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Shalnov [3]
3 years ago
6

What’s the midpoint of the line segment joining A and B A(2,-5);B(6,1)

Mathematics
1 answer:
IRISSAK [1]3 years ago
7 0

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{2}~,~\stackrel{y_1}{-5})\qquad B(\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{6+2}{2}~~,~~\cfrac{1-5}{2} \right)\implies \left( \cfrac{8}{4}~,~\cfrac{-4}{2} \right)\implies (4,-2)

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Question:
pishuonlain [190]
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Which classification describes AMNO with vertices M(2, -3), N(3, 1), and<br> 0(-3, 1)?
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Option D

Step-by-step explanation:

32). Given vertices of the triangle are M(2, -3), N(3, 1) and O(-3. 1).

Distance between two points (x_1,y_1) and (x_2,y_2) is given by the expression,

Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Distance between M(2, -3) and N(3, 1) will be,

MN = \sqrt{(3-2)^2+(1+3)^2}

      = \sqrt{1+16}

      = \sqrt{17}

Distance between M(2, -3) and O(-3, 1),

MO = \sqrt{(2+3)^2+(-3-1)^2}

      = \sqrt{25+16}

      = \sqrt{41}

Distance between N(3, 1) and O(-3, 1),

NO = \sqrt{(3+3)^2+(1-1)^2}

      = 6

Condition for right triangle,

c² = a² + b² [Here c is the longest side of the triangle]

By this property,

MO² = MN² + NO²

(\sqrt{41})^2=(\sqrt{17})^2+6^2

41 = 17 + 36

41 = 51

False.

Therefore, given triangle is not a right triangle.

Since, length of all sides are not equal, given triangle will be a scalene triangle.

Option D is the correct option.

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