The midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9) is 
<u>Solution:</u>
Given, two points are (-6, 6) and (-3, -9)
We have to find the midpoint of the segment formed by the given points.
The midpoint of a segment formed by 
is given by:
Plugging in the values in formula, we get,
Hence, the midpoint of the segment is
to find the x-intercept of a function, we simply set y = 0 and then solve for "x", so, let's first find the equation of it and then set y = 0.
![\bf (\stackrel{x_1}{-12}~,~\stackrel{y_1}{16})~\hspace{10em} slope = m\implies-\cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-16=-\cfrac{2}{3}[x-(-12)] \\\\\\ y-16=-\cfrac{2}{3}(x+12)\implies \stackrel{\stackrel{y}{\downarrow }}{0}-16=-\cfrac{2}{3}x-8\implies -8=-\cfrac{2x}{3} \\\\\\ -24=-2x\implies \cfrac{-24}{-2}=x\implies 12=x \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (12,0) ~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B-12%7D~%2C~%5Cstackrel%7By_1%7D%7B16%7D%29~%5Chspace%7B10em%7D%20slope%20%3D%20m%5Cimplies-%5Ccfrac%7B2%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-16%3D-%5Ccfrac%7B2%7D%7B3%7D%5Bx-%28-12%29%5D%20%5C%5C%5C%5C%5C%5C%20y-16%3D-%5Ccfrac%7B2%7D%7B3%7D%28x%2B12%29%5Cimplies%20%5Cstackrel%7B%5Cstackrel%7By%7D%7B%5Cdownarrow%20%7D%7D%7B0%7D-16%3D-%5Ccfrac%7B2%7D%7B3%7Dx-8%5Cimplies%20-8%3D-%5Ccfrac%7B2x%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20-24%3D-2x%5Cimplies%20%5Ccfrac%7B-24%7D%7B-2%7D%3Dx%5Cimplies%2012%3Dx%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%2812%2C0%29%20~%5Chfill)
Well let's see.
The equation has exactly one solution.
Hope this helps.
This revolves around exact trig values - no easy way to say this, you just need to memorise them. They are there for sin cos and tan, but I will give you the main tan ones below - note this is RADIANS (always work in them when you can, everything is better):
tan0: 0
tanpi/6: 1/sqrt(3)
tanpi/4: 1
tanpi/3: sqrt(3)
tanpi/2: undefined
Now we just need to equate -2pi/3 to something we understand. 2pi/3 is 1/3 of the way round a circle, so -2pi/3 is 1/3 of the way round the circle going backwards (anticlockwise), so on a diagram we already know it's in the third quadrant of the circle (somewhere between pi and 3pi/2 rads).
We also know it is pi/3 away from pi, so we are looking at sqrt(3) or -sqrt(3) because of those exact values.
Now we just need to work out if it's positive or negative. You can look up a graph of tan and it'll show that the graph intercepts y at (0,0) and has a period of pi rads. Therefore between pi and 3pi/2 rads, the values of tan are positive. Therefore, this gives us our answer of sqrt(3).
Answer:
D
when you look at it it is upside down so d is the right answer
