The (x, y) coordinates of vertices P, Q and R of a parallelogram PQRS are (-3, -2), (1, -5) and (9, 1), respectively. If the diagonal SQ intersects the x-axis at (a, 0), then the value of a is

The correct answer is $\frac{29}{9}$.

First, we need to find the coordinates of the fourth vertex, S, of the parallelogram PQRS. We are given the coordinates of three vertices: P(-3, -2), Q(1, -5), and R(9, 1).

In a parallelogram, the diagonals bisect each other. This means the midpoint of diagonal PR is the exact same as the midpoint of diagonal QS.

Let's calculate the midpoint of PR using the midpoint formula:

$\text{Midpoint of}PR=(\frac{-3+9}{2},\frac{-2+1}{2})$

$=(\frac{6}{2},\frac{-1}{2})=(3,-\frac{1}{2})$

Let the coordinates of vertex S be (x, y). The midpoint of QS is:

$\text{Midpoint of}QS=(\frac{1+x}{2},\frac{-5+y}{2})$

Since the midpoints are equal, we can set up the following equations for the x and y coordinates:

$\frac{1+x}{2}=3$ and $\frac{-5+y}{2}=-\frac{1}{2}$

Solving for x:

$1+x=6$

$x=5$

Solving for y:

$-5+y=-1$

$y=4$

So, the coordinates of vertex S are (5, 4).

Next, we need to find the equation of the line representing diagonal SQ, which passes through S(5, 4) and Q(1, -5). First, find the slope (m) of this line:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{-5-4}{1-5}=\frac{-9}{-4}=\frac{9}{4}$

Using the point-slope form with point S(5, 4):

$y-4=\frac{9}{4}(x-5)$

We are given that diagonal SQ intersects the x-axis at the point (a, 0). We can substitute x = a and y = 0 into the line equation to solve for a:

$0-4=\frac{9}{4}(a-5)$

$-4=\frac{9}{4}(a-5)$

$-16=9(a-5)$

$-16=9a-45$

$9a=45-16$

$9a=29$

$a=\frac{29}{9}$