1. Show that ABCD is a trapezoid.
(A: (-2, 5); B: (4, 5); C: (6,2); D: (2, -1)).
2. Write an equation in slope-intercept form for the line that passes through (5,0) and is perpendicular to the line given by y = -5/2x + 6.
find the slopes of AB, BC, CD, DA
2. slope of line=2/5
y=2/5 x+b
0=2/5 * 5 + b
b=-2
y=2/5 x -2
1. To show that ABCD is a trapezoid, we need to check if it satisfies the definition of a trapezoid. A trapezoid is a quadrilateral with exactly one pair of parallel sides.
To determine if ABCD is a trapezoid, we can find the slopes of the line segments AB, BC, CD, and AD:
Slope of AB: (y2 - y1) / (x2 - x1) = (5 - 5) / (4 - (-2)) = 0 / 6 = 0
Slope of BC: (y2 - y1) / (x2 - x1) = (2 - 5) / (6 - 4) = -3 / 2
Slope of CD: (y2 - y1) / (x2 - x1) = (-1 - 2) / (2 - 6) = -3 / -4 = 3 / 4
Slope of AD: (y2 - y1) / (x2 - x1) = (-1 - 5) / (2 - 4) = -6 / (-2) = 3
Now, we can see that the slopes of AB and CD are both 0, which means AB and CD are parallel. Therefore, ABCD satisfies the definition of a trapezoid.
2. To find an equation in slope-intercept form for a line that is perpendicular to y = -5/2x + 6 and passes through (5,0), we need to determine the slope of the perpendicular line.
First, we find the slope of the given line by comparing it to the slope-intercept form: y = mx + b. In this case, the given line can be rewritten as y = -5/2x + 6, which indicates that its slope is -5/2.
To find the slope of the perpendicular line, we use the fact that the product of the slopes of two perpendicular lines is -1. So, the slope of the perpendicular line is the negative reciprocal of -5/2, which is 2/5.
Next, we can use the point-slope form of a line y - y1 = m(x - x1), where (x1, y1) is the given point (5,0) and m is the slope we just found (2/5).
Substituting the values, we get:
y - 0 = (2/5)(x - 5)
y = (2/5)x - (2/5)(5)
y = (2/5)x - 2
Therefore, the equation in slope-intercept form for the line that passes through (5,0) and is perpendicular to y = -5/2x + 6 is y = (2/5)x - 2.