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 prove that lines AD and DC are perpendicular, we need to show that the product of their gradients equals -1.

First, we find the gradient of AD:

Let A(-4; 3) and D(0; -5): $m_{AD} = \frac{-5 - 3}{0 - (-4)} = \frac{-8}{4} = -2$

Now, since we have: $m_{CD} = \frac{1}{2}$

We calculate: $m_{CD} \times m_{AD} = \frac{1}{2} \times -2 = -1$

Thus, AD is perpendicular to DC.

To show that triangle ABC is isosceles, we need to find the lengths of sides AB, BC, and AC:

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

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

Since AB = BC, triangle ABC is isosceles.

To find the equation of line BF, we first need the gradient:

Using points B(3; 4) and F(x_F; 0) (where y = 0 on the x-axis):

The gradient is: $m_{BF} = \frac{0 - 4}{x_F - 3} = \frac{-4}{x_F - 3}$

Using point-slope form: $y - y_1 = m(x - x_1)$

For point B and substituting the gradient: $y - 4 = \frac{-4}{x_F - 3}(x - 3)$

Rearranging gives the equation in the form y = mx + c.

Using trigonometric functions, consider the angle θ such that:

$\tan(\theta) = \frac{1}{m_{AD}} = \frac{1}{-2}$

Thus, $\theta = \arctan(-\frac{1}{2})\approx 26.57°$

To find the equation of the circle, we first calculate the radius (r) using point C(4, -3):

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

The equation of the circle is then: $x^2 + y^2 = r^2 = 25$