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Mathematics · Class 11 Science

Ch 15Locus — Class 11 Mathematics, concept-first.

A locus is the set of all points in a plane that satisfy one specific geometric condition, and only that condition. Two ideas sit inside this one sentence, and both are needed for a full understanding.

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Key concepts

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The Five-Step Method for Finding a Locus Equation

Finding the equation of a locus is not a matter of guesswork; it is a fixed, five-step translation from a geometric sentence to an algebraic one, and the same five steps apply no matter how the condition is phrased.

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Chapter contents

The NCERT structure, section by section. Open a section to see its questions, then read the concept-first solution.

1.1

Definition and Consistency of a Locus

A locus is the set of all points in a plane that satisfy one specific geometric condition, and only that condition.

1.2

The Five-Step Method for Finding a Locus

Every locus problem in this chapter is solved by the same five-step procedure, and learning to apply it mechanically removes almost all of the difficulty from the topic.

1.3

Worked Loci: Distance, Ratio, Angle and Area Conditions

This section applies the five-step method to the four condition-types that recur throughout the exercise: equal distance from two fixed points, a fixed ratio of two distances, a right angle subtended…

+Exercise 1(a)i9 questions
  1. Q1Find the equation of the locus of a point $P$ which is equidistant from the points $A(1,2)$ and $B(3,-4)$.Free
  2. Q2Find the equation of the locus of a point $P(x,y)$ which is equidistant from $A(2,0)$ and $B(-2,0)$.Free
  3. Q3A point $P$ moves such that its distances from the fixed points $A(-3,0)$ and $B(3,0)$ are in the ratio $PA:PB = 1:2$. Find the equation of…Free
  4. Q4Find the equation of the locus of a point $P$ such that $PA:PB = 3:1$, where $A(4,0)$ and $B(-4,0)$ are fixed points.Preview
  5. Q5$A(2,3)$ and $B(2,-3)$ are two fixed points. Find the equation of the locus of a point $P$ such that $\angle APB = 90^\circ$.Preview
  6. Q6Find the equation of the locus of a point $P$ such that the segment joining $A(-5,0)$ and $B(5,0)$ subtends a right angle at $P$.Preview
  7. Q7$A(4,0)$ and $B(-4,0)$ are two fixed points. Find the equation of the locus of a point $P$ such that the area of triangle $PAB$ is $16$ squa…Preview
  8. Q8Find the equation of the locus of a point $P(x,y)$ such that the area of the triangle formed by $P$ with $A(1,0)$ and $B(0,1)$ is $2$ square…Preview
  9. Q9$A(2,1)$ and $B(5,4)$ are two fixed points. Find the equation of the locus of a point $P$ such that the area of triangle $PAB$ is $9$ square…Preview

Sample & Board Papers

Sample papers and previous-year board questions for this subject.

+Show 8 questions8 questions
  1. Q1$(4, 0)$ and $(0, 4)$ are the ends of the hypotenuse of a right-angled triangle. Find the locus of its third vertex.Preview
  2. Q2Find the circumcenter of the triangle whose vertices are $(1, 3)$, $(-3, 5)$, $(5, -1)$.Preview
  3. Q3$A(1, 2)$, $B(2, -3)$ and $C(-2, 3)$ are three points. If a point $P$ moves such that $PA^{2} + PB^{2} = 2PC^{2}$, then show that the equati…Preview
  4. Q4$A(2, 3)$ and $B(-3, 4)$ are two given points. Find the equation of locus of $P$ so that the area of the triangle $PAB$ is $8.5$.Preview
  5. Q5If the distance from P to the points $(2, 3)$ and $(2, -3)$ are in the ratio $2:3$, then find the equation of the locus of P.Preview
  6. Q6The ends of the hypotenuse of a right angled triangle are $(0, 6)$ and $(6, 0)$. Find the equation of the locus of its third vertex.Preview
  7. Q7The ends of the hypotenuse of a right angled triangle are $(0, 6)$ and $(6, 0)$. Find the equation of the locus of its third vertex.Preview
  8. Q8$A(5, 3)$ and $B(3, -2)$ are two fixed points. Find the equation of the locus of $P$, so that the area of triangle $PAB$ is $9$.Preview