Find the gradient of the line joining \( (1, 2) \) and \( (4, 8) \).
\( m = \dfrac{8-2}{4-1} = \dfrac{6}{3} = 2 \).
Graphs
8 exam-style questions on straight-line graphs, gradients, intercepts and quadratic graphs.
3 Easy 3 Medium 2 Hard
Write down the gradient and \( y \)-intercept of \( y = 3x - 5 \).
Gradient \( = 3 \), \( y \)-intercept \( = -5 \).
Find the equation of the line with gradient \( 2 \) that passes through \( (0, 4) \).
\( y = 2x + 4 \).
Find the equation of the line passing through \( (2, 5) \) and \( (4, 11) \).
Gradient \( = \dfrac{11-5}{4-2} = 3 \). \( y - 5 = 3(x-2) \) ⇒ \( y = 3x - 1 \).
Does the point \( (3, 7) \) lie on the line \( y = 2x + 1 \)? Show working.
When \( x=3 \), \( y = 6 + 1 = 7 \). Yes, it lies on the line.
A quadratic graph has equation \( y = x^2 - 4x + 3 \). Find the coordinates of the turning point by completing the square.
\( y = (x-2)^2 - 4 + 3 = (x-2)^2 - 1 \). Turning point: \( (2, -1) \).
Find the equation of the line parallel to \( y = 3x + 1 \) that passes through \( (2, -4) \).
Same gradient \( 3 \). \( y + 4 = 3(x-2) \) ⇒ \( y = 3x - 6 - 4 \) ⇒ \( y = 3x - 10 \).
The line \( L \) passes through \( (1, 4) \) and is perpendicular to \( y = 2x - 3 \). Find the equation of \( L \).
Gradient of given line \( = 2 \), so perpendicular gradient \( = -\dfrac{1}{2} \). \( y - 4 = -\dfrac{1}{2}(x-1) \) ⇒ \( y = -\dfrac{1}{2}x + \dfrac{1}{2} + 4 \) ⇒ \( y = -\dfrac{1}{2}x + \dfrac{9}{2} \).