Here's the solution to the problem: Given the vertices of a triangle P(1, -3), Q(7, 5), and R(-3, -5). i) Express $\vec{PQ}$ and $\vec{PR}$ as column vectors. Step 1: Find vector $\vec{PQ}$. To find the vector from point P to point Q, subtract the coordinates of P from the coordinates of Q. $$ \vec{PQ} = Q - P = \begin{bmatrix} 7 \\ 5 \end{bmatrix} - \begin{bmatrix} 1 \\ -3 \end{bmatrix} = \begin{bmatrix} 7-1 \\ 5-(-3) \end{bmatrix} = \begin{bmatrix} 6 \\ 5+3 \end{bmatrix} $$ $$ \vec{PQ} = \boxed{\begin{bmatrix} 6 \\ 8 \end{bmatrix}} $$ Step 2: Find vector $\vec{PR}$. To find the vector from point P to point R, subtract the coordinates of P from the coordinates of R. $$ \vec{PR} = R - P = \begin{bmatrix} -3 \\ -5 \end{bmatrix} - \begin{bmatrix} 1 \\ -3 \end{bmatrix} = \begin{bmatrix} -3-1 \\ -5-(-3) \end{bmatrix} = \begin{bmatrix} -4 \\ -5+3 \end{bmatrix} $$ $$ \vec{PR} = \boxed{\begin{bmatrix} -4 \\ -2 \end{bmatrix}} $$ ii) Show that triangle PQR is isosceles. To show that a triangle is isosceles, we need to calculate the lengths of its three sides. If at least two sides have equal length, the triangle is isosceles. The length (magnitude) of a vector $\begin{bmatrix} x \\ y \end{bmatrix}$ is given by $\sqrt{x^2 + y^2}$. Step 1: Calculate the length of side PQ. Using $\vec{PQ} = \begin{bmatrix} 6 \\ 8 \end{bmatrix}$: $$ |\vec{PQ}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 $$ The length of PQ is 10 units. Step 2: Calculate the length of side PR. Using $\vec{PR} = \begin{bmatrix} -4 \\ -2 \end{bmatrix}$: $$ |\vec{PR}| = \sqrt{(-4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} $$ The length of PR is $\sqrt{20}$ units, which can be simplified to $2\sqrt{5}$ units. Step 3: Calculate the length of side QR. First, find the vector $\vec{QR}$. Subtract the coordinates of Q from the coordinates of R. $$ \vec{QR} = R - Q = \begin{bmatrix} -3 \\ -5 \end{bmatrix} - \begin{bmatrix} 7 \\ 5 \end{bmatrix} = \begin{bmatrix} -3-7 \\ -5-5 \end{bmatrix} = \begin{bmatrix} -10 \\ -10 \end{bmatrix} $$ Now, calculate the length of $\vec{QR}$: $$ |\vec{QR}| = \sqrt{(-10)^2 + (-10)^2} = \sqrt{100 + 100} = \sqrt{200} $$ The length of QR is $\sqrt{200}$ units, which can be simplified to $10\sqrt{2}$ units. Step 4: Compare the lengths of the