3 years ago

Find the distance between the two points in simplest radical form. (5, -6) and (7, 1)

Mathematics

1 answer:

Naily [24]3 years ago

5 0

Distance: sqr (x1-x2)^2 + (y1-y2)^2
Sqr (5-7)^2 + (-6-1)^2
Sqr (-2)^2 + (-7)^2
Sqr (4) + (49)
Sqr (53)
= 7.28

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Correct answer gets brainliest

Ilia_Sergeevich [38]

Answer:

sides: isosceles
angles: right

Step-by-step explanation:

At first, lets find the distance from one point to the other.

$d(A,B)=\sqrt{(x_{A}-x_{B} )^2 +(y_{A}-y_{B} )^2 } \\d(A,B)=\sqrt{(2-0 )^2 +(-2-4 )^2 } \\d(A,B)=\sqrt{4 +36 } = \sqrt{40 } = \sqrt{4 * 10 } = 2\sqrt{10}$

The same way, you can find that:

$d(A,C)= 2\sqrt{10}\\d(A,B)= 2\sqrt{20}$

Since the triangle has only its two sides equal to each other (AB = AC),

it's an isosceles one.

As long as the angles are concerned, we gotta "cheat". In other words, we see that:

$AB^2 + AC^2 = (\sqrt{40})^2 + (\sqrt{40})^2 = 40 + 40 = 80\\ AB^2 + AC^2 = 80 = (2\sqrt{20})^2 = BC^2\\AB^2 + AC^2 = BC^2$

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