# Geometry - Coordinate Geometry - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic Geometry - Coordinate Geometry. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**CAT 2020 QA Slot 3 | Geometry - Coordinate Geometry**

The vertices of a triangle are (0, 0), (4, 0) and (3, 9). The area of the circle passing through these three points is

- A.
123π/7

- B.
205π/9

- C.
14π/3

- D.
12π/5

Answer: Option B

**Explanation** :

The diagram can be drawn as follows:

Height of the given triangle = 9 and base = 4

∴ Area of the triangle = $\frac{1}{2}\times 9\times 4$ = 18

We know area of a triangle = abc/4R

(where a, b, and c are the sides of the triangle and R is the circumradius)

a = AB = 4 units

b = BC = $\sqrt{{9}^{2}+{1}^{2}}=\sqrt{82}$ units

c = AC = $\sqrt{{3}^{2}+{9}^{2}}=\sqrt{90}$ units

The sides of the triangle are 4, √82 and √90

⇒ 18 = $\frac{4\times \sqrt{82}\times \sqrt{90}}{4R}$

⇒ R = $\frac{\sqrt{205}}{3}$

∴ Area of the circumcircle = πR2 = $\frac{205\mathrm{\pi}}{9}$

Hence, option (b).

Workspace:

**CAT 2020 QA Slot 3 | Geometry - Coordinate Geometry**

The area, in sq. units, enclosed by the lines x = 2, y = |x – 2| + 4, the X-axis and the Y-axis is equal to

- A.
6

- B.
10

- C.
8

- D.
12

Answer: Option B

**Explanation** :

We can plot x = 2 and y = |x - 2| + 4

The required area is sum of the triangle (orange) + rectangle (green).

Area of triangle (orange) = 1/2 × 2 × 2 = 2

Area of rectangle = 2 × 4 = 8

∴ Total required area = 2 + 8 = 10 sq. units.

Hence, option (b).

Workspace:

**CAT 2019 QA Slot 1 | Geometry - Coordinate Geometry**

With rectangular axes of coordinates, the number of paths from (1, 1) to (8, 10) via (4, 6), where each step from any point (x, y) is either to (x, y + 1) or to (x + 1, y), is

Answer: 3920

**Explanation** :

We have to go from (1, 1) to (8, 10) via (4, 6)

Number of paths from (1,1) to (8,10) = [Number of paths from (1,1) to (4,6)] × [Number of paths from (4,6) to (8,10)]

Path from (1,1) to (4,6):

Number of horizontal displacements (∆x) = 4 − 1 = 3 units and

Number of vertical displacements (∆y) = 6 − 1 = 5 units.

Hence, a total of 8 units.

∴ Number of paths from (1,1) to (4,6) = ^{8}C_{3} × ^{5}C_{5} = 56.

Path from (4,6) to (8,10):

Number of horizontal displacements (∆x) = 8 − 4 = 4 units and

Number of vertical displacements (∆y) = 10 − 6 = 4 units.

Hence, a total of 8 units.

∴ Number of paths from (4,6) to (8,10) = ^{8}C_{4} × ^{4}C_{4} = 70.

Total required number of paths = 56 × 70 = 3920.

Hence, 3920.

Workspace:

**CAT 2019 QA Slot 1 | Geometry - Coordinate Geometry**

Let S be the set of all points (x, y) in the x-y plane such that |x| + |y| ≤ 2 and |x| ≥ 1. Then, the area, in square units, of the region represented by S equals

Answer: 2

**Explanation** :

|x| ≥ 1 ⇒ x ≥ 1 and x ≤ −1. This is represented as the shaded region in the figure below.

|x| + |y| ≤ 2 : This will have four subcases depending on which quadrant the point is.

1st quadrant (x > 0; y > 0) : y ≤ 2 − x

2nd quadrant (x < 0; y > 0) : y ≤ 2 + x

3rd quadrant (x < 0; y < 0) : y ≥ −2 − x

4th quadrant (x > 0; y < 0) : y ≥ x − 2

Combining these four, the graph for |x| + |y| ≤ 2 is as shown below.

The intersection of |x| + |y| ≤ 2 and |x| ≥ 1 is as shown as the shaded area in the following image.

We need to find Area ∆ABC + Area ∆DEF

Area ∆ABC = Area ∆DEF = (1/2) × BC × AM = (1/2) × 2 × 1 = 1 square unit.

Required area = 1 + 1 = 2 square units.

Hence, 2.

Workspace:

**CAT 2019 QA Slot 1 | Geometry - Coordinate Geometry**

Let T be the triangle formed by the straight line 3x + 5y − 45 = 0 and the coordinate axes. Let the circumcircle of T have radius of length L, measured in the same unit as the coordinate axes. Then, the integer closest to L is

Answer: 9

**Explanation** :

3x + 5y − 45 = 0 cuts the coordinate axes at C(15,0) and A(0,9) as shown in the image below.

∠ABC = 90°, so we can deduce that AC is the diameter of the circumcircle.

Diameter = √(15^{2} + 9^{2}) = √306.

Radius = (√306)/2 = 8.74 ≈ 9 units.

Hence, 9.

Workspace:

**CAT 2018 QA Slot 2 | Geometry - Coordinate Geometry**

A triangle ABC has area 32 sq units and its side BC, of length 8 units, lies on the line x = 4. Then the shortest possible distance between A and the point (0,0) is

- A.
8 units

- B.
4 units

- C.
$2\sqrt{2}$ units

- D.
$2\sqrt{4}$ units

Answer: Option B

**Explanation** :

The area of the triangle is 32 sq units.

Therefore, 32 = ½ × BC × height =

Givne, BC = 8

⇒ Height = $\frac{2\times 32}{8}$ = 8 cm

Therefore Height of the triangle = 8 cm.

Side BC lies on the line x = 4 or it is parallel to y-axis. Therefore the height of the triangle will be parallel to x axis. In order to minimize the distance of point A from the origin, point B should be (4, 4) and point C should be (4, -4), as shown.

Therefore point A is (-4,0) and the distance of point A from the origin = 4 units.

Hence, option 2.

Workspace:

**CAT 2017 QA Slot 1 | Geometry - Coordinate Geometry**

The area of the closed region bounded by the equation | x | + | y | = 2 in the two-dimensional plane is

- A.
4π

- B.
4

- C.
8

- D.
2π

Answer: Option C

**Explanation** :

The closed region bounded by |ax| + |by| = c in the two- dimensional plane has x-intercepts of ± c/|a| and y-intercepts of ± c/|b| . This is in general a rhombus. In the given question, we have a square which has each of its diagonals as 4.

Area = 1/2 × 4 × 4 = 8.

Hence, option 3.

Workspace:

**CAT 2017 QA Slot 2 | Geometry - Coordinate Geometry**

The points (2, 5) and (6, 3) are two end points of a diagonal of a rectangle. If the other diagonal has the equation y = 3x + c, then c is

- A.
-5

- B.
-6

- C.
-7

- D.
-8

Answer: Option D

**Explanation** :

Let ABCD be the rectangle where A = (2, 5) and C= (6, 3). We know that AC and BD will be the diagonals of the rectangle. It can be represented using the diagram shown below

In the above figure, O is the point of intersection of the 2 diagonals and the mid point of both the diagonals.

Mid Point of AC = $\left(\frac{2+6}{2},\frac{5+3}{2}\right)\equiv $ (4, 4)

Now O also lies on diagonal BD

Hence co-ordinates of O will satisfy the equation

y = 3x + c

∴ 4 = 3 × 4 + c ⇒ c = –8

Hence, option 4.

Workspace:

**CAT 2005 QA | Geometry - Coordinate Geometry**

Consider a triangle drawn on the X-Y plane with its three vertices at (41, 0), (0, 41) and (0, 0), each vertex being represented by its (X, Y) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is

- A.
780

- B.
800

- C.
820

- D.
741

Answer: Option A

**Explanation** :

The points satisfying the equations x + y < 41, y > 0, x > 0 lie inside the triangle.

Integer solutions of x + y < 41 can be found as follows:

If x + y = 40, then

(x, y) could be (1, 39), (2, 38), …, (39, 1) ... (39 solutions)

If x + y = 39, then

(x, y) could be (1, 38), (2, 37), …, (38, 1) ... ( 38 solutions)

If x + y = 38, we get 37 solutions and so on till x + y = 2 ... (1 solution)

Thus, there are 39 × 40/2 = 780 integer solutions to x + y < 41.

The number of points with integer coordinates lying inside the circle = 780

Hence, option 1.

Workspace:

**CAT 2002 QA | Geometry - Coordinate Geometry**

The area of the triangle with the vertices (a, a), (a + 1, a) and (a, a + 2) is

- A.
a

^{3} - B.
1

- C.
0

- D.
None of these

Answer: Option B

**Explanation** :

Area (∆) = $\left|\begin{array}{ccc}{x}_{1}& {y}_{1}& 1\\ {x}_{2}& {y}_{2}& 2\\ {x}_{3}& {y}_{3}& 3\end{array}\right|$

where *x*_{1} = *y*_{1} = *a*,* x*_{2} = *a* + 1, *y*_{2} = *a*, *x*_{3} = *a* and *y*_{3} = *a* + 2

∴ Area (∆) = $\frac{1}{2}\left|\begin{array}{ccc}a& a& 1\\ a+1& a& 1\\ a& a+12& 1\end{array}\right|$

∴ Area (∆) = $\frac{1}{2}${a[a(1) - (a + 2)(1)] - a[(a + 1)(1) - a(1)] + 1[(a + 1)(a + 2) - a^{2}]} = $\frac{1}{2}$ {- 2a - a + 3a + 2} = 1

Hence, option 2.

Workspace:

**CAT 2000 QA | Geometry - Coordinate Geometry**

The area bounded by the three curves |x + y| = 1, |x| = 1, and |y| = 1, is equal to

- A.
4

- B.
3

- C.
2

- D.
1

Answer: Option B

**Explanation** :

We need to find area bounded by |x + y| = 1, |x| = 1, and |y| = 1

i.e. all points P(x, y) in the plane such that −1≤ x + y ≤ 1, −1≤ x ≤ 1, and −1≤ y ≤ 1

−1≤ x + y will be the origin side of line x + y = −1

x + y ≤ 1 will be the origin side of line x + y = 1

−1≤ x ≤ 1 will be the area between the lines x = −1 and x = 1.

−1≤ y ≤ 1 will be the area between the lines y = −1 and y = 1.

Tracing these curves, we get the area shown in the graph below

∴ Shaded Area = Area of 3 squares of side 1 unit = 3 × (1)2 = 3

Hence, option 2.

Workspace:

**CAT 2000 QA | Geometry - Coordinate Geometry**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

There are two straight lines in the x-y plane with equations ax + by = c , dx + ey = f. Do the two straight lines intersect?

- a, b, c, d, e and f are distinct real numbers.
- c and f are non-zero.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option D

**Explanation** :

Statement A implies that a, b, c, d, e, f are distinct real numbers.

But if a/d = b/e = c/f then the lines may be parallel and might not intersect at all.

∴ Statement A alone is not sufficient.

Statement B implies that the equations are not homogenous equations.

∴ Statement B is also not sufficient.

The condition for intersecting lines is a/d ≠ b/e

Even after combining both the statements the above condition is not clear.

∴ We cannot be sure whether the lines are intersecting or parallel.

Hence, option 4.

Workspace:

**Directions: **Answer the questions based on the following information.

A robot moves on a graph sheet with X and Y-axis. The robot is moved by feeding it with a sequence of instructions. The different instructions that can be used in moving it, and their meanings are:

**CAT 1999 QA | Geometry - Coordinate Geometry**

The robot reaches point (6, 6) when a sequence of three instructions is executed, the first of which is a GOTO(x,y) instruction, the second is WALKX(2) and the third is WALKY(4). What are the value of x and y?

- A.
2, 4

- B.
0, 0

- C.
4, 2

- D.
2, 2

Answer: Option C

**Explanation** :

The final point is (6, 6). The previous point is (6, 2) and the one before is (4, 2).

Workspace:

**CAT 1999 QA | Geometry - Coordinate Geometry**

The robot is initially at (x, y), x > 0 and y < 0. The minimum number of instructions needed to be executed to bring it to the origin (0, 0) if you are prohibited from using the GOTO instruction is

- A.
2

- B.
1

- C.
x + y

- D.
0

Answer: Option A

**Explanation** :

Two instructions are needed, one parallel to the X-axis and the other parallel to the Y-axis. i.e. WALKX(–x) and WALKY (–y)

Workspace:

**CAT 1991 QA | Geometry - Coordinate Geometry**

What is the distance between the points A(3, 8) and B(–2, –7)?

- A.
5√2

- B.
5

- C.
5√10

- D.
10√2

Answer: Option A

**Explanation** :

The distance between two points (x_{1}, y_{1}) and (x_{2}, y_{2}) is given as $\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}$

Hence, required distance = $\sqrt{{(-2-3)}^{2}+{(-7-8)}^{2}}$ = 5√10.

Hence, option (c).

Workspace:

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