The points A 0-6- B -5-3 and C 3-1 are the vertices of a triangle which isA- ScaleneB- EquilateralC- None of theseD- Isosceles

The correct option is D IsoscelesAB^2=((0 ( 5))^2 + (6 3)^2)) = (5^2+3^2)=34, BC^2=(( 5 3)^2 + (3 1)^2)) = { ( 8)^2+2^2 }=68, and AC^2=( 3)^2+(5)^2=34 ∴ AB=AC ...

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Mathematics

Distance between two points using Pythagoras theorem

The points A ...

Question

The points A(0, 6), B(-5, 3) and C(3, 1) are the vertices of a triangle which is

Isosceles

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Equilateral

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Scalene

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None of these

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Solution

The correct option is A Isosceles
$A B^{2} = ((0 - (- 5))^{2} + (6 - 3)^{2})) = (5^{2} + 3^{2}) = 34, B C^{2} = ((- 5 - 3)^{2} + (3 - 1)^{2})) = {(- 8)^{2} + 2^{2}} = 68, and A C^{2} = (- 3)^{2} + (5)^{2} = 34$
$∴ A B = A C$ .
Hence, $Δ A B C$ is isoscles.

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The points A(0, 6), B(-5, 3) and C(3, 1) are the vertices of a triangle which is

The correct option is A IsoscelesAB2=((0−(−5))2+(6−3)2)) = (52+32)=34,BC2=((−5−3)2+(3−1)2)) = {(−8)2+22}=68,and AC2=(−3)2+(5)2=34 ∴ AB=AC. Hence, ΔABC is isoscles.

The correct option is A Isosceles
$A B^{2} = ((0 - (- 5))^{2} + (6 - 3)^{2})) = (5^{2} + 3^{2}) = 34, B C^{2} = ((- 5 - 3)^{2} + (3 - 1)^{2})) = {(- 8)^{2} + 2^{2}} = 68, and A C^{2} = (- 3)^{2} + (5)^{2} = 34$
$∴ A B = A C$ .
Hence, $Δ A B C$ is isoscles.