The line j has a slope of \( \frac{2}{5} \) - Junior Cycle Mathematics - Question 7 - 2018

The equation of the line can be determined using the point-slope form. Given the slope ( -\frac{5}{2} ) and that it passes through the point (6, -1), we use:

$y - y_1 = m (x - x_1)$

Substituting the known values:

$y - (-1) = -\frac{5}{2} (x - 6)$

This gives:

$y + 1 = -\frac{5}{2}x + 15$

Rearranging it provides:

$y = -\frac{5}{2}x + 14$

Thus, the equation of line n is:

$y = -\frac{5}{2}x + 14$.

For line k, the slope can be derived from its equation, which is already in the slope-intercept form (y = mx + b). Here, the slope ( m_k = 1 ) and the y-intercept is ( -1 ), thus the point is (0, -1).

For line l, we rearrange the equation ( 2x - 3y = 6 ) to slope-intercept form:

$-3y = -2x + 6 \Rightarrow y = \frac{2}{3}x - 2 \Rightarrow m_l = \frac{2}{3} \text{ and y-intercept at (0, -2).}$

The completed table is:

Line	Slope	Point where the line crosses the y-axis
k	1	(0, -1)
l	\frac{2}{3}	(0, -2)

To find the intersection, we set the equations of line k and line l equal to each other:

1. Line k: ( y = x - 1 )
2. Line l: ( y = \frac{2}{3}x - 2 )

Setting them equal:

$x - 1 = \frac{2}{3}x - 2$

Multiplying through by 3 to eliminate the fraction:

$3(x - 1) = 2x - 6 \Rightarrow 3x - 3 = 2x - 6 \Rightarrow x = -3$

Substituting ( x = -3 ) back into line k's equation to solve for y:

$y = -3 - 1 = -4$

Thus, the point of intersection of the lines k and l is (−3, −4).