The coordinates of the midpoint X of AB can be found by taking the average of the x-coordinates and the y-coordinates of A and B.
X = ((-3 + 5)/2, (4 + 8)/2)
X = (1, 6)
Similarly, the coordinates of the midpoint Z of BC can be found by taking the average of the x-coordinates and the y-coordinates of B and C.
Z = ((5 + 2)/2, (8 - 2)/2)
Z = (7/2, 6/2)
Z = (7/2, 3)
The slope of the line containing the midsegment XZ can be found by taking the difference in y-coordinates over the difference in x-coordinates.
slope = (3 - 6)/(7/2 - 1)
slope = -3/(5/2)
slope = -3 * 2/5
slope = -6/5
Using the point-slope form of a linear equation, we can write the equation of the line containing XZ as:
y - 6 = (-6/5)(x - 1)
Multiplying through by 5 to eliminate the fraction, we get:
5y - 30 = -6x + 6
Finally, rearranging the equation to standard form, we have:
6x + 5y = 36
So, the correct equation is:
6x + 5y = 36
Given ΔABC with A(–3, 4), B(5, 8), and C(2, –2), write the equation of the line containing midsegment line XZ in standard form, where X is the midpoint of line ABand Z is the midpoint of line BC.
(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
1 answer
1 answer