In the diagram below, P(1 ; 1), Q(0 ; -2) and R are the vertices of a triangle and \( \angle PQR=\theta \) The x-intercepts of PQ and PR are M and N respectively - NSC Mathematics - Question 3 - 2016 - Paper 2

Since gradients of perpendicular lines multiply to -1, we first find the gradient of line QR:

Using the equation of QR ( y = -x + 2 ), we get ( m_{QR} = -1 ).

Now, we check:

[ m_{QP} \times m_{QR} = 3 \times (-1) = -3 ]

Since the multiplication does not equal -1, it implies that PQ is not perpendicular to QR. We verify the gradients:

For line PR, which can be derived from the equation ( x + 3y + 6 = 0 ):

Rearranging gives ( y = -\frac{1}{3}x - 2 ) thus ( m_{PR} = -\frac{1}{3} )

Now we check:

[ m_{QR} \times m_{PR} = -1 \times (-\frac{1}{3}) = \frac{1}{3} ]

Thus, ( \angle PQR = 90^\circ ) is proven.

To find the coordinates of point R, we solve for the intersection of the lines PR and QR:

Set the equations equal to each other:

1. ( y = -x + 2 )
2. ( x + 3y + 6 = 0 )

Substituting ( y ) from the first equation into the second gives:

[ x + 3(-x + 2) + 6 = 0 ]

Solving this:

[ x - 3x + 6 + 6 = 0 \implies -2x + 12 = 0 \implies 2x = 12 \implies x = 6 ]

Substituting back to find ( y ):

[ y = -6 + 2 = -4 ]

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

The length of PR can be found using the distance formula:

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

Here, ( P(1, 1) ) and ( R(6, -4) ):

[ d = \sqrt{(6 - 1)^2 + (-4 - 1)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} ]

Thus, the length of PR is ( 5\sqrt{2} ).

The general form of the equation of a circle is:

[ (x - a)^2 + (y - b)^2 = r^2 ]

First, we find the center (( a, b )) using the coordinates of P(1, 1), Q(0, -2), R(6, -4).

The center can be derived from the average of the x-coordinates and y-coordinates:

[ a = \frac{1 + 0 + 6}{3} = \frac{7}{3}, b = \frac{1 + (-2) + (-4)}{3} = -\frac{5}{3} ]

Next, calculate the radius using any of the three points and the center:

Using point P:

[ r^2 = (1 - \frac{7}{3})^2 + (1 - (-\frac{5}{3}))^2 = (\frac{-4}{3})^2 + (\frac{8}{3})^2 = \frac{16}{9} + \frac{64}{9} = \frac{80}{9} ]

Thus, the equation of the circle is:

[ \left(x - \frac{7}{3}\right)^2 + \left(y + \frac{5}{3}\right)^2 = \frac{80}{9} ]

To find the equation of the tangent at point P(1, 1), we first find the slope of the radius.

The slope of the radius from center (( \frac{7}{3}, -\frac{5}{3} )) to point P:

[ m_{radius} = \frac{1 - (-\frac{5}{3})}{1 - \frac{7}{3}} = \frac{8/3}{-4/3} = -2 ]

The slope of the tangent line would then be the negative reciprocal:

[ m_{tangent} = \frac{1}{2} ]

Using point-slope form:

[ y - 1 = \frac{1}{2}(x - 1) ]

Rearranging gives:

[ y = \frac{1}{2}x + \frac{1}{2} ]

To find the angle ( \theta ), we can use the tangent of the angle formed by lines PQ and PR. We compute:

[ \tan(\theta) = \frac{m_{PR} - m_{QP}}{1 + m_{PR} \cdot m_{QP}} = \frac{-\frac{1}{3} - 3}{1 + (-\frac{1}{3})(3)} ]

Calculating the numerator and denominator:

Numerator: ( -\frac{1}{3} - 3 = -\frac{1}{3} - \frac{9}{3} = -\frac{10}{3} ) Denominator: ( 1 + 1 = 2 )

Thus:

[ \tan(\theta) = \frac{-\frac{10}{3}}{2} = -\frac{5}{3} ]

Taking the arctangent, we find:

[ \theta = \tan^{-1}(-\frac{5}{3}) \approx 26.57^{\circ} ]