y-intercept
\(7=4(1)+q \Rightarrow q=\boldsymbol{3}\)
y when x=3
\(y=4(9)+3=\boldsymbol{39}\)
Past Year Paper Questions
Cambridge 0606 — Straight Lines
0606
[10]
The points \(A(-1,4)\), \(B(3,2)\), \(C(5,k)\) are such that \(BC\perp AB\).
(a) Find the value of \(k\). [3]
(b) Find the equation of the perpendicular bisector of \(AC\). [4]
(c) This perpendicular bisector meets the y-axis at \(D\). Find the area of triangle \(ACD\). [3]
(a) Find the value of \(k\). [3]
(b) Find the equation of the perpendicular bisector of \(AC\). [4]
(c) This perpendicular bisector meets the y-axis at \(D\). Find the area of triangle \(ACD\). [3]
FULL SOLUTION
(a)
\(m_{AB}=\frac{2-4}{3-(-1)}=-\tfrac12\). Since \(BC\perp AB\): \(m_{BC}=2\)
\(m_{BC}=\frac{k-2}{5-3}=\frac{k-2}{2}=2 \Rightarrow k=\boldsymbol{6}\)
\(m_{BC}=\frac{k-2}{5-3}=\frac{k-2}{2}=2 \Rightarrow k=\boldsymbol{6}\)
(b)
Midpoint \(AC=\left(\frac{-1+5}{2},\frac{4+6}{2}\right)=(2,5)\)
\(m_{AC}=\frac{6-4}{5-(-1)}=\tfrac13\). Perp. gradient \(=-3\)
Equation: \(y-5=-3(x-2) \Rightarrow y=-3x+11\)
\(m_{AC}=\frac{6-4}{5-(-1)}=\tfrac13\). Perp. gradient \(=-3\)
Equation: \(y-5=-3(x-2) \Rightarrow y=-3x+11\)
(c)
D: set \(x=0\): \(y=11\), so \(D=(0,11)\).
Area \(=\tfrac12|(-1)(6-11)+5(11-4)+0(4-6)|=\tfrac12|5+35|=\boldsymbol{20}\)
Area \(=\tfrac12|(-1)(6-11)+5(11-4)+0(4-6)|=\tfrac12|5+35|=\boldsymbol{20}\)
0606 Style
Linear Law — Reducing to y = mx + c
[8]
Variables \(x\) and \(y\) are related by \(y = Ax^n\) where \(A\) and \(n\) are constants.
(a) Show that a graph of \(\lg y\) against \(\lg x\) gives a straight line. State the gradient and intercept. [3]
(b) The straight line passes through \((0.5, 2.1)\) and \((1.5, 3.3)\). Find \(A\) and \(n\). [5]
(a) Show that a graph of \(\lg y\) against \(\lg x\) gives a straight line. State the gradient and intercept. [3]
(b) The straight line passes through \((0.5, 2.1)\) and \((1.5, 3.3)\). Find \(A\) and \(n\). [5]
FULL WORKED SOLUTION
(a)
\(\lg y=\lg A+n\lg x\). Gradient \(=n\), \(y\)-intercept \(=\lg A\). ✓
(b) Gradient
\(n=\dfrac{3.3-2.1}{1.5-0.5}=\dfrac{1.2}{1.0}=\boldsymbol{1.2}\)
(b) Intercept
\(2.1=\lg A+1.2(0.5)\Rightarrow\lg A=2.1-0.6=1.5\Rightarrow A=10^{1.5}\approx\boldsymbol{31.6}\)
0606 Style
Midpoint, Gradient & Area
[7]
A(2,5), B(6,1), C(8,7) form a triangle.
(a) Find the coordinates of the centroid G. [2]
(b) Find the area of triangle ABC using the shoelace formula. [3]
(c) The median from A to the midpoint M of BC — find its equation. [2]
(a) Find the coordinates of the centroid G. [2]
(b) Find the area of triangle ABC using the shoelace formula. [3]
(c) The median from A to the midpoint M of BC — find its equation. [2]
FULL WORKED SOLUTION
(a)
Centroid \(=\left(\frac{2+6+8}{3},\frac{5+1+7}{3}\right)=\boldsymbol{\left(\frac{16}{3},\frac{13}{3}\right)}\)
(b)
\(\text{Area}=\frac{1}{2}|2(1-7)+6(7-5)+8(5-1)|=\frac{1}{2}|-12+12+32|=\boldsymbol{16\text{ units}^2}\)
(c)
M = midpoint of BC = \((7,4)\). Gradient AM \(=\frac{4-5}{7-2}=-\frac{1}{5}\).
Median: \(y-5=-\frac{1}{5}(x-2)\Rightarrow\boldsymbol{5y=27-x}\)
Median: \(y-5=-\frac{1}{5}(x-2)\Rightarrow\boldsymbol{5y=27-x}\)