Question

Hi :) how to do 10(iii) ?:)

Answer 1

(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:

$y-8=\dfrac12(x-12)\implies y=\dfrac x2+2\implies\boxed{2y=x+4}$

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

$y-(-1) = -2(x-4)\implies \boxed{y=-2x+7}$

(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:

$2\cdot0=x+4\implies x=-4$

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 x and y :

$y=\dfrac x2+2\text{ and }y=-2x+7$

$\implies\dfrac x2+2=-2x+7\implies 5x=10\implies x=2$

$\implies y=-2\cdot2+7\implies y=3$

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

$y-(-1)=3(x-4)\implies y=3x-13$

We already know AB has slope 1/2. Then the line containing SC has equation

$y-8=\dfrac12(x-12)\implies y=\dfrac x2+2$

S is the intersection of these two lines:

$3x-13=\dfrac x2+2\implies\dfrac{5x}2=15\implies5x=30\implies x=6$

$\implies y=3\cdot6-13\implies y=5$

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.

$AB=\sqrt{(4-10)^2+(-1-2)^2}=3\sqrt5$

$CD=\sqrt{(12-(-4))^2+(8-0)^2}=8\sqrt5$

$AP=\sqrt{(4-2)^2+(-1-3)^2}=2\sqrt5$

so the area is

$\dfrac{(3\sqrt5+8\sqrt5)2\sqrt5}2=\boxed{55}$