The straight line L1 passes through the points with coordinates (4, 6) and (12, 2) - Edexcel - GCSE Maths - Question 19 - 2018 - Paper 2

We can use the point-slope form to find the equation of line L1. Using point (4, 6) and the gradient -1/2, we have:

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

Substituting the values: $y - 6 = -\frac{1}{2}(x - 4)$

Rearranging gives: $y = -\frac{1}{2}x + 8$

Line L2 passes through the origin (0, 0) with a gradient of -3. The equation is given by:

$y = mx + c$

where $c = 0$:
$y = -3x$

To find the coordinates of point P, set the equations of L1 and L2 equal to each other:

$-\frac{1}{2}x + 8 = -3x$

Solving for x: $-\frac{1}{2}x + 3x = 8$ $\frac{5}{2}x = 8$ $x = \frac{8 \times 2}{5} = \frac{16}{5}$

Now substituting $x$ back into the equation of L2 to find $y$: $y = -3\left(\frac{16}{5}\right) = -\frac{48}{5}$

Thus, the coordinates of P are: $P\left(\frac{16}{5}, -\frac{48}{5}\right)$