In the diagram below, P(1; 1), Q(0; -2) and R are the vertices of a triangle and PQR = θ - NSC Mathematics - Question 3 - 2016 - Paper 2

To show that PQR is a right angle, we will determine the gradients of both lines PR and QR and check if the product of the gradients is -1, indicating perpendicular lines.

The gradient of QR can be found from its equation:
$x + 3y + 6 = 0 \implies y = -\frac{1}{3}x - 2$
Thus, the gradient of QR is:

$m_{QR} = -\frac{1}{3}$

Next, for the line PR, we rewrite its equation: $y = -x + 2 \implies m_{PR} = -1$

Now, we check the product of the gradients:

$m_{QP} \cdot m_{QR} = 3 \cdot -\frac{1}{3} = -1$

Since the product of the gradients is -1, it follows that PQR = 90°.

To find the coordinates of R, we solve the system of equations defined by the lines PR and QR:

1. From PR:
   y = -x + 2
2. From QR:
   x + 3y + 6 = 0

Substituting the expression for y from PR into QR:

$x + 3(-x + 2) + 6 = 0$
Expanding and simplifying:

$x - 3x + 6 + 6 = 0 \implies -2x + 12 = 0 \implies 2x = 12 \implies x = 6$

Now substituting x back into PR to solve for y:

$y = -6 + 2 = -4$

Thus, the coordinates of R are (6, -4).

To calculate the length of PR, we use the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Using points P(1, 1) and R(6, -4):

$PR = \sqrt{(6 - 1)^2 + (-4 - 1)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$

So, the length of PR is $5\sqrt{2}$.

Let the center of the circle be (a, b) and its radius be r. Using the points P(1, 1), Q(0, -2), and R(6, -4), we can set up three equations:

1. From P: $(1 - a)^2 + (1 - b)^2 = r^2$
2. From Q: $(0 - a)^2 + (-2 - b)^2 = r^2$
3. From R: $(6 - a)^2 + (-4 - b)^2 = r^2$

Subtract the equations pairwise to express a and b values, and eventually identify r from one of the equations.

The slope of the radius at P is found using implicit differentiation on the circle equation. Denote the slope at P: $m = -\frac{d(y)}{d(x)}$ Using point P and radius slope, we will find the tangent equation:
$y - 1 = m(x - 1)$ This represents the equation of the tangent line at point P.

The angle θ can be calculated using the formula for the tangent between the gradients of lines PT and RQ. Finding these gradients, we apply the inverse tangent function:

$\tan(\theta) = \frac{m_{PT} - m_{RQ}}{1 + m_{PT} m_{RQ}}$ The resulting angle θ can be determined using the arctangential function, where it is processed through geometric relations, giving θ ≈ 26.57° or similar, depending on determined values.