Question
Find the third vertex of triangle whose centroid is origin and two vertices are (2, 4, 6) and (0, -2, -5).
✓
Answer
Let the third or unknown vertex of $\triangle\text{ABC}$ be A(x, y, z).
Other vertices of triangle are B(2,4, 6) and C(0, -2, -5).
The centroid is G(0, 0, 0).
$\therefore(0,0,0)=\Big(\frac{2+0+\text{x}}{3},\frac{4-2+\text{y}}{3},\frac{6-5+\text{z}}{3}\Big)$
On comparing coordinates, we get,
$\frac{2+\text{x}}{3}=0,\frac{2+\text{y}}{3}=0\ \text{and}\ \frac{1+\text{z}}{3}=0$
$\Rightarrow\text{x}=-2,\ \text{y}=-2\ \text{and}\ \text{z}=-1$
Other vertices of triangle are B(2,4, 6) and C(0, -2, -5).
The centroid is G(0, 0, 0).
$\therefore(0,0,0)=\Big(\frac{2+0+\text{x}}{3},\frac{4-2+\text{y}}{3},\frac{6-5+\text{z}}{3}\Big)$
On comparing coordinates, we get,
$\frac{2+\text{x}}{3}=0,\frac{2+\text{y}}{3}=0\ \text{and}\ \frac{1+\text{z}}{3}=0$
$\Rightarrow\text{x}=-2,\ \text{y}=-2\ \text{and}\ \text{z}=-1$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Differentiate the following function with respect to $\text{x}:$$(\text{x}^3+\text{x}^2+1)\sin\text{x}$
→Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Find the derivative of cos x from first principle.
The focus of a parabolic mirror as shown in is at a distance of 5 cm from its vertex. If the mirror is 45 cm deep, find the distance AB
Find the equation of the parabola that satisfies the given conditions: Vertex $(0, 0)$ Focus $(3, 0)$
→The perpendicular from the origin to a line meets it at the point (-2, 9), find the equation of the line.
Sketch the graphs of the following function:
$\theta\text{(x)}=\sin\Big(\frac{\pi}{2}-\frac{\pi}{4}\Big),0\leq\text{x}\leq4\pi$
→$\theta\text{(x)}=\sin\Big(\frac{\pi}{2}-\frac{\pi}{4}\Big),0\leq\text{x}\leq4\pi$
Solve the following system of equations in R.
$10\leq-5(\text{x}-2)<20$
→$10\leq-5(\text{x}-2)<20$
Solve the inequality $\frac{(2 x-1)}{3} \geq \frac{(3 x-2)}{4}-\frac{(2-x)}{5}$ for real x.
→Find the coordinates of the centre and radius of each of the following circles:$\text{x}^2+\text{y}^2-\text{ax}-\text{by}=0$
→