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

To find the equation of line n, we start by determining the slope. The slope of line n, which is perpendicular to line j, can be calculated using the negative reciprocal:

m_n = -rac{1}{m_j} = -rac{5}{2}

Using the point-slope form of the equation of a line, we have:

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

Substituting the point (6, -1):

y - (-1) = -rac{5}{2}(x - 6)

Thus, the equation of line n is:

y = -rac{5}{2}(x - 6) - 1

For line k:

The equation is:
$y = x - 1$
Here, the slope is $1$ and it crosses the y-axis at $b = -1$. Thus, the point is $(0, -1)$.

For line l:

Rearranging the equation:
$2x - 3y = 6$
we solve for y:

-3y = -2x + 6 \\ y = rac{2}{3}x - 2
The slope is rac{2}{3} and it crosses the y-axis at $b = -2$, leading to the point $(0, -2)$.

To find the intersection, we set the equations equal to each other:

From line k: $y = x - 1$

From line l: $2x - 3y = 6$

Substituting $y$ from line k into line l:

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

Expanding and simplifying:

$2x - 3x + 3 = 6 \\ -x + 3 = 6 \\ -x = 3 \\ x = -3$

Substituting $x = -3$ back into line k:

$y = -3 - 1 = -4$

Thus, the point of intersection is $(-3, -4)$.