Coordinate Geometry, Loci and Analytic Regions Previous Year Questions (PYQs) for CAT: 3+ Solved Questions with Step-by-Step Solutions

Solve 3+ Coordinate Geometry, Loci and Analytic Regions previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Coordinate Geometry, Loci and Analytic Regions

CAT QA Geometry

Coordinate Geometry, Loci and Analytic Regions

A 3-step journey from points and lines to equations, curves, and bounded areas.

chapter PYQs

①

📍 t1 — Coordinate Geometry of Lines and Polygons

You learn distance, midpoint, line equations, parallelogram coordinates, and coordinate-area methods.

3 own-course PYQs Selected topic Moderate-high utility

②

⭕ t2 — Analytic Circles and Loci

You move from straight-line geometry to circle equations and distance-based moving points.

1 own-course PYQ

③

🧩 t3 — Inequality Regions and Coordinate Areas

You learn how inequalities shade regions and how boundaries combine to create areas.

2 own-course PYQs

End goal: convert coordinate information into distances, equations, intersections, and areas without depending on a perfect diagram.

Topic Hero: Coordinate Geometry of Lines and Polygons

Geometry → Coordinate Geometry → t1

Coordinate Geometry of Lines and Polygons

One-line hook: coordinates let you solve geometry by calculation, not by eye.

What you'll learn here

Distance, midpoint, and slope as geometry tools.
Line equation and x-axis / y-axis intersection methods.
Parallelogram vertex shortcuts using vectors or diagonals.
Coordinate triangle side lengths and inradius.
Polygon area through coordinate formulas.

Coordinate Geometry, Loci and Analytic Regions: Solved Questions with Step-by-Step Explanations (3 Problems)

Question 1 · Quantitative Ability MCQ

Let ABCD be a parallelogram such that the coordinates of its three vertices A, B, C are (1, 1), (3, 4) and (-2, 8), respectively. Then, the coordinates of the vertex D are

A.
(0, 11)
B.
(4, 5)
C.
(-3, 4)
D.
(-4, 5)

Correct Answer:

Step-by-Step Solution

In a parallelogram $A B C D$ , the diagonals $A C$ and $B D$ bisect each other. This means the midpoint of diagonal $A C$ is exactly the same as the midpoint of diagonal $B D$ . First, find the midpoint of $A C$ using the coordinates $A (1, 1)$ and $C (- 2, 8)$ . The midpoint is $(\frac{1 + ( - 2 )}{2}, \frac{1 + 8}{2}) = (- 0.5, 4.5)$ . Let the coordinates of $D$ be $(x, y)$ . The midpoint of $B D$ , with $B (3, 4)$ , is $(\frac{3 + x}{2}, \frac{4 + y}{2})$ . Equating the two midpoints gives $\frac{3 + x}{2} = - 0.5 ⟹ 3 + x = - 1 ⟹ x = - 4$ , and $\frac{4 + y}{2} = 4.5 ⟹ 4 + y = 9 ⟹ y = 5$ . Thus, the coordinates of $D$ are $(- 4, 5)$ .

Question 2 · Quantitative Ability NAT

The coordinates of the three vertices of a triangle are: (1, 2), (7, 2), and (1, 10). Then the radius of the incircle of the triangle is

Correct Answer:

Step-by-Step Solution

This is an incircle-radius-of-a-coordinate-triangle question. You can recognise the pattern

because three coordinate vertices of a triangle are given and the inradius is asked. Here

the coordinates reveal a right triangle, which gives a fast shortcut instead of the general

area-over-semi-perimeter method.

Step 1: Identify the shape of the triangle.

Vertices: A(1,2), B(7,2), C(1,10).

A and B share the same y-coordinate (2), so AB is a horizontal segment with length

|7 - 1| = 6.

A and C share the same x-coordinate (1), so AC is a vertical segment with length

|10 - 2| = 8.

Since AB is horizontal and AC is vertical, they are perpendicular to each other, so the

triangle is right-angled at A, with legs 6 and 8.

Step 2: Find the hypotenuse.

BC = sqrt(AB^2 + AC^2) = sqrt(6^2 + 8^2) = sqrt(36 + 64) = sqrt(100) = 10.

Step 3: Use the right-triangle inradius shortcut.

For a right triangle with legs a and b and hypotenuse c, the inradius is:

r = (a + b - c) / 2

This comes from the general formula r = Area / semi-perimeter, and simplifies neatly

whenever the triangle has a right angle.

r = (6 + 8 - 10) / 2 = 4 / 2 = 2

Answer: the inradius is 2.

Common trap: using the general formula r = Area / s without noticing the right angle is

more error-prone since it needs computing Area separately by the shoelace method and the

full semi-perimeter with a square root for the hypotenuse. Recognising the right angle

first (from matching x- or y-coordinates) makes this a much faster, safer calculation.

Question 3 · Quantitative Ability NAT

The area of the quadrilateral bounded by the Y-axis, the line $x = 5$ , and the lines $∣ x - y ∣ - ∣ x - 5∣ = 2$ , is

Correct Answer:

Step-by-Step Solution

This is the "Absolute Value Boundaries Form a Quadrilateral" pattern: an absolute-value equation splits into two lines once you fix the sign of each expression, and together with two given boundary lines they enclose a quadrilateral.

Step 1 — Recognize the trigger.

The region is bounded by the y-axis ( $x = 0$ ), the line $x = 5$ , and the curve $∣ x - y ∣ - ∣ x - 5∣ = 2$ . Since we are confined between $x = 0$ and $x = 5$ , we only need the equation for $0 \leq x \leq 5$ .

Step 2 — Remove the second absolute value using the known sign of $(x - 5)$ .

For $x \leq 5$ , $x - 5 \leq 0$ , so $∣ x - 5∣ = 5 - x$ . The equation becomes:

$∣ x - y ∣ - (5 - x) = 2 \Rightarrow ∣ x - y ∣ = 7 - x$

Step 3 — Split by the sign of $(x - y)$ .

If $y \leq x$ : $x - y = 7 - x \Rightarrow 2 x - y = 7 \Rightarrow y = 2 x - 7$
If $y > x$ : $y - x = 7 - x \Rightarrow y = 7$

So the boundary consists of the line $y = 2 x - 7$ (for the part where $y \leq x$ ) and the line $y = 7$ (for the part where $y > x$ ), both restricted to $0 \leq x \leq 5$ .

Step 4 — Find the four vertices of the quadrilateral.

On $x = 0$ : line $y = 2 x - 7$ gives $y = - 7$ , giving point $(0, - 7)$ .
On $x = 0$ : line $y = 7$ gives point $(0, 7)$ .
On $x = 5$ : line $y = 2 x - 7$ gives $y = 3$ , giving point $(5, 3)$ .
On $x = 5$ : line $y = 7$ gives point $(5, 7)$ .

The quadrilateral has vertices $(0, - 7)$ , $(5, 3)$ , $(5, 7)$ , $(0, 7)$ — a trapezium with the two parallel (vertical) sides on $x = 0$ and $x = 5$ .

Step 5 — Compute the area.

Left side length (on $x = 0$ ): from $y = - 7$ to $y = 7$ , length $= 14$ .

Right side length (on $x = 5$ ): from $y = 3$ to $y = 7$ , length $= 4$ .

Distance between the parallel sides $= 5$ .

$Area = \frac{1}{2} (14 + 4) (5) = \frac{1}{2} (18) (5) = 45$

Trap avoided: forgetting to restrict to $0 \leq x \leq 5$ before removing $∣ x - 5∣$ would give the wrong line equations entirely.

Answer: Area $= 45$ .

Coordinate Geometry, Loci and Analytic Regions Previous Year Questions (PYQs) for CAT: 3+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Coordinate Geometry, Loci and Analytic Regions

Coordinate Geometry, Loci and Analytic Regions

📍 t1 — Coordinate Geometry of Lines and Polygons

⭕ t2 — Analytic Circles and Loci

🧩 t3 — Inequality Regions and Coordinate Areas

Topic Hero: Coordinate Geometry of Lines and Polygons

Coordinate Geometry of Lines and Polygons

What you'll learn here

Coordinate Geometry, Loci and Analytic Regions: Solved Questions with Step-by-Step Explanations (3 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More previous year questions (pyqs) in this unit

A question from this chapter

Login to view the complete previous-year questions and solutions

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course