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.
For this case we have a square whose sides are known and equal to 60 ft. We want to find the diagonal of the square. For this, we use the Pythagorean theorem. We have then: Answer: from home to second base it is about:
