The equation of the line l is $x - 3y - 6 = 0$ - Junior Cycle Mathematics - Question i - 2014

To verify whether the point (1,–2) is on the line l, we substitute $x = 1$ and $y = -2$ into the original equation:

1. Substitute into $x - 3y - 6 = 0$: $$1 - 3(-2) - 6 = 1 + 6 - 6 = 1 \neq 0$$ Therefore, the left-hand side (LHS) does not equal the right-hand side (RHS), confirming that the point (1,–2) is not on the line l.

Since line k is parallel to line l, it has the same slope:

• Slope of line k: $m = \frac{1}{3}$.

To find the equation of line k which passes through the point (1,–2), we can use the point-slope form:

1. Using the point-slope formula: $$y - y_1 = m(x - x_1)$$ Substituting $m = \frac{1}{3}$, $x_1 = 1$, and $y_1 = -2$ gives: $$y + 2 = \frac{1}{3}(x - 1)$$ Simplifying: $$y + 2 = \frac{1}{3}x - \frac{1}{3}$$ Thus: $$y = \frac{1}{3}x - \frac{7}{3}$$ Alternatively, in standard form, we can arrange it as: $$x - 3y - 7 = 0.$$ Therefore, the equation of the line k is either $y = \frac{1}{3}x - \frac{7}{3}$ or $x - 3y - 7 = 0$.