## Thursday, July 2, 2015

### Links to practice problems for distance between two points and area of a triangle defined by three points, one of which is (0, 0)

Here is the link to the practice problems.

More practice
Here are four points. Distances should be given in simplified square root form and approximated to the nearest thousandth,

(0, 0), (6, 3), (10, 1), (-4, -5)

a) Distance from (0, 0) to (6, 3)
b) Distance from (0, 0) to (10, 1)
c) Distance from (0, 0) to (-4, 5)
d) Distance from (6, 3) to (10, 1)
e) Distance from (6, 3) to (-4, -5)
f) Distance from (10, 1) to (-4, -5)

g) the area of the triangle defined by the points (0, 0), (6, 3) and (10, 1)
h) the area of the triangle defined by the points (6, 3), (10, 1) and
(-4, -5)

#### 1 comment:

Prof. Hubbard said...

a) Distance from (0, 0) to (6, 3)
sqrt((0-6)^2 + (0-3)^2) = sqrt(36+9) = sqrt(45) =
3*sqrt(5) ~= 6.708

b) Distance from (0, 0) to (10, 1)
sqrt((0-10)^2 + (0-1)^2) =
sqrt(101) ~= 10.050

c) Distance from (0, 0) to (-4, 5)
sqrt((0-[-4])^2 + (0-5)^2) = sqrt(16+25) =
sqrt(41) ~= 6.403

d) Distance from (6, 3) to (10, 1)
sqrt((6-10)^2 + (3-1)^2) = sqrt(16+4) = sqrt(20) =
2*sqrt(5) ~= 4.472

e) Distance from (6, 3) to (-4, -5)
sqrt((6-[-4])^2 + (3-[-5])^2) = sqrt(100+64) = sqrt(164) =
2*sqrt(41) ~= 12.806

f) Distance from (10, 1) to (-4, -5)
sqrt((10-[-4])^2 + (0-[-5])^2) = sqrt(196+25) =
sqrt(221) = ~= 14.866

g) the area of the triangle defined by the points (0, 0), (6, 3) and (10, 1)
1/2 * |6*1 - 3*10| = 1/2 * |6 - 30| = 1/2 |24| =
12

h) the area of the triangle defined by the points (6, 3), (10, 1) and (-4, -5)
We need one of the points to be (0, 0), so we will subtract (6, 3) from every point first.

(6-6, 3-3) = (0, 0)
(10-6, 1-3) = (4, -2)
(-4-6, -5-3) = (-10, -8)

1/2 * |4*-8 - -2*-10| = 1/2 * |-32 - 20| = 1/2 |-52| =
26