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)
6x – 5y = –24
2x + 5y = 32
2x – 5y = –28
6x + 5y = 36
To find the equation of the midsegment line, we first need to find the midpoints of segments AX and AC.
Midpoint of segment AX:
X( ( -3 + 5 ) / 2 , ( 4 + 8 ) / 2 )
X( 1 , 6 )
Midpoint of segment AC:
Z( ( -3 + 2 ) / 2 , ( 4 - 2 ) / 2 )
Z( -0.5, 1 )
Now, we can find the equation of the line passing through points X and Z.
Slope of line XZ:
m = (6 - 1) / (1 - (-0.5))
m = 5 / 1.5
m = 10 / 3
Using point-slope form:
y - y1 = m(x - x1)
y - 6 = (10/3)(x - 1)
y - 6 = (10/3)x - 10/3
3(y - 6) = 10x - 10
3y - 18 = 10x - 10
3y - 10x = 8
10x - 3y = -8
Therefore, the equation of the line containing the midsegment of triangle ABC is:
10x - 3y = -8.