The points A(2, 3, 5),B(-1, 5, -1),and C(4, - | vedclass

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

Question

(D) Let the points be $A(2, 3, 5)$,$B(-1, 5, -1)$,and $C(4, -3, 2)$.Calculate the squared distances between the points using the distance formula $d^2

Vedclass · Accepted Answer

an isosceles right angled triangle

Vedclass · Answer

(D) Let the points be $A(2, 3, 5)$,$B(-1, 5, -1)$,and $C(4, -3, 2)$.Calculate the squared distances between the points using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$:$AB^2 = (-1 - 2)^2 + (5 - 3)^2 + (-1 - 5)^2 = (-3)^2 + (2)^2 + (-6)^2 = 9 + 4 + 36 = 49$.$BC^2 = (4 - (-1))^2 + (-3 - 5)^2 + (2 - (-1))^2 = (5)^2 + (-8)^2 + (3)^2 = 25 + 64 + 9 = 98$.$AC^2 = (4 - 2)^2 + (-3 - 3)^2 + (2 - 5)^2 = (2)^2 + (-6)^2 + (-3)^2 = 4 + 36 + 9 = 49$.Since $AB^2 = AC^2 = 49$,the triangle is isosceles because $AB = AC = 7$.Check for the right-angled condition: $AB^2 + AC^2 = 49 + 49 = 98 = BC^2$.Since the sum of the squares of two sides equals the square of the third side,the triangle is a right-angled triangle.Thus,the triangle is an isosceles right-angled triangle.

The points $A(2, 3, 5)$,$B(-1, 5, -1)$,and $C(4, -3, 2)$ form:

Similar Questions

The extremities of a diagonal of a parallelogram are the points $(3, -4)$ and $(-6, 5)$. If the third vertex is $(-2, 1)$,then the fourth vertex is:

The circumcentre of the triangle formed by the points $A(3,4,5)$,$B(2,3,1)$,and $C(-1,6,1)$ is:

Let $A (2, 3, 5)$,$B (-1, 3, 2)$ and $C (\lambda, 5, \mu)$ be the vertices of a $\Delta ABC$. If the median through $A$ is equally inclined to the coordinate axes,then

Three vertices of a parallelogram $ABCD$ are $A(1, 2, 3)$,$B(-1, -2, -1)$ and $C(2, 3, 2)$. Find the fourth vertex $D(x, y, z)$.

If the three vertices of a parallelogram are $(-1, 0), (3, 1),$ and $(2, 2)$ respectively,find the coordinates of the fourth vertex.

Vedclass Test Series

Exam Paper Generator

Online Exam Module