Gradient of Lines - SS2 Mathematics Past Questions and Answers - page 1
Find the gradient of the line joining the points (1,3) and (5,3)
3
0
2
6
\(\frac{\mathrm{\Delta}y}{\mathrm{\Delta}x} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{3 - 3}{5 - 1} = \frac{0}{4} = 0\)
Determine the equation of a line with gradient \(\frac{1}{2}\) and \(y\)-intercept \(6\)
2y = x + 12
\(y = \frac{1}{2}x + 6\)
y = 3x
A and B
\[y = mx + c\]
\[y = \frac{1}{2}x + 6\]
\(2y = x + 12\) is the required equation
Find the equation of a straight line with a gradient of -1, passing the point \((3,0)\)
\[y - y_{1} = m(x - x_{1})\]
\[y - 0 = - 1(x - 3)\]
\(y = - x + 3\) is the required equation
Find the equation of a straight line passing through the pair of points \(( - 4,3)\) and \((2, - 5)\)
\[\frac{y - y_{1}}{x - x_{1}} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\]
\[\frac{y - 3}{x - ( - 4)} = \frac{- 5 - 3}{2 - ( - 4)}\]
\[\frac{y - 3}{x + 4} = \frac{- 8}{2 + 4}\]
\[\frac{y - 3}{x + 4} = \frac{- 8}{6}\]
\[6(y - 3) = - 8(x + 4)\]
\[6y - 18 = - 8x - 32\]
\[6y = - 8x - 32 + 18\]
\[6y = - 8x - 14\]
\[y = \frac{- 8}{6}x - \frac{14}{6}\]
\(y = \frac{- 4}{3}x - \frac{7}{3}\) is the required equation
Determine the gradient of the linear equations \(y = 3x + 2\)
Comparing \(y = mx + c\) with \(y = 3x + 2\)
\(m = 3\), \(y\)-intercept \(= 2\)
\(x\)-intercept\(= \frac{- c}{m} = \ \frac{- 2}{3}\)