In a triangle ABC,if the midpoints of sides A | vedclass

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

Question

(D) Let the vertices of the triangle be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.Given the midpoints of $AB, BC, CA$ are $M_1(3,0,

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(D) Let the vertices of the triangle be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.Given the midpoints of $AB, BC, CA$ are $M_1(3,0,0)$,$M_2(0,4,0)$,and $M_3(0,0,5)$ respectively.Using the midpoint formula:$x_1+x_2=6, x_2+x_3=0, x_3+x_1=0$Solving these,we get $x_1=3, x_2=3, x_3=-3$.Similarly for $y$ coordinates:$y_1+y_2=0, y_2+y_3=8, y_3+y_1=0$Solving these,we get $y_1=-4, y_2=4, y_3=4$.Similarly for $z$ coordinates:$z_1+z_2=0, z_2+z_3=0, z_3+z_1=10$Solving these,we get $z_1=5, z_2=-5, z_3=5$.Thus,the vertices are $A(3, -4, 5)$,$B(3, 4, -5)$,and $C(-3, 4, 5)$.Now,calculate the squares of the side lengths:$AB^2 = (3-3)^2 + (4-(-4))^2 + (-5-5)^2 = 0^2 + 8^2 + (-10)^2 = 64 + 100 = 164$.$BC^2 = (-3-3)^2 + (4-4)^2 + (5-(-5))^2 = (-6)^2 + 0^2 + 10^2 = 36 + 100 = 136$.$CA^2 = (3-(-3))^2 + (-4-4)^2 + (5-5)^2 = 6^2 + (-8)^2 + 0^2 = 36 + 64 = 100$.Finally,$AB^2+BC^2+CA^2 = 164 + 136 + 100 = 400$.

Column $I$	Column $II$
$(A)$ The centroid of the triangle formed by $(2, 3, -1)$,$(5, 6, 3)$,$(2, -3, 1)$ is	$(p)$ $(2, 2, 2)$
$(B)$ The circumcentre of the triangle formed by $(1, 2, 3)$,$(2, 3, 1)$,$(3, 1, 2)$ is	$(q)$ $(3, 1, 4)$
$(C)$ The orthocentre of the triangle formed by $(2, 1, 5)$,$(3, 2, 3)$,$(4, 0, 4)$ is	$(r)$ $(1, 1, 0)$
$(D)$ The incentre of the triangle formed by $(0, 0, 0)$,$(3, 0, 0)$,$(0, 4, 0)$ is	$(s)$ $(3, 2, 1)$

In a triangle $ABC$,if the midpoints of sides $AB, BC, CA$ are $(3,0,0), (0,4,0), (0,0,5)$ respectively,then $AB^2+BC^2+CA^2=$

Similar Questions

If the circumcenter of the triangle formed by the points $(1,2,3), (3,-1,5)$ and $(4,0,-3)$ is $(\alpha, \beta, \gamma)$,then $|\alpha|+|\beta|=$

If a triangle $ABC$ with two vertices $A(5,4,6)$ and $B(1,-1,3)$ has its centroid at $\left(\frac{10}{3}, 2, \frac{11}{3}\right)$,then the third vertex $C$ is

If $G(4, 3, 3)$ is the centroid of the triangle $ABC$ whose vertices are $A(a, 3, 1)$,$B(4, 5, b)$,and $C(6, c, 5)$,then the values of $a, b, c$ are:

If $A(3, 5), B(-5, -4), C(7, 10)$ are the vertices of a parallelogram,taken in order,then the coordinates of the fourth vertex are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module