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

If the centroid of $\Delta ABC$ is $G(3, 3)$ and the vertices are $B(-1, 2)$ and $C(2, -3)$,then find the area of $\Delta ABC$.

Find the circumcentre of $\Delta ABC$ with vertices $A(-1, 1)$,$B(0, -4)$,and $C(-1, -5)$.

$P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$ are the given points. If $\overline{PQ}$ is parallel to the $X$-axis,then $PQ = \dots$

The midpoint of the line segment joining $(1, 1)$ and $(3, 3)$ is:

Prove that,if the midpoints of the sides of a quadrilateral are joined in order,we get a parallelogram.

Vedclass Test Series

Exam Paper Generator

Online Exam Module