Given ΔABC with A(–3, 4), B(5, 8), and C(2, –2), write the equation of the line containing midsegment in standard form, where X is the midpoint of and Z is the midpoint of . (1 point) Responses 6x – 5y = –24 6 x – 5 y = –24 2x + 5y = 32 2 x + 5 y = 32 2x – 5y = –28 2 x – 5 y = –28 6x + 5y = 36 6 x + 5 y = 36
To find the midpoint of a line segment, we can use the midpoint formula:
Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Let's find the midpoint of segment AB, which we'll call X:
X = ((-3 + 5)/2, (4 + 8)/2)
X = (1, 6)
Now let's find the midpoint of segment BC, which we'll call Z:
Z = ((5 + 2)/2, (8 + -2)/2)
Z = (7/2, 6/2)
Z = (7/2, 3)
Now we have two points, X(1, 6) and Z(7/2, 3). We can use these two points to find the equation of the line containing the midsegment.
First, let's find the slope of the line:
slope = (y₂ - y₁)/(x₂ - x₁)
slope = (3 - 6)/(7/2 - 1)
slope = (-3)/(1/2)
slope = -6
Now we can use the point-slope form of the equation of a line, with point X(1, 6):
(y - y₁) = m(x - x₁)
(y - 6) = -6(x - 1)
y - 6 = -6x + 6
y = -6x + 12
Finally, we can rewrite this equation in standard form Ax + By = C:
6x + y = 12
Therefore, the equation of the line containing the midsegment is 6x + y = 12.
However, this option is not among the given possible answers.