Here is a sketch of the line L - Edexcel - GCSE Maths - Question 14 - 2021 - Paper 2

To find the coordinates of point R, we first need to determine the distance between points P and Q.

1. Calculate the distance PQ: Using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ where P(-6, 0) and Q(0, 3).

   $PQ = \sqrt{(0 - (-6))^2 + (3 - 0)^2} = \sqrt{(6)^2 + (3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$

2. Find the distances based on the ratio: Since PQ : QR = 2 : 3, we find QR:

   • Let the total distance be divided into 5 equal parts (2 + 3).

   • Find the lengths:

     • Length of PQ: $PQ = \frac{2}{5} \times (3\sqrt{5}) = \frac{6\sqrt{5}}{5}$
       • Length of QR: $QR = \frac{3}{5} \times (3\sqrt{5}) = \frac{9\sqrt{5}}{5}$

3. Calculate the coordinates of R: We can find R by moving from Q to R at the coordinates Q(0, 3).

   To find coordinates of R, we need to find the direction vector of PQ, which is:

   • Direction from P to Q: $(0 - (-6), 3 - 0) = (6, 3)$

   • Normalize this direction vector: $V = \frac{(6, 3)}{\sqrt{(6^2 + 3^2)}} = \frac{(6, 3)}{3\sqrt{5}}$

   • Moving from Q(0, 3) towards R in the same direction:

     Let's move by the length QR:

     $R = Q + \frac{9}{5} \cdot \frac{(6, 3)}{3\sqrt{5}} = (0, 3) + (\frac{108}{15}, \frac{54}{15}) = (\frac{108}{15}, \frac{129}{15})$

So, the coordinates of point R are approximately: $R(\frac{108}{15}, \frac{129}{15})$.