The parallelogram PQRS has vertices P(-2, 3), Q(1, 4), R(2, 6) and S(-1,5). Find the coordinates of the point of intersection of the diagonals.

Answer Details

To find the point of intersection of the diagonals of a parallelogram, we need to find the midpoint of the segment that connects the two diagonals. This midpoint is the point of intersection of the diagonals. Let's start by finding the equation of the line containing the diagonal PR. The slope of PR is: $$m_{PR}=\frac{y_R-y_P}{x_R-x_P}=\frac{6-3}{2-(-2)}=\frac{3}{4}$$ Using point-slope form, the equation of line PR is: $$y-3=\frac{3}{4}(x+2)$$ Simplifying, we get: $$y=\frac{3}{4}x+\frac{15}{4}$$ Now let's find the equation of the line containing the diagonal QS. The slope of QS is: $$m_{QS}=\frac{y_S-y_Q}{x_S-x_Q}=\frac{5-4}{-1-1}=\frac{-1}{2}$$ Using point-slope form, the equation of line QS is: $$y-4=\frac{-1}{2}(x-1)$$ Simplifying, we get: $$y=-\frac{1}{2}x+\frac{9}{2}$$ To find the point of intersection of the diagonals, we need to find the midpoint of the segment that connects the two diagonals. Let's call this midpoint M. The midpoint M is the average of the coordinates of the endpoints of the segment, which are the intersection points of the diagonals with their respective sides. To find the intersection point of PR with side QS, we solve the system of equations formed by the equations of lines PR and QS: $$\begin{cases}y=\frac{3}{4}x+\frac{15}{4}\\y=-\frac{1}{2}x+\frac{9}{2}\end{cases}$$ Solving for x and y, we get: $$\begin{cases}x=\frac{6}{5}\\y=\frac{39}{10}\end{cases}$$ So the intersection point of PR with side QS is (\(\frac{6}{5}, \frac{39}{10}\)). To find the intersection point of QS with side PR, we solve the system of equations formed by the equations of lines QS and PR: $$\begin{cases}y=-\frac{1}{2}x+\frac{9}{2}\\y=\frac{3}{4}x+\frac{15}{4}\end{cases}$$ Solving for x and y, we get: $$\begin{cases}x=-\frac{6}{5}\\y=\frac{21}{10}\end{cases}$$ So the intersection point of QS with side PR is (-\(\frac{6}{5}\), \(\frac{21}{10}\)). The midpoint M is the average of these two points, so we have: $$M=(\frac{\frac{6}{5}-\frac{6}{5}}{2}, \frac{\frac{39}{10}+\frac{21}{10}}{2})=(0, \frac{3}{2})$$ Therefore, the coordinates of the point of intersection of the diagonals is (0, \(\frac{3}{2}\)). So the answer is option (C), \((0, 4\frac{1}{2})\).