In the diagram, ABCD is a quadrilateral having vertices A(−4; 3), B(3; 4), C(4; −3) and D(0; −5) - NSC Mathematics - Question 3 - 2017 - Paper 2

To show that AD is perpendicular to DC, we need to calculate the gradients of both lines. The gradient m_{AD} can be calculated using points A(-4, 3) and D(0, -5):

$m_{AD} = \frac{-5 - 3}{0 - (-4)} = \frac{-8}{4} = -2$

Next, we check the product of the gradients:

$m_{DC} \times m_{AD} = \left( \frac{1}{2} \right) \times (-2) = -1$

Since the product is -1, AD is perpendicular to DC.

To prove that ΔABC is isosceles, we need to find the lengths of AB, BC, and AC.

1. Length of AB: $AB = \sqrt{(3 - (-4))^2 + (4 - 3)^2} = \sqrt{(7)^2 + (1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}$

2. Length of BC: $BC = \sqrt{(4 - 3)^2 + (-3 - 4)^2} = \sqrt{(1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}$

Since AB = BC, ΔABC is an isosceles triangle.

To find the equation of line BF using point B(3, 4) and its gradient with respect to F, we first need to determine the slope m_{BF}. Since BF || DE, we have:

$y - 4 = m_{BF}(x - 3)$

We previously calculated m_{DC} = \frac{1}{2}, therefore for BF:

$y - 4 = \frac{1}{2}(x - 3)$

Expanding:

$y = \frac{1}{2}x - \frac{3}{2} + 4 = \frac{1}{2}x + \frac{5}{2}$

To find the angle θ, we will use the tangent function. Since m_{AD} = -2, we have:

$\tan(θ) = -2 \implies θ = \tan^{-1}(-2)$

Calculating this gives:

$θ \approx 116.57°$

The general equation of a circle with centre at the origin is:

$x^2 + y^2 = r^2$

To find r, we calculate the distance from the origin to C(4, -3):

$r = \sqrt{(4 - 0)^2 + (-3 - 0)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

Thus, the equation of the circle is:

$x^2 + y^2 = 25$