To draw the median of the triangle from vertex A, the mid point of BC must be determined. The median of the vertex A is given at (-1/2, 1). See explanation below.
<h3>How you would draw the median of the triangle from vertex A?</h3>
Recall that B = (3, 7)
and C = (-4, -5).
- Note that when you are given coordinates in the format above, B or C = (x, y)
- Hence the mid point of line BC is point D₁ which is derived as:
D₁ 
, ![(\frac{7-5}{2}) ]](https://tex.z-dn.net/?f=%28%5Cfrac%7B7-5%7D%7B2%7D%29%20%5D)
- hence, the Median of the Vertex A = (-1/2, 1).
Connecting D' and A gives us the median of the vertex A. See attached graph.
<h3>What is the length of the median from C to AB?</h3>
Recall that
A → (4, 2); and
B → (3, 7)
Hence, the Midpoint will be
, ![(\frac{2+7}{2} )]](https://tex.z-dn.net/?f=%28%5Cfrac%7B2%2B7%7D%7B2%7D%20%29%5D)
→ 
Recall that
C → (-4, 5)
Hence,
= ![\sqrt{[(-4 -\frac{7}{2} })^{2} + (-5-\frac{9}{2} )^{2} ]](https://tex.z-dn.net/?f=%5Csqrt%7B%5B%28-4%20-%5Cfrac%7B7%7D%7B2%7D%20%7D%29%5E%7B2%7D%20%20%2B%20%28-5-%5Cfrac%7B9%7D%7B2%7D%20%29%5E%7B2%7D%20%5D)
Simplified, the above becomes
= √(586)/2)
= 24.2074/2
= 12.1037
The length of the Median from C to AB ≈ 12
Learn more about Vertex at;
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Your answer would be 3160.
This problem can be solved by writing a conditional equation.
(8*1.5)+6= d
12+6= d
d= 18GB
Check- 18-12= 6GB
to find the equation of a linear equation given two points:
1) find the slope y=mx+b
here's the equation: (y2-y1)/(x2-x1)
y2- y value of the 2nd point y1- y value of 1st point
x2- x value of 2nd point x1- x value of 1st point
1st point- (0, 12) 2nd point- (1,3)
so your slope equation is this: (3-12) / (1-0) or -9/1 or -9
m is -9
2) find the y intercept y=mx+b
b is whatever y is when x is 0. There's a long way to find it, but you already have it, because one of your points is (0,12). They told you what y was when x was 0.
b is 12.
your linear equation is y=-9x+12
to find the equation of an exponential equation given two points:
1) find a y=a(b)<span>×
</span>if one of your points has a 0 as the x value, the y value is a. you do have a point like this: (0,12) so...
<span>a is 12
</span>2) find b y=a(b)<span>×
put what you know a is in the equation: y=12(b)</span><span>×
then put the x and y values of the other point in for x and y in the equation
so (1,3) 1 is x and 3 is y
3=12(b)^1
when the exponent is 1, it disappears:
3=12b
simplify: b= 3/12 or b=1/4
then put all that in the equation:
y=12(3/4)</span><span>×</span>
