In the diagram, ABCD is a quadrilateral having vertices A(7; 1), B(2; 9), C(-3; -4) and D(8; -11) - NSC Mathematics - Question 3 - 2018 - Paper 2

The gradient m = (\frac{1}{2}) and we use the point-slope form to find the equation:

$y - y_1 = m(x - x_1)$ Using point A(7, 1): $y - 1 = \frac{1}{2}(x - 7)$ Expanding this gives: $y - 1 = \frac{1}{2}x - \frac{7}{2}$ Bringing 1 to the right: $y = \frac{1}{2}x - \frac{5}{2}$

Thus, the equation of AC is: $y = \frac{1}{2}x - \frac{5}{2}$.

To determine if M, the midpoint of BD, lies on AC, we first find the coordinates of M. The coordinates of B(2, 9) and D(8, -11) can be used:

$M = \left( \frac{x_B + x_D}{2}, \frac{y_B + y_D}{2} \right) = \left( \frac{2 + 8}{2}, \frac{9 - 11}{2} \right) = \left( 5, -1 \right)$.

Now, substitute x = 5 into the equation of AC: $y = \frac{1}{2}(5) - \frac{5}{2} = \frac{5}{2} - \frac{5}{2} = 0$.

Since M(5, -1) does not satisfy the equation (it gives y = 0), M does not lie on AC.

To prove BD is perpendicular to AC, the product of their gradients must equal -1. We found:

• Gradient of AC, m_{AC} = (\frac{1}{2})

Now, let's find the gradient of BD: $m_{BD} = \frac{y_D - y_B}{x_D - x_B} = \frac{-11 - 9}{8 - 2} = \frac{-20}{6} = -\frac{10}{3}$.

Now, confirm:

\text{(not equal to -1, BD and AC are not perpendicular)}$$.

The inclination θ of line BD can be found using the gradient:

$\tan(θ) = m_{BD} = -\frac{10}{3}$.

To find θ, we take the arctangent: $θ = \tan^{-1}\left(-\frac{10}{3}\right)$.

This gives us the inclination of BD.

To find the angle ∠CBD, we know:

• m_{AC} = (\frac{1}{2})
• m_{BD} = -(\frac{10}{3})

Using the formula for the angle between two lines: $\tan(∠) = \frac{|m_1 - m_2|}{1 + m_1 m_2}$ Substituting our values gives: $\tan(∠CBD) = \frac{\left|\frac{1}{2} + \frac{10}{3}\right|}{1 + \frac{1}{2} \cdot -\frac{10}{3}}$ which can be calculated further.

To find the length of AC, use the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Thus, between A(7, 1) and C(-3, -4): $d = \sqrt{(-3 - 7)^2 + (-4 - 1)^2} = \sqrt{(-10)^2 + (-5)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5}.$

So the length of AC is $5\sqrt{5}$.

To calculate the area of quadrilateral ABCD, we can use the shoelace formula: $\text{Area} = \frac{1}{2} | x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) |$

Substituting points A(7, 1), B(2, 9), C(-3, -4), D(8, -11):

Calculating gives: $\text{Area} = \frac{1}{2} | 7(9) + 2(-4) + (-3)(-11) + 8(1) - (1(2) + 9(-3) + (-4)(8) + (-11)(7)) |$ This can be evaluated for the final area.