Gradient of Lines - SS2 Mathematics Past Questions and Answers - page 1

The gradient (m) of a line joining two points (x1, y1) and (x2, y2) is given by the formula: m = (y2 - y1) / (x2 - x1) In this case, the points are (1, 3) and (5, 3). Therefore: x1 = 1, y1 = 3 x2 = 5, y2 = 3 Substituting these values into the formula: m = (3 - 3) / (5 - 1) m = 0 / 4 m = 0 Therefore, the gradient of the line is 0.

The equation of a line is generally expressed in the slope-intercept form as \\(y = mx + c\\), where: * \\(m\\) is the gradient (slope) of the line. * \\(c\\) is the y-intercept (the point where the line crosses the y-axis). In this question, we are given: * Gradient, \\(m = \\frac{1}{2}\\) * y-intercept, \\(c = 6\\) Substituting these values into the slope-intercept form, we get \\(y = \\frac{1}{2}x + 6\\). Option A is also correct but a transformation of B.

\[y - y_{1} = m(x - x_{1})\]

\[y - 0 = - 1(x - 3)\]

\(y = - x + 3\) is the required equation

\[\frac{y - y_{1}}{x - x_{1}} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\]

\[\frac{y - 3}{x - ( - 4)} = \frac{- 5 - 3}{2 - ( - 4)}\]

\[\frac{y - 3}{x + 4} = \frac{- 8}{2 + 4}\]

\[\frac{y - 3}{x + 4} = \frac{- 8}{6}\]

\[6(y - 3) = - 8(x + 4)\]

\[6y - 18 = - 8x - 32\]

\[6y = - 8x - 32 + 18\]

\[6y = - 8x - 14\]

\[y = \frac{- 8}{6}x - \frac{14}{6}\]

\(y = \frac{- 4}{3}x - \frac{7}{3}\) is the required equation

Comparing \(y = mx + c\) with \(y = 3x + 2\)

\(m = 3\), \(y\)-intercept \(= 2\)

\(x\)-intercept\(= \frac{- c}{m} = \ \frac{- 2}{3}\)