Perhaps the easiest way to find the midpoint between two given points is to average their coordinates: add them up and divide by 2.
A) The midpoint C' of AB is
.. (A +B)/2 = ((0, 0) +(m, n))/2 = ((0 +m)/2, (0 +n)/2) = (m/2, n/2) = C'
The midpoint B' is
.. (A +C)/2 = ((0, 0) +(p, 0))/2 = (p/2, 0) = B'
The midpoint A' is
.. (B +C)/2 = ((m, n) +(p, 0))/2 = ((m+p)/2, n/2) = A'
B) The slope of the line between (x1, y1) and (x2, y2) is given by
.. slope = (y2 -y1)/(x2 -x1)
Using the values for A and A', we have
.. slope = (n/2 -0)/((m+p)/2 -0) = n/(m+p)
C) We know the line goes through A = (0, 0), so we can write the point-slope form of the equation for AA' as
.. y -0 = (n/(m+p))*(x -0)
.. y = n*x/(m+p)
D) To show the point lies on the line, we can substitute its coordinates for x and y and see if we get something that looks true.
.. (x, y) = ((m+p)/3, n/3)
Putting these into our equation, we have
.. n/3 = n*((m+p)/3)/(m+p)
The expression on the right has factors of (m+p) that cancel*, so we end up with
.. n/3 = n/3 . . . . . . . true for any n
_____
* The only constraint is that (m+p) ≠ 0. Since m and p are both in the first quadrant, their sum must be non-zero and this constraint is satisfied.
The purpose of the exercise is to show that all three medians of a triangle intersect in a single point.
Answer:
2(x² - 2)(x² - 8)
Step-by-step explanation:
Given
2
- 20x² + 32 ← factor out 2 from each term
= 2( - 10x² + 32)
let u = x², then
- 10x² + 32 = u² - 10u + 32 = (u - 2)(u - 8)
change u back to x²
- 10x² + 32 = (x² - 2)(x² - 8) and
2 - 20x² + 32 = 2(x² - 2)(x² - 8) ← in factored form
Replace each letter for the value it's worth, so:
= 10 * -4 * 5
= -200
Answer: First find the GCF ( greatest Common Factor):-
GCF of 16, 24 and 48 is 8
GCF of the x terms is x^3
GCF of the y terms is y^4
So GCF is 8x^3y^4
Factoring:-
= 8x^3y^4 ( 2x^6y^4 - 16x^2y^2 + 3) answer
Step-by-step explanation:
