In the diagram, A(-2; 10), B(k; k) and C(4; -2) are the vertices of ΔABC - NSC Mathematics - Question 3 - 2021 - Paper 2

To find the gradient of line AB, we use the angle of inclination.

The gradient is given by:

$m_{AB} = \tan(81.87°)$

Calculating this gives us:

$m_{AB} \approx 7$

Substituting the gradient obtained into the equation of a line, we have:

$y = \frac{1}{4}x + c$

Given that point B is (-4; -4), we can find c:

$-4 = \frac{1}{4}(-4) + c \Rightarrow -4 = -1 + c \Rightarrow c = -3$

Thus, the equation of BE is:

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

From the evaluation, substituting k = -4 into the coordinates of B gives:

$B(-4; -4)$

To find the angle ∠Â, we can use the relationship between the gradients:

Given that:

$m_{AC} = -2$

The gradient of AB is 7:

Using the formula for the angle between two lines:

$\theta = \arctan \left( \frac{m_1 - m_2}{1 + m_1 m_2} \right)$

Substituting in the values:

$\theta = \arctan \left( \frac{7 - (-2)}{1 + 7(-2)} \right) \approx 34.7°$

To find the intersection M of the diagonals, we average the x and y coordinates of A and C:

$M = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right) = \left( \frac{-2 + 4}{2}, \frac{10 + (-2)}{2} \right) = \left( 1, 4 \right)$

Using the distance formula, we set the length ET equal to 4√17:

$\sqrt{(12 - p)^2 + (0 - r)^2} = 4\sqrt{17}$

Squaring both sides and solving for the coordinates leads to:

Letting p = 16 gives coordinates of T as T(16; 0).

Thus:

$T(16; 0)$

The center of the circle is E(12; 0), and the radius is the distance EB:

Calculating r:

$r = ext{Distance}(E, B) = \sqrt{(12 - (-4))^2 + (0 - (-4))^2} = \sqrt{16^2 + 4^2} = \sqrt{272}$

The equation of the circle is:

$(x - 12)^2 + (y - 0)^2 = 272$

The gradient of the radius at B is:

From E to B, using the coordinates E(12;0) and B(-4; -4):

$m_{EB} = \frac{-4 - 0}{-4 - 12} = \frac{-4}{-16} = \frac{1}{4}$

The tangent's gradient is the negative reciprocal:

$m_{tangent} = -4$

Using the point-slope form of a line's equation:

$y - k = -4(x - k)$

Thus, the equation of the tangent line is:

$y = -4x + (4k + k) = -4x + 5k$