The following points A2a- 4a- B2a- 6a and C 2 a -3 a - 5 a - a -0 are the vertices ofA- An acute angled triangleB- A right angled triangleC- An isosceles triangleD- None of these

The correct options are A An acute angled triangle C An isosceles triangle AB = √((2a 2a)^2 + (4a 6a)^2) = 2a BC = √((√(3)a)^2 + a^2) = 2a CA = √((√(3)a)^2 ...

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Byju's Answer

Standard XI

Mathematics

Relative Position of a Point with Respect to a Line

The following...

Question

The following points A(2a, 4a), B(2a, 6a) and C $(2 a + \sqrt{3} a, 5 a)$ , (a > 0) are the vertices of

An acute angled triangle

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A right angled triangle

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An isosceles triangle

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None of these

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Solution

The correct options are
A
An acute angled triangle

C
An isosceles triangle

AB = $\sqrt{(2 a - 2 a)^{2} + (4 a - 6 a)^{2}}$ = 2a

BC = $\sqrt{(\sqrt{3} a)^{2} + a^{2}}$ = 2a

CA = $\sqrt{(\sqrt{3} a)^{2} + (- a)^{2}}$ = 2a

Since AB = BC = CA, hence triangle is equilateral.

Therefore, it is an acute angled triangle.

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The following points A(2a, 4a), B(2a, 6a) and C(2a+√3a,5a), (a > 0) are the vertices of

The correct options are A An acute angled triangle C An isosceles triangle AB = √(2a−2a)2+(4a−6a)2 = 2a BC = √(√3a)2+a2 = 2a CA = √(√3a)2+(−a)2 = 2a Since AB = BC = CA, hence triangle is equilateral. Therefore, it is an acute angled triangle.

The following points A(2a, 4a), B(2a, 6a) and C $(2 a + \sqrt{3} a, 5 a)$ , (a > 0) are the vertices of