In Delta ABC, if A(0, 0), B(4, 0) and C(0, 3) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The coordinates of the vertices are $A(0, 0)$,$B(4, 0)$,and $C(0, 3)$.Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:$AB = \sqr

Vedclass · Accepted Answer

right angled

Vedclass · Answer

(A) The coordinates of the vertices are $A(0, 0)$,$B(4, 0)$,and $C(0, 3)$.Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:$AB = \sqrt{(4-0)^2 + (0-0)^2} = \sqrt{16} = 4$ units.$BC = \sqrt{(0-4)^2 + (3-0)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$ units.$AC = \sqrt{(0-0)^2 + (3-0)^2} = \sqrt{3^2} = 3$ units.Since $AB^2 + AC^2 = 4^2 + 3^2 = 16 + 9 = 25$ and $BC^2 = 5^2 = 25$,we have $AB^2 + AC^2 = BC^2$.This satisfies the Pythagorean theorem,indicating that $\Delta ABC$ is a right-angled triangle at vertex $A$. Since $AB 
eq AC$,it is not isosceles.

In $\Delta ABC,$ if $A(0, 0), B(4, 0)$ and $C(0, 3)$ then $\Delta ABC$ is a $\ldots \ldots \ldots$ triangle.

Similar Questions

Show that the centroid of the triangle with vertices $(a, b-c)$,$(b, c-a)$,and $(c, a-b)$ lies on the $X$-axis.

The distance between the points $A(0, 6)$ and $B(0, -2)$ is

Find $k$ if $(1, 7), (2, 4)$ and $(k, 5)$ form a right triangle.

State whether the following statement is true or false. Justify your answer.
The points $A(-1, 0)$,$B(3, 1)$,$C(2, 2)$,and $D(-2, 1)$ are the vertices of a parallelogram.

If the midpoints of the sides $\overline{AB}$,$\overline{BC}$,and $\overline{CA}$ in $\Delta ABC$ are respectively $D(1, 2)$,$E(2, 1)$,and $F(3, 3)$,then find the centroid of $\Delta ABC$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module