Question

If A(0, 3), B(5, 0) and C(-5, 0) are the vertices of ∆ABC, then the triangle is __________

(a) Right-angled

(b) Isosceles

(c) Scalene

(d) Equilateral

I had been asked this question in final exam.

The doubt is from Geometry in division Coordinate Geometry of Mathematics – Class 10

Answer 1

Right answer is (b) Isosceles

The explanation is: Distance between (0, 3) and (5, 0) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(5-0)^2 + (0-3)^2} \)

= \( \sqrt {5^2 + -3^2 } \)

= \( \sqrt {25 + 9} \)

= √34

Distance between (5, 0) and (-5, 0) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(-5-5)^2 + (0-0)^2} \)

= \( \sqrt {-10^2} \)

= 10

Distance between (0, 3) and (-5, 0) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(-5-0)^2 + (0-3)^2} \)

= \( \sqrt {-5^2 + (-3)^2} \)

= \( \sqrt {25 + 9} \)

= √34

Since, the two sides of the triangle are equal.

Hence, the triangle will be isosceles triangle.