Answer: Choice C
4/(3*sqrt(3))
which is the same as writing [tex]\frac{4}{3\sqrt{3}}[/tex]
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Explanation:
We have the line [tex]\left(\sqrt{2}\right)x+y +2 = 0[/tex] in which points B and C are on this line. In other words, all of segment BC is on that line.
Point A is somewhere off that line as the diagram below indicates. We need to find the line through A that is perpendicular to the given line.
Recall that for any line of the standard form [tex]Px+Qy+R = 0\\\\[/tex], any perpendicular line will look like this: [tex]Qx-Py+S = 0\\\\[/tex]. Note the swap and sign change for the P,Q coefficients.
So anything perpendicular to [tex]\left(\sqrt{2}\right)x+y +2 = 0[/tex] will be of the form [tex]x-\left(\sqrt{2}\right)y+S = 0[/tex]. The sub-goal here is to find what S is equal to.
If you plugged in the origin point (x,y) = (0,0), then you should find that S = 0.
Therefore, the equation [tex]x-\left(\sqrt{2}\right)y = 0[/tex] is perpendicular to the given line and it passes through point A(0,0).
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The reason why we computed that perpendicular line is to help determine the point D as shown in the diagram below. This point is at the intersection of the green perpendicular lines. This is at the base of the triangle such that BC is the base.
To find the location of point D, you need to solve this system of equations
[tex]\begin{cases}\left(\sqrt{2}\right)x+y +2 = 0\\\\x-\left(\sqrt{2}\right)y = 0\end{cases}[/tex]
Let's isolate x in the second equation
[tex]x-\left(\sqrt{2}\right)y = 0\\\\x=\left(\sqrt{2}\right)y\\\\[/tex]
Then substitute that into the other equation. Solve for y.
[tex]\left(\sqrt{2}\right)x+y +2 = 0\\\\\left(\sqrt{2}\right)*\left(\sqrt{2}\right)y+y +2 = 0\\\\2y+y +2 = 0\\\\3y +2 = 0\\\\3y = -2\\\\y = -2/3\\\\[/tex]
Now we can find x like so
[tex]x=\left(\sqrt{2}\right)y\\\\x=\left(\sqrt{2}\right)*(-2/3)\\\\x=\frac{-2\sqrt{2}}{3}\\\\[/tex]
The point D is located at the exact coordinates of
[tex]D=(x,y) = \left(\frac{-2\sqrt{2}}{3},\frac{-2}{3}\right)\\\\[/tex]
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Next, we use the distance formula to compute the distance between point A and point D.
I'll use lowercase d to represent distance.
[tex]A = (x_1,y_1) = (0,0)\\\\D = (x_2,y_2) = \left(\frac{-2\sqrt{2}}{3},\frac{-2}{3}\right)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{\left(0-\left(\frac{-2\sqrt{2}}{3}\right)\right)^2 + \left(0-\left(\frac{-2}{3}\right)\right)^2}\\\\d = \sqrt{\left(\frac{2\sqrt{2}}{3}\right)^2 + \left(\frac{2}{3}\right)^2}\\\\d = \sqrt{\frac{8}{9} + \frac{4}{9}}\\\\d = \sqrt{\frac{12}{9}}\\\\d = \sqrt{\frac{4}{3}}\\\\d = \frac{\sqrt{4}}{\sqrt{3}}\\\\d = \frac{2}{\sqrt{3}}\\\\[/tex]
This represents the exact length of segment AD.
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Since triangle ABC is equilateral, this means triangles ADB and ADC are identical mirrored copies of each other. Those two smaller triangles are equal in area and combine to form triangle ABC back again. Furthermore, they are right triangles. This last fact mentioned will allow us to use the pythagorean theorem like so
[tex]a^2+b^2 = c^2\\\\\left(\text{AD}\right)^2+\left(\text{DC}\right)^2=\left(\text{AC}\right)^2\\\\\left(\frac{2}{\sqrt{3}}\right)^2+\left(\frac{x}{2}\right)^2=\left(x\right)^2\\\\[/tex]
where x represents the lengths of AB, BC and AC.
After skipping a bit of steps, solving that said equation should lead to these two solutions
[tex]x = -\frac{4}{3}, \ x = \frac{4}{3}[/tex]
Because x is a side length, we ignore the negative solution.
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We can now compute the area of triangle ABC
[tex]\text{area of triangle} = \frac{1}{2}*\text{base}*\text{height}\\\\\text{area of triangle ABC} = \frac{1}{2}*\text{BC}*\text{AD}\\\\\text{area of triangle ABC} = \frac{1}{2}*\frac{4}{3}*\frac{2}{\sqrt{3}}\\\\\text{area of triangle ABC} = \frac{1*4*2}{2*3*\sqrt{3}}\\\\\text{area of triangle ABC} = \frac{8}{6\sqrt{3}}\\\\\text{area of triangle ABC} = \frac{4}{3\sqrt{3}}\\\\[/tex]
which points us to choice C as the final answer.
When writing this on a keyboard, we would write "4/(3*sqrt(3))" without quotes. The use of parenthesis is needed to ensure we divide by all of 3*sqrt(3). Otherwise, the computer might mistake it for [tex]\frac{4}{3}\sqrt{3}[/tex] which is entirely different.
