
Steve S. answered 04/02/14
Tutor
5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus
Consider the points (1,6) and (3,-2).
(a.) State the midpoint of the line segment with the given endpoints.
M((1+3)/2,(6+(-2))/2) = M(2,2)
(b.) Find the distance between the points. Show some work. Give the exact answer (involving a radical), simplify the radical as much as possible, and also state an approximation to three decimal places.
Graph the points and draw vertical and horizontal lines through them forming right triangles. Use either right triangle to substitute into the Pythagorean Theorem:
d^2 = (3-1)^2 + (6-(-2))^2 = 4 + 144 = 148
= 2*74 = 2*2*37 = 2^2*37
Take square root of both sides (distance is always positive):
d = 2√(37) ≈ 12.16552506059644 ≈ 12.166
(c.)Find the slope-intercept equation of the line passing through the two given points. Show work.
Define a general third point on the line: (x,y). Then
slope, m = Δy/Δx = (y-6)/(x-1) = (-2-6)/(3-1) = -4
Multiply by (x-1):
y-6 = -4(x-1) <== Point-Slope Form
Solve for y:
y-6 = -4x + 4
y = -4x + 10 <== Slope-Intercept Form
(d) If a line is perpendicular to your line in part (c), what would the slope of that line be? (no work/explanation required)
-1/(-4) = 1/4
(a.) State the midpoint of the line segment with the given endpoints.
M((1+3)/2,(6+(-2))/2) = M(2,2)
(b.) Find the distance between the points. Show some work. Give the exact answer (involving a radical), simplify the radical as much as possible, and also state an approximation to three decimal places.
Graph the points and draw vertical and horizontal lines through them forming right triangles. Use either right triangle to substitute into the Pythagorean Theorem:
d^2 = (3-1)^2 + (6-(-2))^2 = 4 + 144 = 148
= 2*74 = 2*2*37 = 2^2*37
Take square root of both sides (distance is always positive):
d = 2√(37) ≈ 12.16552506059644 ≈ 12.166
(c.)Find the slope-intercept equation of the line passing through the two given points. Show work.
Define a general third point on the line: (x,y). Then
slope, m = Δy/Δx = (y-6)/(x-1) = (-2-6)/(3-1) = -4
Multiply by (x-1):
y-6 = -4(x-1) <== Point-Slope Form
Solve for y:
y-6 = -4x + 4
y = -4x + 10 <== Slope-Intercept Form
(d) If a line is perpendicular to your line in part (c), what would the slope of that line be? (no work/explanation required)
-1/(-4) = 1/4