a) Use the slope formula to find the slopes of each line segment:
`m_(AB) = (y_A - y_B)/(x_A - x_B) = (0- 3)/(-2-2) = 3/4`
`m_(AC) = (y_A - y_C)/(x_A - x_C) = (0- (-1))/(-2-5) =- 1/7`
`m_(BC) = (y_B - y_C)/(x_B - x_C) = (3- (-1))/(2-5) = -4/3`
b) The slope with the largest magnitude is the slope of BC, `m_(BC) = -4/3` , so BC is the steepest line segment.
c) ABC is a right triangle because the segments AB and BC are perpendicular to each other, which means angle B is the right angle.
The segments AB and BC are perpendicular because their slopes are negative reciprocals of each other, that is, their product equals negative 1:
`m_(AB) * m_(BC) = (3/4) * (-4/3) = -1`
d) A line through C parallel to segment AB will have the slope equal to that of AB: `m = 3/4`
Since this line passes through the point C with coordinates (5, -1), this is its equation in point-slope form:
`y+1 = 3/4(x - 5)`
In slope-intercept form, this equation will be
`y = 3/4x - 15/4 - 1`
`y=3/4x - 19/4`
e) The line through B perpendicular to segment AC will have the slope that is negative reciprocal of the slope of AC, or
`m = -1/m_(AC) =-1/(-1/7) = 7`
Since this line passes through point B with the coordinates (2, 3),
its equation in point-slope form will be
y - 3 = 7(x - 2)
In slope-intercept form, this equation is
y = 7x - 14 + 3
y = 7x - 11