Question

A(0, 0), B(3, 0), C(3, 4) are the vertices of a ...... triangle.select a proper option (a), (b), (c) or (d) from given options so that the statement becomes correct.
(a) right angled
(b) equilateral
(c) isosceles
(d) acute angled

Answer 1

use distance formula to find side length of triangle.
if two points $(x_1,y_1)$ and $(x_2,y_2)$ is given
then, distance between them = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Let A = (0,0) , B = (3, 0) and C = (0,4)
AB = $\sqrt{(3-0)^2+(0-0)^2}=3$
BC = $\sqrt{(0-3)^2+(4-0)^2}=5$
CA = $\sqrt{(0-0)^2+(0-4)^2}=4$
we see , AB ≠ BC ≠ CA
but AB² = 9 , BC² = 25 and CA² = 16
we see , BC² = AB² + CA²
from Pythagoras theorem,
ABC is right angled triangle where A is right angle .

hence option (a) is correct.

Answer 2

Given points are A(0,0), B(3,0) and C(3,4).

= > AB^2 = (3 - 0)^2 + (0 - 0)^2 = 9

= > BC^2 = (3 - 3)^2 + (4 - 0)^2 = 16

= > AC^2 = (3 - 0)^2 + (4 - 0)^2 = 25

Here, AB^2 + BC^2 = AC^2.

Therefore, the above vertices are points of right angled triangle.

Hope this helps!