y = mx + c, Gradients, Midpoints & Parallel and Perpendicular Lines
Every non-vertical straight line on a coordinate grid can be written in the form y = mx + c, where m is the gradient (rate of change of y with respect to x) and c is the y-intercept (where the line crosses the y-axis at (0, c)). Parallel lines always have the exact same gradient (m₁ = m₂), whereas on the Higher Tier, two perpendicular lines intersecting at 90° have gradients that are negative reciprocals of each other (m₁ x m₂ = -1, or m₂ = -1 / m₁). Examiners frequently combine perpendicular gradients with midpoints to find the equation of a perpendicular bisector, or with circle geometry (x² + y² = r²) to find the equation of a tangent to a circle at a given point.
- Gradient formula between two points
- Given two points (x₁, y₁) and (x₂, y₂), the gradient m is the change in y divided by the change in x: m = (y₂ - y₁) / (x₂ - x₁). Lines that slope upwards from left to right have a positive gradient; lines that slope downwards from left to right have a negative gradient; horizontal lines (y = k) have gradient 0; vertical lines (x = k) have an undefined gradient.
- Equation of a Straight Line (y = mx + c)
- In y = mx + c, m is the gradient and c is the y-intercept (the line crosses the y-axis at (0, c)). To find where the line crosses the x-axis (the x-intercept), substitute y = 0 into the equation and solve for x. Alternative point-gradient formula: y - y₁ = m(x - x₁).
- Midpoint & Length of a Line Segment
- Given endpoints A(x₁, y₁) and B(x₂, y₂): 1. Midpoint M is the mean of the coordinates: M = ((x₁ + x₂) / 2, (y₁ + y₂) / 2). 2. Length (distance) of AB is found using Pythagoras' Theorem on the horizontal and vertical changes: Length AB = √((x₂ - x₁)² + (y₂ - y₁)²).
Quadratic Parabola Plotter, Roots, Turning Point & Quadratic Formula
Adjust coefficients a, b, and c in y = ax² + bx + c to see how the parabola, roots (x-intercepts), and turning point move on the graph.
y = x² - 2x - 3
(-2)² − 4(1)(-3) = 16 → 2 real roots (crosses x-axis twice)x = 1.00b in brackets when squaring: (-2)² = 4 (always positive!). Typing -2² into a calculator without brackets gives -4 and loses the accuracy mark.Key Concepts & Exam Definitions (6)
- Gradient formula between two points
- Given two points (x₁, y₁) and (x₂, y₂), the gradient m is the change in y divided by the change in x: m = (y₂ - y₁) / (x₂ - x₁). Lines that slope upwards from left to right have a positive gradient; lines that slope downwards from left to right have a negative gradient; horizontal lines (y = k) have gradient 0; vertical lines (x = k) have an undefined gradient.
- Equation of a Straight Line (y = mx + c)
- In y = mx + c, m is the gradient and c is the y-intercept (the line crosses the y-axis at (0, c)). To find where the line crosses the x-axis (the x-intercept), substitute y = 0 into the equation and solve for x. Alternative point-gradient formula: y - y₁ = m(x - x₁).
- Midpoint & Length of a Line Segment
- Given endpoints A(x₁, y₁) and B(x₂, y₂): 1. Midpoint M is the mean of the coordinates: M = ((x₁ + x₂) / 2, (y₁ + y₂) / 2). 2. Length (distance) of AB is found using Pythagoras' Theorem on the horizontal and vertical changes: Length AB = √((x₂ - x₁)² + (y₂ - y₁)²).
- Parallel Lines (m₁ = m₂)
- Two straight lines are parallel if and only if they have the exact same gradient (m₁ = m₂). To prove two lines are parallel, rearrange both equations into the form y = mx + c and state that their gradients are equal.
- Perpendicular Lines (m₁ x m₂ = -1) (Higher Tier)
- Two lines are perpendicular (meet at a right angle, 90°) if the product of their gradients is -1: m₁ x m₂ = -1, meaning m₂ = -1 / m₁ (the negative reciprocal). For example, the perpendicular to y = (4/3)x + 1 has gradient m = -3/4.
- Equation of a Tangent to a Circle x² + y² = r² (Higher Tier)
- A circle with centre (0, 0) and radius r has equation x² + y² = r². To find the equation of the tangent at point P(x₁, y₁) on the circle: 1. Calculate the gradient of the radius OP from (0, 0) to (x₁, y₁): m_OP = (y₁ - 0) / (x₁ - 0) = y₁ / x₁. 2. Since the tangent is perpendicular to the radius, m_tangent = -x₁ / y₁. 3. Substitute (x₁, y₁) into y = m_tangent(x) + c to find c.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
State the formulas for the gradient, midpoint, and length of the line segment joining (x₁, y₁) and (x₂, y₂).
Find the gradient and y-intercept coordinates of the line 3x + 4y - 20 = 0.
Write down the negative reciprocal (perpendicular gradient) for each gradient: (a) m = 5, (b) m = -1/2, (c) m = 3/7.
Where does the line 2y - 5x = 10 cross the x-axis and the y-axis?
What are the 4 steps to find the equation of the perpendicular bisector of AB?
What circle theorem links the radius of a circle and the tangent at a point P on the circumference?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Point A has coordinates (-2, 7) and Point B has coordinates (4, -5). Find the equation of the perpendicular bisector of the line segment AB, giving your answer in the form y = mx + c. (Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Find the midpoint M of AB: M = ((-2 + 4)/2, (7 + (-5))/2) = (2/2, 2/2) = (1, 1) [1 mark]. 2. Find the gradient of the line segment AB: m_AB = (-5 - 7) / (4 - (-2)) = -12 / 6 = -2 [1 mark]. 3. Find the gradient of the perpendicular bisector by taking the negative reciprocal of -2: m_perp = +1/2 (or 0.5) [1 mark]. 4. Substitute m = 1/2 and the midpoint (1, 1) into y = mx + c: 1 = (1/2)(1) + c -> c = 1/2, so the equation is y = (1/2)x + 1/2 (or y = 0.5x + 0.5) [1 mark].
- Q2
A circle has the equation x² + y² = 25. Point P(3, -4) lies on the circumference of the circle. Find the equation of the tangent to the circle at point P, giving your answer in the form ax + by + c = 0 where a, b, and c are integers. (Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Find the gradient of the radius OP from the centre (0, 0) to P(3, -4): m_radius = (-4 - 0) / (3 - 0) = -4/3 [1 mark]. 2. Since the tangent to a circle is perpendicular to the radius at the point of contact, take the negative reciprocal to find the gradient of the tangent: m_tangent = +3/4 [1 mark]. 3. Substitute m = 3/4 and x = 3, y = -4 into y = mx + c: -4 = (3/4)(3) + c -> -4 = 9/4 + c -> c = -16/4 - 9/4 = -25/4, so y = (3/4)x - 25/4 [1 mark]. 4. Multiply every term by 4 and rearrange into ax + by + c = 0: 4y = 3x - 25 -> 3x - 4y - 25 = 0 [1 mark].
- Q3
Line L₁ has the equation 2y + 3x = 11. Line L₂ passes through the points (1, -4) and (7, -13). Determine algebraically whether L₁ and L₂ are parallel, perpendicular, or neither. You must show your working clearly. [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Rearrange the equation of Line L₁ into the form y = mx + c to find its gradient: 2y = -3x + 11 -> y = -(3/2)x + 5.5, so gradient m₁ = -3/2 (or -1.5) [1 mark]. 2. Calculate the gradient of Line L₂ using the two given points (1, -4) and (7, -13): m₂ = (-13 - (-4)) / (7 - 1) [1 mark]. 3. Simplify m₂: m₂ = -9 / 6 = -3/2 (or -1.5) [1 mark]. 4. Conclusion: Because both lines have the exact same gradient (m₁ = m₂ = -1.5) and different y-intercepts, L₁ and L₂ are parallel [1 mark].
- Q4
The straight line L is perpendicular to the line y = 2x - 3 and passes through the point (4, 5). Line L crosses the x-axis at point A and the y-axis at point B. Calculate the exact area of the right-angled triangle OAB, where O is the origin (0, 0). (Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. The gradient of y = 2x - 3 is 2, so the perpendicular gradient of Line L is m = -1/2 (or -0.5) [1 mark]. 2. Substitute m = -0.5 and (4, 5) into y = mx + c: 5 = -0.5(4) + c -> 5 = -2 + c -> c = 7, so the equation of Line L is y = -0.5x + 7 [1 mark]. 3. Find the intercepts A and B: on the y-axis (x = 0), y = 7 so B is (0, 7); on the x-axis (y = 0), 0 = -0.5x + 7 -> 0.5x = 7 -> x = 14 so A is (14, 0) [1 mark]. 4. Calculate the area of right-angled triangle OAB: Area = ½ x base x height = ½ x 14 x 7 = 49 square units [1 mark].
Common Exam Questions (4)
What is the biggest trap when asked for the gradient and y-intercept of an equation like 2y + 6x = 10?
Never read off the coefficient of x before making 1y the subject. In 2y + 6x = 10, the gradient is not 6. You must first rearrange the equation into the exact form y = mx + c: subtract 6x from both sides (2y = -6x + 10) and divide every term by 2 (y = -3x + 5). Now you can see that the true gradient is m = -3 and the y-intercept is (0, 5).
How do you find the gradient of a line perpendicular to a given line? (Higher Tier)
Use the 'Flip and Change Sign' rule (negative reciprocal). If the first line has gradient m₁ = a / b, the perpendicular line has gradient m₂ = -b / a (so that m₁ x m₂ = -1). For example: if m₁ = 3 (which is 3/1), then m₂ = -1/3; if m₁ = -2/5, then m₂ = +5/2.
How do you find the equation of a perpendicular bisector of the line segment joining A and B? (Higher Tier)
A perpendicular bisector cuts the line segment AB at 90° (perpendicular) through its exact middle (bisector): 1. Find the midpoint M of AB by averaging the x-coordinates and averaging the y-coordinates: M = ((x₁ + x₂)/2, (y₁ + y₂)/2). 2. Find the gradient of AB using m_AB = (y₂ - y₁) / (x₂ - x₁). 3. Take the negative reciprocal to find the perpendicular gradient m_perp = -1 / m_AB. 4. Substitute m_perp and the midpoint M into y = mx + c (or y - y₁ = m(x - x₁)).
How do you find the equation of the tangent to a circle x² + y² = r² at a point P(x₁, y₁)? (Higher Tier)
Use the circle theorem that a tangent is always perpendicular (at 90°) to the radius at the point of contact. 1. Find the gradient of the radius joining the centre (0, 0) to the point P(x₁, y₁): m_radius = y₁ / x₁. 2. Take the negative reciprocal to get the gradient of the tangent: m_tangent = -x₁ / y₁. 3. Substitute m_tangent and the coordinates of P(x₁, y₁) into y = mx + c.
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