If the points $P(6, 2)$ and $Q(-2, 1)$, and $R$, are the vertices of a $\triangle PQR$, and $R$ is a point on the locus $y = x^2 - 3x + 4$, then find the equation of the locus of the centroid of $\triangle PQR$.

If the points P(6, 2) and Q(-2, 1), and R, are the vertices of a triangle PQR, and R is a point on the locus y

Exercise 6.1 · Q12

P(6, 2)

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Concept understanding — Locus of a Point

A point is not a thing but a place — a position located, with reference to a coordinate system, by a unique ordered pair $(x,y)$ : $x$ is the (signed) distance from the $y$ -axis and $y$ the (signed) distance from the $x$ -axis. The locus of a moving point is the path traced out by it as it moves subject to a stated geometric condition; equivalently, an equation in $x,y$ has infinitely many real solution-pairs $(x,y)$ , and the collection of all the points corresponding to those solutions is the locus of the equation. (The plural of locus is loci.)

Some standard loci: a point moving so as to stay equidistant from two fixed points $A,B$ traces the perpendicular bisector of $AB$ ; a point equidistant from two fixed lines traces the angle bisector; a point at a fixed distance $r$ from a fixed point $O$ traces a circle of radius $r$ centred at $O$ . A point on the rim of a circle rolling along a straight line traces a cycloid.

Procedure for finding the equation of a locus.

Assign coordinates $(h,k)$ to the moving point $P$ whose locus is wanted.
Translate the given geometric condition(s) into equation(s) relating $h,k$ and any parameters/known quantities.
Eliminate the parameter(s), so that only $h,k$ and known constants remain.
Replace $h$ by $x$ and $k$ by $y$ ; the resulting equation in $x,y$ is the equation of the locus.

Tip

When the given coordinates involve a trigonometric parameter $\theta$ (e.g. $(a\sec\theta,\,b\tan\theta)$ ), eliminate $\theta$ using a Pythagorean identity: $\sin^2\theta+\cos^2\theta=1$ , $\sec^2\theta-\tan^2\theta=1$ , or $\csc^2\theta-\cot^2\theta=1$ . When the parameter is a real number $t$ (e.g. $(ct,\,c/t)$ ), eliminate it by taking the product, ratio, or another algebraic combination of the two coordinate equations that cancels $t$ . …

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Express $R$ 's coordinates through the centroid formula, then substitute into the parabola $R$ lies on.

Step 1. Set up $R$ . Let $R=(p,q)$ lie on $y=x^2-3x+4$ , so $q=p^2-3p+4$ .

Step 2. Apply the centroid formula. For $\triangle PQR$ with $P(6,2)$ , $Q(-2,1)$ , $R(p,q)$ , the centroid $(x,y)$ satisfies

x=\frac{6+(-2)+p}{3}=\frac{4+p}{3}, \qquad y=\frac{2+1+q}{3}=\frac{3+q}{3}.

Step 3. Solve for $p,q$ .

p=3x-4, \qquad q=3y-3.

Step 4. Substitute into $q=p^2-3p+4$ .

$3y-3=(3x-4)^2-3(3x-4)+4.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$(a\cos\theta, b\sin\theta)$
(a) $x^2+y^2=a^2$
(b) $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$
(c) $y^2=4ax$
(d) $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
›Reveal solutionSolution

With $x=a\cos\theta$ , $y=b\sin\theta$ , using $\cos^2\theta+\sin^2\theta=1$ eliminates $\theta$ to give the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ .

Let $x=a\cos\theta$ and $y=b\sin\theta$ . Then $\cos\theta=\dfrac{x}{a}$ and $\sin\theta=\dfrac{y}{b}$ .

Using the Pythagorean identity $\cos^2\theta+\sin^2\theta=1$ :

$\left(\dfrac{x}{a}\right)^2+\left(\dfrac{y}{b}\right)^2=1 \;\Rightarrow\; \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

…

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CBSE 2024Set ANNUAL1 markMCQ
$(8,-5)$
(a) $2$
(b) $0$
(c) $3$
(d) $1$
›Reveal solutionSolution

Substituting $(8,-5)$ into the locus equation gives $k=3$ .

The locus is $\dfrac{x^2}{16}-\dfrac{y^2}{25}=k$ . Since $(8,-5)$ lies on it, put $x=8,\ y=-5$ : …

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CBSE 2023Set ANNUAL1 markMCQ
$3x^2 + 3y^2 - 8x - 12y + 17 = 0$
(a) (1, 2)
(b) (0, 0)
(c) (0, -1)
(d) (-2, 3)
›Reveal solutionSolution

Substitute each option into the equation; the point that makes it exactly zero lies on the curve.

Test $(1,2)$ : $3(1)^2+3(2)^2-8(1)-12(2)+17 = 3+12-8-24+17 = 0$ . ✓ It satisfies the equation.

Test $(0,0)$ : $0+0-0-0+17=17\ne 0$ . ✗

…

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CBSE 2020Set ANNUAL1 markMCQ
$(8, -5)$
(a) 0
(b) 1
(c) 2
(d) 3
›Reveal solutionSolution

Direct substitution of the point $(8,-5)$ into the locus equation gives $k=3$ .

A point lies on a locus if its coordinates satisfy the locus equation. Substituting $x=8$ , $y=-5$ : …

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Q.If the points P(6,2)P(6, 2)P(6,2) and Q(−2,1)Q(-2, 1)Q(−2,1), and RRR, are the vertices of a △PQR\triangle PQR△PQR, and RRR is a point on the locus y=x2−3x+4y = x^2 - 3x + 4y=x2−3x+4, then find the equation of the locus of the centroid of △PQR\triangle PQR△PQR.

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Q.If the points $P(6, 2)$ and $Q(-2, 1)$ , and $R$ , are the vertices of a $\triangle PQR$ , and $R$ is a point on the locus $y = x^2 - 3x + 4$ , then find the equation of the locus of the centroid of $\triangle PQR$ .

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