1.

The points A(– 4, 0), B(4, 0) and C(0, 3) are the vertices of a triangle. Is the triangle isosceles? Show by calculations.

Answer»

Given that the vertices of a triangle ABC are A(–4, 0), B(4, 0) and C(0, 3). 

The side length of AB = \(\sqrt{(4-(-4))^2+(0-0)^2}\) = \(\sqrt{(4+4)^2+0}\) = 8 unit. 

(By distance formula between two points) 

The side length of BC = \(\sqrt{(0-4)^2+(3-0)^2}\) = \(\sqrt{16+9}\) = \(\sqrt{25}\) = 5 unit. 

The side length of AC = \(\sqrt{(0-(-4))^2+(3-0)^2}\) = \(\sqrt{16+9}\) = \(\sqrt{25}\) = 5 unit. 

Hence, BC = AC = 5 unit in triangle ABC. 

∴ Triangle ABC is an isosceles triangle. 

Hence Proved



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