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 25 - 2019 - Paper 2

The gradient of a line perpendicular to another line is the negative reciprocal of the original line's gradient. Therefore, for line M:

1. Calculate the gradient of line M: $m_M = -\frac{1}{m_L} = -\frac{1}{-\frac{3}{2}} = \frac{2}{3}$

To find the y-intercept (the point B) of line L, set $x = 0$ in line L's equation:

1. Substitute: $3(0) + 2y = 17$
2. This simplifies to: $2y = 17$
3. Solving for $y$ gives: $y = \frac{17}{2} = 8.5$

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

To find point C, we need to solve the equations of lines L and M together:

1. The equation of line M passing through point A (0, 2): $y - 2 = \frac{2}{3}(x - 0)$ Simplifying gives: $y = \frac{2}{3}x + 2$

2. Set the two equations equal to find point C: $\frac{2}{3}x + 2 = -\frac{3}{2}x + \frac{17}{2}$ Rearranging gives: $\frac{2}{3}x + \frac{3}{2}x = \frac{17}{2} - 2$ Simplifying further, we can multiply by 6 to eliminate fractions: $4x + 9x = 51 - 12$ Giving: $13x = 39$ Thus $x = 3$. Substitute $x = 3$ back into either line equation to find $y$: $\frac{2}{3}(3) + 2 = 4$ Therefore, C is $(3, 4)$.

To compute the area of triangle ABC using the coordinates A(0, 2), B(0, 8.5), and C(3, 4), we can use the formula for the area of a triangle defined by vertices at points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:

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

1. Plugging in the points: $\text{Area} = \frac{1}{2} \left| 0(8.5-4) + 0(4-2) + 3(2-8.5) \right|$ $= \frac{1}{2} \left| 0 + 0 + 3(-6.5) \right|$ $= \frac{1}{2} \left| -19.5 \right|$ $= \frac{19.5}{2} = 9.75$

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