Given ∆ABC with A(-3, 4), B(5, 8), and C(2, -2), write the equation of the line containing midsegment XZ in standard form, where X is the midpoint of AB and Z is the midpoint of BC.
To find the midsegment, we need to find the coordinates of the midpoint of AB (X) and BC (Z).
The coordinates of the midpoint of AB (X) are:
X = ((x1 + x2)/2, (y1 + y2)/2)
X = ((-3 + 5)/2, (4 + 8)/2)
X = (2/2, 12/2)
X = (1, 6)
The coordinates of the midpoint of BC (Z) are:
Z = ((x2 + x3)/2, (y2 + y3)/2)
Z = ((5 + 2)/2, (8 + -2)/2)
Z = (7/2, 6/2)
Z = (7/2, 3)
Now we need to find the equation of the line passing through X and Z.
First, let's find the slope of the line passing through X and Z.
m = (y2 - y1)/(x2 - x1)
m = (3 - 6)/(7/2 - 1)
m = (-3)/(7/2 - 2/2)
m = (-3)/(5/2)
m = -6/5
Now that we have the slope, we can find the equation of the line using the formula y - y1 = m(x - x1), where (x1, y1) = (1, 6).
y - 6 = (-6/5)(x - 1)
To convert the equation to standard form, we need to multiply through by 5 to clear the fraction and rearrange the terms:
5y - 30 = -6x + 6
6x + 5y = 6 + 30
6x + 5y = 36
Therefore, the equation of the line containing midsegment XZ in standard form is 6x + 5y = 36.