# Question #9f8d5

Sep 21, 2015

B (-11, 0), D (13, 12), and E (1, 6).

#### Explanation:

The steps:
1. find equation of line (AC)
2. find coordinates of point E.
3. find equation of line (BD) passing through E.
4. find coordinates of B and D (note, 3 and 4 goes together)

First, find the equation of the line that passes through A and C.
A is at (-1, 10)
C is at (3, 2)
So the equation of the line (in the form $y = a x + b$ where $a$ and $b$ are the unknown) is:
(1): $10 = - a + b$
and
(2): $2 = 3 a + b$

Then solve for $b$ and $a$. (1) minus (2) gives:
$10 - 2 = - a + b - 3 a - b$
which becomes:
$8 = - 4 a$
i.e. $a = - 2$
Plugging this back in (1), we get:
$10 = 2 + b$
i.e. $b = 8$
So the equation of the line (AC) is $y = - 2 x + 8$.

With this, we can find the coordinates of point E,
because it is the middle of the segment [AC].
$\Delta x = {x}_{C} - {x}_{A} = 3 - \left(- 1\right) = 4$ divided by two is 2. Then ${x}_{E} = {x}_{A} + \frac{\Delta x}{2} = - 1 + 2 = 1$

Similarly,
$\Delta y = {y}_{C} - {y}_{A} = 2 - 10 = - 8$ divided by two is -4. Then
${y}_{E} = {y}_{A} + \frac{\Delta y}{2} = 10 - 4 = 6$
Thus, point E has coordinates (1, 6).

Now, we want the equation of the line (BD), again in the form $y = a ' x + b '$.
We know that the two diagonals are perpendicular to each other because it is one of the properties of a rhombus (something you should know).
So we know that the product of their slopes should be -1 (that's also something you should know).
That is to say,
$a \cdot a ' = - 1$ but since we know $a = - 2$, we have:
$a ' = \frac{1}{2}$

But remember that this line also passes through E, so we have:
$6 = \frac{1}{2} \cdot 1 + b '$
and solve for $b '$:
$b ' = 6 - \frac{1}{2} = \frac{11}{2}$

So the equation of line (BD) is:
$y = \frac{1}{2} x + \frac{11}{2}$

We can now get the coordinates of B.
We need to use the extra piece of information given in the text.
They say "the point B lies on the x-axis".
That is to say, "the coordinates of point B is (${x}_{B}$, 0)".
Applying this to the equation of line (BD):
$0 = \frac{1}{2} {x}_{B} + \frac{11}{2}$
and solve for ${x}_{B}$, we get:
${x}_{B} = - 11$
Point B has coordinates (-11, 0).

Now, we can get the coordinates of point D.
Since E is the midpoint of segment [BD],
D is twice as far from B than E.
$\Delta x = {x}_{E} - {x}_{B} = 1 - \left(- 11\right) = 12$
multiplied by 2 is 24. So,
${x}_{D} = {x}_{B} + 2 \Delta x = - 11 + 24 = 13$.

Similarly,
$\Delta y = {y}_{E} - {y}_{B} = 6 - 0 = 6$
multiplied by 2 is 12. So,
${y}_{D} = {y}_{B} + 2 \Delta y = 0 + 12 = 12$
Point D has coordinates (13, 12).