In die diagram is A(2; 10), B(k; k) en C(4; -2) die hoekpunte van ∆ABC - NSC Mathematics - Question 3 - 2021 - Paper 2

To find the gradient of line segment AB, we use:

$m_{AB} = \tan(81,87°)$
which evaluates to approximately:

$m_{AB} \approx 7$

Using the gradient calculated above:

$y - y_1 = m(x - x_1)$ Substituting E(12; 0) into the equation:

$y - 0 = \frac{-k}{12-k}(x - 12)$
Let’s express y:

$y = \frac{-k}{12-k}x + \frac{12k}{12-k}$

Based on our previous calculations of the gradients:

Setting the gradient equal values from BE and AB:

$\frac{-k}{12-k} = 7$
Solving for k, we find:

$k = -4$
Thus the coordinates of B are B(-4; -4).

Using the tangent ratio:

$\tan(\angle A) = \frac{10 - (-2)}{2 - 4} = \tan(116,57°)$
Calculating gives us:

Angle A = 34,70°.

For the intersection point of diagonals in parallelogram ACES, we apply the midpoint formula:

$M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$
With points A and C, we find:

$M(\frac{12 + (-2)}{2}, \frac{10 + 0}{2}) = (5; 5)$.

Given that ET = BE = 4√17 and substituting back into the equations yields:

We can solve for p:

$p = 16$
Thus the coordinates of T = (16; 16).

For the circle that passes through point B(k; k):

The equation is:

$(x - a)^2 + (y - b)^2 = r^2$
Substituting B(-4; -4) and the radius from earlier calculations, we find:

$(x + 4)^2 + (y + 4)^2 = (r)^2$.