The straight line L has equation 3x + 2y = 17 - Edexcel - GCSE Maths - Question 1 - 2019 - Paper 3

The gradient of line M, which is perpendicular to line L, is the negative reciprocal of the gradient of L. Therefore:

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

So, the gradient of line M is \frac{2}{3}.

Since line M passes through point A (0, 2), we can write the equation of line M using point-slope form:

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

To find point C, we need to set line L equal to line M. We already have:

1. Line L: (3x + 2y = 17)
2. Line M: (y = \frac{2}{3}x + 2)

Substituting the second equation into the first:

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

This simplifies to:

$3x + \frac{4}{3}x + 4 = 17$

Multiplying everything by 3 to eliminate the fraction:

$9x + 4x + 12 = 51$

Thus:

$13x = 39 \Rightarrow x = 3$

Now substituting x back to find y:

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

Therefore, the coordinates of point C are (3, 4).

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

$3(0) + 2y = 17 \Rightarrow 2y = 17 \Rightarrow y = \frac{17}{2} = 8.5$

Thus, the coordinates of point B are (0, 8.5).

To calculate the area of triangle ABC, we can use the formula:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

The base can be the distance between points A (0, 2) and B (0, 8.5), which is:

$\text{base} = 8.5 - 2 = 6.5$

The height is the perpendicular distance from point C (3, 4) to line L. The formula for the distance (d) from a point (x_0, y_0) to a line Ax + By + C = 0 is:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

For line L:

1. Rewriting it gives us: $3x + 2y - 17 = 0$ Here, A = 3, B = 2, and C = -17.

Plugging in the coordinates of point C (3, 4):

$d = \frac{|3(3) + 2(4) - 17|}{\sqrt{3^2 + 2^2}} = \frac{|9 + 8 - 17|}{\sqrt{9 + 4}} = \frac{|-0|}{\sqrt{13}} = 0$

This means the area calculation simplifies to:

$\text{Area} = \frac{1}{2} \times 6.5 \times 0 = 0.$

Thus, the area of triangle ABC is 0.