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 .%0D%0A%0D%0A(1 point)%0D%0AResponses%0D%0A%0D%0A6x – 5y = –24%0D%0A6 x – 5 y = –24%0D%0A%0D%0A2x + 5y = 32%0D%0A2 x + 5 y = 32%0D%0A%0D%0A2x – 5y = –28%0D%0A2 x – 5 y = –28%0D%0A%0D%0A6x + 5y = 36
To find the equation of the line containing the midsegment of ΔABC, we first need to find the coordinates of the midpoints X and Z.
The coordinates of the midpoint X can be found by taking the average of the x-coordinates and the average of the y-coordinates of points A and B:
X = ((-3 + 5) / 2, (4 + 8) / 2)
X = (1, 6)
The coordinates of the midpoint Z can be found by taking the average of the x-coordinates and the average of the y-coordinates of points A and C:
Z = ((-3 + 2) / 2, (4 - 2) / 2)
Z = (-0.5, 1)
Now that we have the coordinates of the midpoints X and Z, we can find the slope of the line passing through them. The slope formula is given by:
m = (y2 - y1) / (x2 - x1)
m = (1 - 6) / (-0.5 - 1)
m = -5 / (-1.5)
m = 5/1.5
m = 10/3
Now we can use the point-slope form of a line to find the equation. We will use the point (X) and the slope (m) we just found:
y - y1 = m(x - x1)
y - 6 = (10/3)(x - 1)
To write the equation in standard form, we distribute and rearrange the terms:
y - 6 = (10/3)x - 10/3
3(y - 6) = 10x - 10
3y - 18 = 10x - 10
10x - 3y = 8
Therefore, the equation of the line containing the midsegment of ΔABC in standard form is 10x - 3y = 8.