What is the image of (8, -6) after a reflection over the y=x Submit Answer

Mathematics

1 answer:

scoundrel [369]2 years ago

4 0

$\: ( \: \: \: 8 \: \: \: , \: \: \: - 6 \: \: \: )$

$y = x$

$\: ( \: \: \: - 6 \: \: \: , \: \: \: 8 \: \: \: )$

<h2>Just switch x and y</h2>

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5. Show that the following points are collinear. a) (1, 2), (4, 5), (8,9)

Irina-Kira [14]

Label the points A,B,C

A = (1,2)
B = (4,5)
C = (8,9)

Let's find the distance from A to B, aka find the length of segment AB.

We use the distance formula.

$A = (x_1,y_1) = (1,2) \text{ and } B = (x_2, y_2) = (4,5)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-4)^2 + (2-5)^2}\\\\d = \sqrt{(-3)^2 + (-3)^2}\\\\d = \sqrt{9 + 9}\\\\d = \sqrt{18}\\\\d = \sqrt{9*2}\\\\d = \sqrt{9}*\sqrt{2}\\\\d = 3\sqrt{2}\\\\$

Segment AB is exactly $3\sqrt{2}$ units long.

Now let's find the distance from B to C

$B = (x_1,y_1) = (4,5) \text{ and } C = (x_2, y_2) = (8,9)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(4-8)^2 + (5-9)^2}\\\\d = \sqrt{(-4)^2 + (-4)^2}\\\\d = \sqrt{16 + 16}\\\\d = \sqrt{32}\\\\d = \sqrt{16*2}\\\\d = \sqrt{16}*\sqrt{2}\\\\d = 4\sqrt{2}\\\\$

Segment BC is exactly $4\sqrt{2}$ units long.

Adding these segments gives

$AB+BC = 3\sqrt{2}+4\sqrt{2} = 7\sqrt{2}$

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Now if A,B,C are collinear then AB+BC should get the length of AC.

AB+BC = AC

Let's calculate the distance from A to C

$A = (x_1,y_1) = (1,2) \text{ and } C = (x_2, y_2) = (8,9)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-8)^2 + (2-9)^2}\\\\d = \sqrt{(-7)^2 + (-7)^2}\\\\d = \sqrt{49 + 49}\\\\d = \sqrt{98}\\\\d = \sqrt{49*2}\\\\d = \sqrt{49}*\sqrt{2}\\\\d = 7\sqrt{2}\\\\$