Problem of the Week Problem E and Solution Parallelogram Askew

Problem

If vertices $$P$$, $$Q$$, and $$S$$ are located at $$(0,30)$$, $$(k,50)$$ and $$(40,0)$$, respectively, and the area of $$PQRS$$ is $$1340\text{ units}^2$$, determine the coordinates of $$Q$$ and $$R$$.

Solution

Solution 1:

In this solution we will use a method known commonly as "completing the rectangle".

Since $$PQRS$$ is a parallelogram, $$PQ=SR$$ and $$PQ$$ is parallel to $$SR$$. We can use this to find the coordinates of $$R$$. To get from $$P$$ to $$Q$$, we go up 20 units and right $$k$$ units. Therefore, to get from $$S$$ to $$R$$ we do the same. Therefore, $$R$$ is located at $$(40+k,20)$$.

Enclose $$PQRS$$ in rectangle $$OTUV$$ such that $$OT$$ is on the positive $$y$$-axis passing through $$P$$, $$TU$$ is parallel to the positive $$x$$-axis passing through $$Q$$, $$UV$$ is parallel to the positive $$y$$-axis passing through $$R$$, and $$OV$$ lies along the positive $$x$$-axis passing through $$S$$. The $$y$$-coordinate of $$Q$$ is the distance from the $$x$$-axis to $$TU$$ and also the height, $$UV$$, of rectangle $$OTUV$$. It follows that $$OT = UV=50$$ units. Therefore, the coordinates of $$T$$ are $$(0,50)$$. Similarly, the $$x$$-coordinate of $$R$$ is the distance from the $$y$$-axis to $$UV$$ and also the width, $$OV$$, of rectangle $$OTUV$$. It follows that $$TU = OV=(40+k)$$ units. Therefore, the coordinates of $$V$$ are $$(40+k,0)$$ and the coordinates of $$U$$ are $$(40+k, 50)$$.

We can now put the information together using areas to determine the value of $$k$$. \begin{aligned} \text{Area }OTUV&=\text{Area }\triangle PTQ+\text{Area }\triangle QUR+\text{Area }\triangle RVS+\text{Area }\triangle SOP+\text{Area }PQRS\\ UV\times OV&=\frac{PT\times TQ}{2}+\frac{QU\times UR}{2}\ +\frac{RV\times VS}{2}+\frac{OS\times OP}{2}\ +\ 1340\\ 50\times (40+k)&=\frac{(50-30)\times k}{2}+\frac{((40+k)-k)\times (50-20)}{2}+\frac{20\times ((40+k)-40)}{2}+\frac{40\times 30}{2}+1340\\ 50\times (40+k)&=\frac{20\times k}{2}+\frac{40\times 30}{2}\ +\frac{20\times k}{2}+\frac{40\times 30}{2}\ +\ 1340\\ 2000+50k&=10k+600+10k+600+1340\\ 2000+50k&=20k+2540\\ 30k&=540\\ k&=18\end{aligned}

Therefore, the value of $$k$$ is $$18$$ and coordinates of $$Q$$ and $$R$$ are $$Q(18,50)$$ and $$R(58,20)$$.

Solution 2:

In this solution we will use linear equations, intersections, and lengths to find $$k$$.

Since $$PQRS$$ is a parallelogram, $$PQ=SR$$ and $$PQ$$ is parallel to $$SR$$. We can use this to find the coordinates of $$R$$. To get from $$P$$ to $$Q$$, we go up 20 units and right $$k$$ units. Therefore, to get from $$S$$ to $$R$$ we do the same. Therefore, $$R$$ is located at $$(40+k,20)$$.

Construct a line perpendicular to $$PS$$ that passes through $$Q$$ and meets $$PS$$ at $$W$$.

We are going to find the coordinates of $$W$$ in terms of $$k$$.

The line through $$PS$$ has a slope of $$-\frac{3}{4}$$ and a $$y$$-intercept of 30. Therefore, the equation of this line is $y=-\frac{3}{4}x + 30 \tag{1}$

The line though $$QW$$ is perpendicular to $$PS$$ and so has slope $$\frac{4}{3}$$. The equation of this line is $$4x - 3y = C$$. Substituting the coordinates of $$Q$$ into this equation, we get $$4k - 3(50) = C$$ or $$C = 4k -150$$. Therefore, the line through $$QW$$ has equation $4x - 3y = 4k - 150\tag{2}$

$$W$$ is the intersection point of the lines with equations $$(1)$$ and $$(2)$$.

Substituting equation $$(1)$$ into equation $$(2)$$, we get: \begin{align*} 4x - 3\left(-\frac{3}{4}x + 30\right) &= 4k -150\\ 4x + \frac{9}{4}x - 90 &= 4k -150\\ 16x + 9x - 360 &= 16k -600\\ 25x &= 16k - 240\\ x &= 0.64k - 9.6 \tag{3}\end{align*}

Substituting equation $$(3)$$ into equation $$(1)$$ we get: \begin{aligned} y &= -\frac{3}{4}(0.64k - 9.6) + 30\\ y &= -0.48k + 37.2\end{aligned}

Therefore, the point $$W$$ has coordinates $$( 0.64k - 9.6, -0.48k + 37.2)$$.

We will now find two expressions for the length of $$QW$$.

Using the distance formula we know $QW = \sqrt{(0.64k - 9.6 - k)^2 + (-0.48k + 37.2 - 50)^2}\tag{4}$

Another way to find the length $$QW$$ is using the area of the parallelogram.

The length of $$PS = \sqrt{(30-0)^2 + (0-40)^2} = \sqrt{2500} = 50$$, since $$PS > 0$$.

$$PS$$ is the base of the parallelogram and $$QW$$ is the height. Therefore, \begin{align*} PS \times QW &= 1340\\ 50QW &= 1340\\ QW &= 26.8 \tag{5} \end{align*}

Now equating equations $$(4)$$ and $$(5)$$, we can solve for $$k$$. \begin{aligned} \sqrt{(0.64k - 9.6 - k)^2 + (-0.48k + 37.2 - 50)^2} &= 26.8\\ (-0.36k - 9.6 )^2 + (-0.48k - 12.8)^2 &= 718.24\\ 0.1296k^2 + 6.912k + 92.16 + 0.2034k^2 + 12.288k + 163.84 &= 718.24\\ 0.36k^2 + 19.2k - 462.24 & = 0 \end{aligned}

Using the quadratic formula, we find $$k=18$$ or $$k=-\frac{214}{3}$$. Since $$Q(k,50)$$ is in the first quadrant, we must have $$k > 0$$ and so $$k=18$$.

Therefore, the coordinates of $$Q$$ and $$R$$ are $$Q(18,50)$$ and $$R(58,20)$$.