The points A, B, and C have co-ordinates as follows: A(3,5) B(-6,2) C(4,-4) (a) Plot A, B, and C on the diagram - Leaving Cert Mathematics - Question 3 - 2014

To find the equation of the line AB, we need the slope and a point on the line. The slope (m) can be calculated using the coordinates of points A(3,5) and B(-6,2):

$m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 5}{-6 - 3} = \frac{-3}{-9} = \frac{1}{3}$

Now, using point-slope form, we can derive the line's equation:

• Starting with $y - y_1 = m(x - x_1)$:
• Plugging in point A: $y - 5 = \frac{1}{3}(x - 3)$
• Distributing and rearranging:

3y - x = 12$$ Thus, the equation of line AB is: $$3y - x = 12$$

To find the area of triangle ABC, we can use the formula for the area of a triangle with vertices at points (x1, y1), (x2, y2), and (x3, y3). Given the coordinates A(3,5), B(-6,2), and C(4,-4):

The formula is:

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

Substituting the coordinates, we get:

$A = \frac{1}{2} \left| 3(2 - (-4)) + (-6)(-4 - 5) + 4(5 - 2) \right|$

Calculating individually:

• $3(6) = 18$
• $-6(-9) = 54$
• $4(3) = 12$

So,

$A = \frac{1}{2} \left| 18 + 54 + 12 \right| = \frac{1}{2} \left| 84 \right| = 42 \, \text{units}^2$ Thus, the area of triangle ABC is 42 square units.