(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:
[tex]y-8=\dfrac12(x-12)\implies y=\dfrac x2+2\implies\boxed{2y=x+4}[/tex]
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
[tex]y-(-1) = -2(x-4)\implies \boxed{y=-2x+7}[/tex]
(ii) D lies on the x axis, so its y-coordinate is 0. Find the x-coordinate by plugging y = 0 into the equation for the line containing CD:
[tex]2\cdot0=x+4\implies x=-4[/tex]
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 x and y :
[tex]y=\dfrac x2+2\text{ and }y=-2x+7[/tex]
[tex]\implies\dfrac x2+2=-2x+7\implies 5x=10\implies x=2[/tex]
[tex]\implies y=-2\cdot2+7\implies y=3[/tex]
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
[tex]y-(-1)=3(x-4)\implies y=3x-13[/tex]
We already know AB has slope 1/2. Then the line containing SC has equation
[tex]y-8=\dfrac12(x-12)\implies y=\dfrac x2+2[/tex]
S is the intersection of these two lines:
[tex]3x-13=\dfrac x2+2\implies\dfrac{5x}2=15\implies5x=30\implies x=6[/tex]
[tex]\implies y=3\cdot6-13\implies y=5[/tex]
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.
[tex]AB=\sqrt{(4-10)^2+(-1-2)^2}=3\sqrt5[/tex]
[tex]CD=\sqrt{(12-(-4))^2+(8-0)^2}=8\sqrt5[/tex]
[tex]AP=\sqrt{(4-2)^2+(-1-3)^2}=2\sqrt5[/tex]
so the area is
[tex]\dfrac{(3\sqrt5+8\sqrt5)2\sqrt5}2=\boxed{55}[/tex]
