Question

A variable plane which remains at a constant distance $3p$ from the origin cuts the coordinate axes at $A,\text{ B, C}$. Show that the locus of the centroid of triangle $ABC$ is $\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{y}^{2}}}+\dfrac{1}{{{z}^{2}}}=\dfrac{1}{{{p}^{2}}}$ .

Answer 1

 Hint: We have to prove that $\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{y}^{2}}}+\dfrac{1}{{{z}^{2}}}=\dfrac{1}{{{p}^{2}}}$ . First take $O$ as the origin then take the equation of the plane passing through $A$ , $B$ and $C$. After that, apply the condition of distance and find the centroid of the triangle. At last substitute the point of centroid in the condition of distance. Try it, you will get the answer.


Complete step-by-step Solution:
A locus is a set of points, in geometry, which satisfies a given condition or situation for a shape or a figure. The plural of the locus is loci. The area of the loci is called region. The word locus is derived from the word location. Before the 20th century, the geometric shapes were considered as an entity or place where points can be located or can be moved. But in the modern maths, the entities are considered as the set of points which satisfies the given condition.
A locus is a curve other shape made by all the points satisfying a particular equation of the relation between the coordinates, or by a point, line, or moving surface. All the shapes such as circle, ellipse, parabola, hyperbola, etc. are defined by the locus as a set of points.
In real-life you must have heard about the word ‘location’. The word location is derived from the word locus itself. Locus defines the position of something. When an object is situated somewhere or something happens at a place, is described by locus. For example, the area has become a locus of opposition to the government.
The locus of points defines a shape in geometry. Suppose, a circle is the locus of all the points which are equidistant from the center. In the same, the other shapes such as ellipse, parabola, hyperbola, etc. are defined by the locus of the points.
The locus is defined only for curved shapes. These shapes can be regular or irregular.
Locus is not described for the shapes having vertex or angles inside them.
Let $O$ be the origin and $OA=a$ , $OB=b$ , $OC=c$ .
Equation of plane passing through $A$ , $B$ and $C$is,
$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$
Now from given condition,
\[\begin{align}
  & \dfrac{\left| 0+0+0-1 \right|}{\sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}}=3p \\
 & \sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}=\dfrac{1}{3p} \\
 & \dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}=\dfrac{1}{9{{p}^{2}}} \\
\end{align}\]
\[\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}=\dfrac{1}{9{{p}^{2}}}\] .............. (1)
Now $A,\text{ B, C}$ are $(a,0,0),(0,b,0),(0,0,c)$ respectively.
Let $({{x}_{1}},{{y}_{1}},{{z}_{1}})$ be the centroid of the triangle $ABC$ .
${{x}_{1}}=\dfrac{a+0+0}{3}=\dfrac{a}{3}$ , ${{y}_{1}}=\dfrac{0+b+0}{3}=\dfrac{b}{3}$ , ${{z}_{1}}=\dfrac{0+0+c}{3}=\dfrac{c}{3}$
$a=3{{x}_{1}}\text{, }b=3{{y}_{1}},\text{ c}=3{{z}_{1}}$
Now let us put the values $a=3{{x}_{1}}\text{, }b=3{{y}_{1}},\text{ c}=3{{z}_{1}}$ in (1) we get,
\[\begin{align}
  & \dfrac{1}{{{(3{{x}_{1}})}^{2}}}+\dfrac{1}{{{(3{{y}_{1}})}^{2}}}+\dfrac{1}{{{(3{{z}_{1}})}^{2}}}=\dfrac{1}{9{{p}^{2}}}\text{ } \\
 & \dfrac{1}{{{x}_{1}}^{2}}+\dfrac{1}{{{y}_{1}}^{2}}+\dfrac{1}{{{z}_{1}}^{2}}=\dfrac{1}{{{p}^{2}}}\text{ } \\
\end{align}\]
So the locus of centroid of triangle $ABC$ is \[\dfrac{1}{{{x}_{1}}^{2}}+\dfrac{1}{{{y}_{1}}^{2}}+\dfrac{1}{{{z}_{1}}^{2}}=\dfrac{1}{{{p}^{2}}}\text{ }\].
Hence proved.

Note: Read the question carefully. Don’t confuse yourself. Your concept regarding locus should be clear. Also, take care that while simplifying no terms are missed. Do not make any silly mistake. While solving take care that no signs are missed.