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  • The points (–4, 0), (4, 0), (0, 3) are the vertices of a (A) right triangle (B) isosceles triangle (C) equilateral triangle (D) scalene triangle

The points (–4, 0), (4, 0), (0, 3) are the vertices of a

(A) right triangle

(B) isosceles triangle

(C) equilateral triangle

(D) scalene triangle

Answers (1)

The given points are (–4, 0), (4, 0), (0, 3)

\\ \mathrm{AB}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}} \\ \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(-4,0) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(4,0) \\ \mathrm{AB}=\sqrt{\left(4-(-4)^{2}+(0-0)^{2}\right.} \\ \mathrm{AB}=\sqrt{(8)^{2}} \\ \mathrm{AB}=8 \\ \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(4,0) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(0,3) \\ \mathrm{BC}=\sqrt{(0-4)^{2}+(3-0)^{2}} \\ \mathrm{BC}=\sqrt{16+9}=\sqrt{25}=5 \\ \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(0,3) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(-4,0) \\ \mathrm{AC}=\sqrt{(0+4)^{2}+(3-0)^{2}} \\ \mathrm{AC}=\sqrt{16+9}=\sqrt{25}=5

In this triangle BC = AC.

Hence it is an isosceles triangle.

Posted by

infoexpert21

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