The straight line L has equation 3x + 2y = 17 The point A has coordinates (0, 2) The straight line M is perpendicular to L and passes through A - Edexcel - GCSE Maths - Question 3 - 2019 - Paper 2

The gradient of line M, which is perpendicular to line L, will be the negative reciprocal of the gradient of line L.

$$m_M = -\frac{1}{m_L} = -\frac{1}{-\frac{3}{2}} = \frac{2}{3}.$$

To find the y-coordinate of point B where line L crosses the y-axis (x=0), substitute x=0 into the equation of line L:

$$3(0) + 2y = 17 \Rightarrow 2y = 17 \Rightarrow y = \frac{17}{2}.$$

Thus, point B is (0, 8.5).

To find point C where lines L and M intersect, we can set up two equations:

From line L:

$$3x + 2y = 17$$

From line M (which passes through A (0, 2)) with the gradient of \frac{2}{3):

$$y - 2 = \frac{2}{3}(x - 0) \Rightarrow y = \frac{2}{3}x + 2.$$

Now, substituting this into the equation of line L:

$$3x + 2(\frac{2}{3}x + 2) = 17$$

Expanding yields:

$$3x + \frac{4}{3}x + 4 = 17 \Rightarrow \left(3 + \frac{4}{3}\right)x = 13 \Rightarrow \frac{13}{3}x = 13 \Rightarrow x = 3.$$

Substituting x = 3 back to find y:

$$y = \frac{2}{3}(3) + 2 = 2 + 2 = 4.$$

Thus, point C is (3, 4).

The coordinates of points A, B, and C are A(0, 2), B(0, 8.5), and C(3, 4).

The formula for the area of a triangle given vertices at (x1, y1), (x2, y2), and (x3, y3) is:

$$\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|.$$

Substituting the coordinates:

$$\text{Area} = \frac{1}{2} \left| 0(8.5-4) + 0(4-2) + 3(2-8.5) \right| \Rightarrow = \frac{1}{2} \left| 0 + 0 - 19.5 \right| = \frac{1}{2} \times 19.5 = 9.75.$$

Therefore, the area of triangle ABC is 9.75 square units.