(i) AB and DC are parallel, so the lines containing these sides have the same slope.
The line containing AB has slope (2 - (-1))/(10 - 4) = 3/6 = 1/2.
Use the point-slope formula to find the equation of the line containing CD:
![y-8=\dfrac12(x-12)\implies y=\dfrac x2+2\implies\boxed{2y=x+4}](https://tex.z-dn.net/?f=y-8%3D%5Cdfrac12%28x-12%29%5Cimplies%20y%3D%5Cdfrac%20x2%2B2%5Cimplies%5Cboxed%7B2y%3Dx%2B4%7D)
AP is perpendicular to both AB and CD, and perpendicular lines have slopes that are negative reciprocals of one another. This means the line containing AP has slope -1/(1/2) = -2. Using the point-slope formula again, the equation of the line containing AP is
![y-(-1) = -2(x-4)\implies \boxed{y=-2x+7}](https://tex.z-dn.net/?f=y-%28-1%29%20%3D%20-2%28x-4%29%5Cimplies%20%5Cboxed%7By%3D-2x%2B7%7D)
(ii) D lies on the <em>x</em> axis, so its <em>y</em>-coordinate is 0. Find the <em>x</em>-coordinate by plugging <em>y</em> = 0 into the equation for the line containing CD:
![2\cdot0=x+4\implies x=-4](https://tex.z-dn.net/?f=2%5Ccdot0%3Dx%2B4%5Cimplies%20x%3D-4)
So D is the point (-4, 0).
P is the intersection of the lines containing CD and AP, so set them equal and solve for <em>x</em> and <em>y </em>:
![y=\dfrac x2+2\text{ and }y=-2x+7](https://tex.z-dn.net/?f=y%3D%5Cdfrac%20x2%2B2%5Ctext%7B%20and%20%7Dy%3D-2x%2B7)
![\implies\dfrac x2+2=-2x+7\implies 5x=10\implies x=2](https://tex.z-dn.net/?f=%5Cimplies%5Cdfrac%20x2%2B2%3D-2x%2B7%5Cimplies%205x%3D10%5Cimplies%20x%3D2)
![\implies y=-2\cdot2+7\implies y=3](https://tex.z-dn.net/?f=%5Cimplies%20y%3D-2%5Ccdot2%2B7%5Cimplies%20y%3D3)
So P is the point (2, 3).
(iii) For ASBC to be a parallelogram, we need to find the coordinates of S such that the line containing AS is parallel to the line containing BC, and the line containing SC is parallel to the line containing AB.
BC has slope (8 - 2)/(12 - 10) = 6/2 = 3. Then the line containing AS has equation
![y-(-1)=3(x-4)\implies y=3x-13](https://tex.z-dn.net/?f=y-%28-1%29%3D3%28x-4%29%5Cimplies%20y%3D3x-13)
We already know AB has slope 1/2. Then the line containing SC has equation
![y-8=\dfrac12(x-12)\implies y=\dfrac x2+2](https://tex.z-dn.net/?f=y-8%3D%5Cdfrac12%28x-12%29%5Cimplies%20y%3D%5Cdfrac%20x2%2B2)
S is the intersection of these two lines:
![3x-13=\dfrac x2+2\implies\dfrac{5x}2=15\implies5x=30\implies x=6](https://tex.z-dn.net/?f=3x-13%3D%5Cdfrac%20x2%2B2%5Cimplies%5Cdfrac%7B5x%7D2%3D15%5Cimplies5x%3D30%5Cimplies%20x%3D6)
![\implies y=3\cdot6-13\implies y=5](https://tex.z-dn.net/?f=%5Cimplies%20y%3D3%5Ccdot6-13%5Cimplies%20y%3D5)
So S is the point (6, 5), and *not* (2, -7) as the answer key suggests. In fact, (2, -7) is located lower than the point A and slightly to the left; if you draw that point and connect it to the other three, there's no way to get a parallelogram.
(iv) Find the lengths of AB, CD, and AP. Then the area is (AB + CD)*AP/2.
![AB=\sqrt{(4-10)^2+(-1-2)^2}=3\sqrt5](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%284-10%29%5E2%2B%28-1-2%29%5E2%7D%3D3%5Csqrt5)
![CD=\sqrt{(12-(-4))^2+(8-0)^2}=8\sqrt5](https://tex.z-dn.net/?f=CD%3D%5Csqrt%7B%2812-%28-4%29%29%5E2%2B%288-0%29%5E2%7D%3D8%5Csqrt5)
![AP=\sqrt{(4-2)^2+(-1-3)^2}=2\sqrt5](https://tex.z-dn.net/?f=AP%3D%5Csqrt%7B%284-2%29%5E2%2B%28-1-3%29%5E2%7D%3D2%5Csqrt5)
so the area is
![\dfrac{(3\sqrt5+8\sqrt5)2\sqrt5}2=\boxed{55}](https://tex.z-dn.net/?f=%5Cdfrac%7B%283%5Csqrt5%2B8%5Csqrt5%292%5Csqrt5%7D2%3D%5Cboxed%7B55%7D)