Question

Given ΔABC 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.
6x – 5y = –24
2 x + 5 y = 32
2 x – 5 y = –28
6x + 5y = 36

Answer 1

To find the equation of the line containing midsegment XZ, we first need to find the coordinates of X and Z.

The midpoint of AB, X, can be found by finding the average of the x-coordinates and the average of the y-coordinates of A and B:
X(x-coordinate) = (x-coordinate of A + x-coordinate of B)/2 = (-3 + 5)/2 = 1/2
X(y-coordinate) = (y-coordinate of A + y-coordinate of B)/2 = (4 + 8)/2 = 12/2 = 6
So, X is (1/2, 6).

The midpoint of BC, Z, can be found by finding the average of the x-coordinates and the average of the y-coordinates of B and C:
Z(x-coordinate) = (x-coordinate of B + x-coordinate of C)/2 = (5 + 2)/2 = 7/2
Z(y-coordinate) = (y-coordinate of B + y-coordinate of C)/2 = (8 + (-2))/2 = 6/2 = 3
So, Z is (7/2, 3).

Now, we can use the coordinates of X and Z to find the equation of the line containing XZ.

First, we find the slope of the line using the formula: slope = (change in y)/(change in x)
slope = (y-coordinate of Z - y-coordinate of X)/(x-coordinate of Z - x-coordinate of X)
slope = (3 - 6)/(7/2 - 1/2) = -3/(6/2) = -3/3 = -1

Next, we can use the slope-intercept form of a line (y = mx + b) to find the equation. We use the point-slope form using the coordinates of X:
y - y-coordinate of X = slope(x - x-coordinate of X)
y - 6 = -1(x - 1/2)
y - 6 = -x + 1/2
y = -x + 1/2 + 6
y = -x + 13/2

To convert this equation to standard form, we multiply both sides of the equation by 2 to eliminate fractions:
2y = -2x + 13

Finally, we rearrange the equation to match the standard form (Ax + By = C) by moving all the terms to one side:
2x + 2y = 13

So, the equation of the line containing midsegment XZ in standard form is 2x + 2y = 13.