The line l contains the points A(4, 5) and B(2, 0) - Leaving Cert Mathematics - Question 4 - 2016

To draw the line k, we can rewrite it in slope-intercept form.

Starting from:

$x + 2y = 8$

We isolate y:

$2y = -x + 8$

$y = -\frac{1}{2}x + 4$

Now, we plot the y-intercept at (0, 4) and another point by letting x = 8:

$y = -\frac{1}{2}(8) + 4 = 0$

So, we plot the second point at (8, 0). Next, we connect these points to illustrate the line k.

To find the coordinates where lines l and k intersect, we can solve the equations:

1. From the equation of line l: $5x - 2y + 10 = 0$ Rearranged: $2y = 5x + 10$ $y = \frac{5}{2}x + 5$

2. From the equation of line k: $x + 2y = 8 \implies 2y = 8 - x \implies y = 4 - \frac{1}{2}x$

Set the two equations for y equal to each other:

$\frac{5}{2}x + 5 = 4 - \frac{1}{2}x$

Solve for x:

$\frac{5}{2}x + \frac{1}{2}x = 4 - 5$

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

Substituting back to find y:

$y = 4 - \frac{1}{2}\left(-\frac{1}{3}\right) = 4 + \frac{1}{6} = \frac{24}{6} + \frac{1}{6} = \frac{25}{6}$

Thus, the coordinates of intersection l ∩ k are:

$\left(-\frac{1}{3}, \frac{25}{6}\right)$