The position vectors of points \( P, Q, R \) are given as: \[ \vec{OP} = 7\mathbf{i} - 10\mathbf{j}, \quad \vec{OQ} = \mathbf{i} + 2\mathbf{j}, \quad \vec{OR} = 10\mathbf{j} - 3\mathbf{i}. \]
First, find vector \(\vec{PQ}\): \[ \vec{PQ} = \vec{OQ} - \vec{OP} = \begin{bmatrix} 1 \\ 2 \end{bmatrix} - \begin{bmatrix} 7 \\ -10 \end{bmatrix} = \begin{bmatrix} -6 \\ 12 \end{bmatrix}. \]
Next, find vector \(\vec{QR}\): \[ \vec{QR} = \vec{OR} - \vec{OQ} = \begin{bmatrix}-3 \\ 10\end{bmatrix} - \begin{bmatrix}1 \\ 2\end{bmatrix} = \begin{bmatrix}-4 \\ 8\end{bmatrix}. \]
Check if \(\vec{PQ}\) and \(\vec{QR}\) are scalar multiples: \[ \vec{PQ} = k \vec{QR}. \]
That is: \[ \begin{bmatrix} 6\\ 12 \end{bmatrix} =k \begin{bmatrix} -4\\ 8 \end{bmatrix} \] Comparing components: \[ -6 = -4k \quad \Rightarrow \quad k = \tfrac{3}{2}, \] \[ 12 = 8k \quad \Rightarrow \quad k = \tfrac{3}{2}. \]
Since both values of \(k\) are equal, \(\vec{PQ}\) is parallel to \(\vec{QR}\). Thus, points \(P, Q, R\) are collinear.
The points with coordinates \((7, 10)\) and \((-3, 8)\) are the end points of a diameter of a cycle centre A. Determine the equation of the circle in the form \(x^2 + y^2 + ax + by + c = 0\) where \(a\), \(b\) and \(c\) are constants.
The centre \(A\) is the midpoint: \[ \text{Centre} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{7 + (-3)}{2}, \frac{10 + 8}{2} \right) = (2, 9). \]
The radius is: \[ r^2 = (7 - 2)^2 + (10 - 9)^2 = 25 + 1 = 26. \]
Equation of the circle: \[ (x - a)^2 + (y - b)^2 = r^2 \] \[ (x - 2)^2 + (y - 9)^2 = 26. \]
Equation of the circle: \[ (x - 2)^2 + (y - 9)^2 = 26 \]
Expand: \[ (x^2 - 4x + 4) + (y^2 - 18y + 81) = 26 \]
Simplify: \[ x^2 + y^2 - 4x - 18y + 85 = 26 \]
\[ x^2 + y^2 - 4x - 18y + 59 = 0 \]
Therefore, the equation of the circle in the form \[ x^2 + y^2 + ax + by + c = 0 \] is \[ x^2 + y^2 - 4x - 18y + 59 = 0. \]