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)

answers in the comments.

## 1 comment:

a) Distance from (0, 0) to (6, 3)

sqrt((0-6)^2 + (0-3)^2) = sqrt(36+9) = sqrt(45) =

3*sqrt(5) ~= 6.708b) Distance from (0, 0) to (10, 1)

sqrt((0-10)^2 + (0-1)^2) =

sqrt(101) ~= 10.050c) Distance from (0, 0) to (-4, 5)

sqrt((0-[-4])^2 + (0-5)^2) = sqrt(16+25) =

sqrt(41) ~= 6.403d) Distance from (6, 3) to (10, 1)

sqrt((6-10)^2 + (3-1)^2) = sqrt(16+4) = sqrt(20) =

2*sqrt(5) ~= 4.472e) Distance from (6, 3) to (-4, -5)

sqrt((6-[-4])^2 + (3-[-5])^2) = sqrt(100+64) = sqrt(164) =

2*sqrt(41) ~= 12.806f) Distance from (10, 1) to (-4, -5)

sqrt((10-[-4])^2 + (0-[-5])^2) = sqrt(196+25) =

sqrt(221) = ~= 14.866g) 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| =

12h) 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| =

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