Coordinate Geometry, Lines and Triangles Previous Year Questions (PYQs) for XAT: 5+ Solved Questions with Step-by-Step Solutions

Solve 5+ Coordinate Geometry, Lines and Triangles previous year questions for XAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Coordinate Geometry, Lines and Triangles

Chapter Roadmap

Coordinate Geometry, Lines and Triangles

Topic 1: Coordinate Geometry & Line Equations

Cartesian plane, distance, slopes, collinearity, line equations. (Current Focus)

Topic 2: Triangle Properties & Area Calculations

Centers of triangles, area formulations, bisectors, inequalities.

The Cartesian Bridge: Geometry Meets Algebra

The Cartesian Bridge

The core philosophy of coordinate geometry is translation.

1. Geometric Object

\to

Algebraic Equation

2. Geometric Condition

e.g., perpendicular, intersecting

\to

Algebraic Constraint (

m_{1} m_{2} = - 1

)

By assigning coordinates to points, we convert visual intuition into rigorous algebraic proofs.

Coordinate Geometry, Lines and Triangles: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Consider a right-angled triangle ABC, right angled at B. Two circles, each of radius r, are drawn inside the triangle in such a way that one of them touches AB and BC, while the other one touches AC and BC. The two circles also touch each other(see the image below).
If AB= 18 cm and BC= 24 cm, then nd the value of r.

A.
3 cm
B.
4 cm
C.
3.5 cm
D.
4.5 cm
E.
None of the remaining options is correct.

Correct Answer:

Step-by-Step Solution

Key idea: This is a coordinate geometry problem disguised as a circle tangent problem. By placing the right angle at the origin, the conditions "touches AB and BC" and "touches AC and BC" translate directly into distance equations for the centers of the circles.

Step 1: Set up the coordinate system.

Let B be at the origin (0, 0). Since the triangle is right-angled at B, let BC lie on the x-axis and AB lie on the y-axis.

Given AB = 18 and BC = 24, the vertices are B(0, 0), C(24, 0), and A(0, 18).

The hypotenuse AC has the equation: x/24 + y/18 = 1, which simplifies to 3x + 4y - 72 = 0.

Step 2: Locate the center of the first circle (O1).

Circle 1 touches AB (the y-axis) and BC (the x-axis). Since it is inside the triangle and has radius r, its center must be at O1(r, r).

Step 3: Locate the center of the second circle (O2).

Circle 2 touches BC (the x-axis), so its y-coordinate is r. Let its center be O2(x2, r).

It also touches AC. The perpendicular distance from O2 to the line 3x + 4y - 72 = 0 must be r.

Distance = |3x2 + 4r - 72| / sqrt(3^2 + 4^2) = |3x2 + 4r - 72| / 5.

Since O2 is inside the triangle, 3x2 + 4r < 72, so we can drop the absolute value:

(72 - 3x2 - 4r) / 5 = r

72 - 3x2 - 4r = 5r

3x2 = 72 - 9r => x2 = 24 - 3r.

So O2 is at (24 - 3r, r).

Step 4: Use the condition that the circles touch each other.

Since both circles have radius r and touch each other externally, the distance between their centers O1 and O2 must be 2r.

Both centers have the same y-coordinate (y = r), so the distance is simply the difference in their x-coordinates:

x2 - r = 2r

(24 - 3r) - r = 2r

24 - 4r = 2r

6r = 24 => r = 4.

Answer: B

Question 2 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

ABC is a triangle with BC=5. D is the foot of the perpendicular from A on BC. E is a point on CD such that BE=3. The value of $A B^{2} - A E^{2} + 6 C D$ is:

A.
5
B.
10
C.
14
D.
18
E.
21

Correct Answer:

Step-by-Step Solution

Key idea: This is a Pythagorean-difference problem. When two triangles share a common altitude, the difference of the squares of their hypotenuses equals the difference of the squares of their bases. The altitude then cancels out.

Step 1: Set up the geometry.

Triangle $A B C$ with altitude $A D$ on $B C$ . Point $E$ lies on segment $C D$ .

Given: $B C = 5$ , $B E = 3$ .

Step 2: Express segments in terms of $B D$ .

Let $B D = u$ . Then $D C = 5 - u$ .

Since $E$ lies on $C D$ , the order along $B C$ is $B$ - $D$ - $E$ - $C$ , so $B E = B D + D E$ .

Hence $D E = B E - B D = 3 - u$ .

Step 3: Apply Pythagoras to the right triangles.

Let $A D = h$ .

$A B^{2} = h^{2} + B D^{2} = h^{2} + u^{2}$

$A E^{2} = h^{2} + D E^{2} = h^{2} + (3 - u)^{2}$

Step 4: Compute the difference.

$A B^{2} - A E^{2} = u^{2} - (3 - u)^{2} = u^{2} - (9 - 6 u + u^{2}) = 6 u - 9$ .

Step 5: Add $6 C D$ .

$6 C D = 6 (5 - u) = 30 - 6 u$ .

Therefore, $A B^{2} - A E^{2} + 6 C D = (6 u - 9) + (30 - 6 u) = 21$ .

The variable $u$ cancels, giving a constant answer.

Answer: E

Question 3 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

ABC is a triangle and the coordinates of A, B and C are(a, b-2c),(a, b+4c) and(-2a,3c) respectively where a, b and c are positive numbers.
The area of the triangle ABC is:

A.
6abc
B.
9abc
C.
6bc
D.
9ac
E.
None of the above

Correct Answer:

Step-by-Step Solution

Key idea: This is a coordinate area problem. When two vertices share the same $x$ -coordinate, the side joining them is vertical, and we can compute the area as $\frac{1}{2} \times base \times height$ without the full shoelace formula.

Step 1: Observe the coordinates.

$A = (a, b - 2 c)$ , $B = (a, b + 4 c)$ , $C = (- 2 a, 3 c)$ with $a, b, c > 0$ .

Points $A$ and $B$ share $x = a$ , so side $A B$ is vertical.

Step 2: Compute the base $A B$ .

Length of $A B = ∣ (b + 4 c) - (b - 2 c) ∣ = ∣6 c ∣ = 6 c$ .

Step 3: Compute the height from $C$ to line $A B$ .

Line $A B$ is the vertical line $x = a$ . The perpendicular distance from $C = (- 2 a, 3 c)$ to this line is $∣ - 2 a - a ∣ = 3 a$ .

Step 4: Compute the area.

$Area = \frac{1}{2} \times 6 c \times 3 a = 9 a c$ .

Answer: D

Question 4 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

ABCD is a rectangle, where the coordinates of C and D are(- 2,0) and(2,0), respectively.

If the area of the rectangle is 24, which of the following is a possible equation representing the line AB?

A.
$4 x + 6 y = 24$
B.
None of the other options is correct.
C.
$y = 6$
D.
$x = 6$
E.
$x + y = 12$

Correct Answer:

Step-by-Step Solution

Key idea: This is a coordinate geometry problem involving rectangle properties. When one side of a rectangle lies on an axis, the opposite side is a horizontal or vertical line, and its equation is simply $x = k$ or $y = k$ .

Step 1: Identify the base side.

$C = (- 2, 0)$ and $D = (2, 0)$ . Both have $y = 0$ , so side $C D$ lies on the x-axis.

Length of $C D = ∣2 - (- 2) ∣ = 4$ units.

Step 2: Use area to find the height.

Area of rectangle $= base \times height$

$24 = 4 \times height$

$height = 6$ units.

Step 3: Determine the line containing AB.

Since $A B C D$ is a rectangle and $C D$ is horizontal, side $A B$ must be parallel to $C D$ , hence also horizontal.

The perpendicular distance between $A B$ and $C D$ is the height $= 6$ .

So the y-coordinate of $A B$ is either $0 + 6 = 6$ or $0 - 6 = - 6$ .

Therefore, the equation of line $A B$ is $y = 6$ or $y = - 6$ .

Step 4: Match with the options.

Among the options, $y = 6$ appears directly.

Answer: C

Question 5 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

The problem below consists of a question and two statements numbered 1& 2.
You have to decide whether the data provided in the statements are su cient to answer the question.
In a cricket match, three slip elders are positioned on a straight line. The distance between 1st slip and 2nd slip is the same as the distance between 2nd slip and the 3rd slip. The player X, who is not on the same line of slip elders, throws a ball to the 3rd slip and the ball takes 5 seconds to reach the player at the 3rd slip. If he had thrown the ball at the same speed to the 1st slip or to the 2nd slip, it would have taken 3 seconds or 4 seconds, respectively. What is the distance between the 2nd slip and the player X?
1. The ball travels at a speed of 3.6 km/hour.
2. The distance between the 1st slip and the 3rd slip is 2 meters.

A.
Statement 1 alone is su cient to answer the question, but statement 2 alone is not su cient.
B.
Statement 2 alone is su cient to answer the question, but statement 1 alone is not su cien
C.
Each statement alone is su cient
D.
Both statements together are su cient, but neither of them alone is su cient
E.
Statements 1& 2 together are not su cient

Correct Answer:

Step-by-Step Solution

Key idea: This is a Data Sufficiency problem combining kinematics ( $d = v t$ ) with the Apollonius theorem for a triangle whose base is bisected. The key is to derive a relation between the unknown speed $v$ and the half-spacing $d$ , then check whether each statement pins down the target distance $P S_{2} = 4 v$ .

Step 1: Model the geometry.

Let the slips be $S_{1}, S_{2}, S_{3}$ on a line with $S_{1} S_{2} = S_{2} S_{3} = d$ . Let player $X$ be at $P$ with $P S_{2} = x$ (the target).

The ball's speed is $v$ , so:

$P S_{3} = 5 v$ , $P S_{1} = 3 v$ , $P S_{2} = 4 v$ .

Step 2: Apply Apollonius in $△ P S_{1} S_{3}$ .

$P S_{2}$ is the median to side $S_{1} S_{3}$ , so:

$P S_{1}^{2} + P S_{3}^{2} = 2 (P S_{2}^{2} + S_{2} S_{1}^{2})$

$(3 v)^{2} + (5 v)^{2} = 2 ((4 v)^{2} + d^{2})$

$9 v^{2} + 25 v^{2} = 32 v^{2} + 2 d^{2}$

$34 v^{2} = 32 v^{2} + 2 d^{2}$

$2 v^{2} = 2 d^{2} \Rightarrow v = d$ (both positive).

Step 3: Identify the target.

We need $P S_{2} = 4 v$ . Since $v = d$ , knowing either $v$ or $d$ determines the answer.

Step 4: Evaluate Statement 1.

$v = 3.6$ km/h $= 3.6 \times \frac{5}{18} = 1$ m/s.

Then $P S_{2} = 4 \times 1 = 4$ m. Statement 1 alone is sufficient.

Step 5: Evaluate Statement 2.

$S_{1} S_{3} = 2$ m $\Rightarrow 2 d = 2 \Rightarrow d = 1$ m.

Since $v = d$ , $v = 1$ m/s, so $P S_{2} = 4$ m. Statement 2 alone is sufficient.

Answer: C

Coordinate Geometry, Lines and Triangles Previous Year Questions (PYQs) for XAT: 5+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Coordinate Geometry, Lines and Triangles

Chapter Roadmap

The Cartesian Bridge: Geometry Meets Algebra

The Cartesian Bridge

Coordinate Geometry, Lines and Triangles: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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