Given AABC with A(-2, 2), B(2,4), and C(4, 4), write the equation of the line containing
midsegmenty7 in standard form, where X is the midpoint of AB and Z is the midpoint of BC
x-y=3
y-x=3
x + y = -3
x+ y=3
X = (0,3)
Z = (3,4)
the slope of XZ is 1/3
None of the choices has that slope, so I suspect a typo
For sure I don't know what "midsegmenty7" means
To find the equation of the line containing the midsegment, we first need to find the coordinates of the midpoints X and Z.
The midpoint of a line segment is calculated by taking the average of the x-coordinates and the average of the y-coordinates.
Midpoint X:
Coordinates of A: (-2, 2)
Coordinates of B: (2, 4)
X-coordinate of midpoint X: (x₁ + x₂)/2 = (-2 + 2)/2 = 0/2 = 0
Y-coordinate of midpoint X: (y₁ + y₂)/2 = (2 + 4)/2 = 6/2 = 3
Therefore, the coordinates of midpoint X of AB are (0, 3).
Midpoint Z:
Coordinates of B: (2, 4)
Coordinates of C: (4, 4)
X-coordinate of midpoint Z: (x₁ + x₂)/2 = (2 + 4)/2 = 6/2 = 3
Y-coordinate of midpoint Z: (y₁ + y₂)/2 = (4 + 4)/2 = 8/2 = 4
Therefore, the coordinates of midpoint Z of BC are (3, 4).
Now that we know the coordinates of midpoint X and midpoint Z, we can find the equation of the line passing through them.
Slope of the line = (y₂ - y₁)/(x₂ - x₁) = (4 - 3)/(3 - 0) = 1/3
Using the slope-intercept form of a line (y = mx + b), we can substitute one of the midpoints and the slope to find the y-intercept.
Using midpoint X (0, 3) and the slope 1/3:
3 = (1/3)(0) + b
3 = 0 + b
b = 3
So the equation of the line is y = (1/3)x + 3.
Converting it into standard form, we can multiply both sides of the equation by 3 to eliminate the fraction and rearrange the terms:
3y = x + 9
x - 3y = -9
Therefore, the equation of the line containing the midsegment is x - 3y = -9.
To find the equation of the line containing the midsegment, we first need to find the coordinates of point X and point Z.
The coordinates of point 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-coordinate of X = (x-coordinate of A + x-coordinate of B) / 2 = (-2 + 2) / 2 = 0 / 2 = 0
y-coordinate of X = (y-coordinate of A + y-coordinate of B) / 2 = (2 + 4) / 2 = 6 / 2 = 3
Therefore, point X has coordinates (0, 3).
Similarly, the coordinates of point Z can be found by taking the average of the x-coordinates and the average of the y-coordinates of points B and C.
x-coordinate of Z = (x-coordinate of B + x-coordinate of C) / 2 = (2 + 4) / 2 = 6 / 2 = 3
y-coordinate of Z = (y-coordinate of B + y-coordinate of C) / 2 = (4 + 4) / 2 = 8 / 2 = 4
Therefore, point Z has coordinates (3, 4).
Now we can find the equation of the line passing through point X and Z.
First, let's find the slope of the line using the formula:
slope = (y2 - y1) / (x2 - x1)
slope = (4 - 3) / (3 - 0) = 1 / 3
Since the line passes through point X(0, 3), we can use the point-slope form of the equation:
y - y1 = m(x - x1)
y - 3 = (1/3)(x - 0)
y - 3 = 1/3x
To convert this equation to standard form, we need to eliminate fractions:
Multiply both sides of the equation by 3 to get rid of the fraction:
3(y - 3) = 1(x)
3y - 9 = x
Rearrange the equation to have x and y on the same side and move the constant to the other side:
x - 3y = 9
Therefore, the equation of the line containing the midsegment is x - 3y = 9, which is in standard form.