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

The best way to prepare for Geometry - Coordinate Geometry is by going through the previous year **Geometry - Coordinate Geometry omet questions**.
Here we bring you all previous year Geometry - Coordinate Geometry omet questions along with detailed solutions.

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**XAT 2023 QADI | Geometry - Coordinate Geometry omet Question**

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

Answer: Option D

**Explanation** :

The length of AC = (b + 4c) - (b - 2c) = 6c

Altitude from C to AB = a - (-2a) = 3a

∴ Area of ABC = 1/2 × b × h = 1/2 × 3a × 6c = 9ac

Hence, option (d).

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**XAT 2023 QADI | Geometry - Coordinate Geometry omet Question**

ABC is a triangle with BC = 5. D is thefoot of the perpendicular from A on BC. E is a point on CD such that BE = 3.

**The value of AB ^{2} - AE^{2} + 6CD is:**

- (a)
5

- (b)
10

- (c)
14

- (d)
18

- (e)
21

Answer: Option E

**Explanation** :

Given, BC = 5 and BE = 3

In △ABD ⇒ AB^{2} = AD^{2} + BD^{2} ...(1)

In △AED ⇒ AE^{2} = AD^{2} + DE^{2} ...(2)

(1) - (2)

⇒ AB^{2} - AE^{2} = BD^{2} - DE^{2}

⇒ AB^{2} - AE^{2} = x^{2} - (3 - x)^{2}

⇒ AB^{2} - AE^{2} = x^{2} - 9 - x^{2} + 6x

⇒ AB^{2} - AE^{2} = - 9 + 6x

We need to find AB^{2} - AE^{2} + 6CD

= (-9 + 6x) + 6 × (5 - x)

= -9 + 6x + 30 - 6x

= 21

Hence, option (e).

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**IIFT 05 Dec 2021 QA | Geometry - Coordinate Geometry omet Question**

A triangular park named ABC, is required to be protected by green fencing. The length of the side BC is 293. If the length of side AB is a perfect square, the length of the side AC is a power of two (2), and the length of side AC is twice the length of side AB.

Determine how much fencing is required to cover the triangular park.

- (a)
1079

- (b)
1024

- (c)
1096

- (d)
1061

Answer: Option D

**Explanation** :

ABC is a triangle and BC = 293

Let the length of AB = x^{2}

The length of AC = 2^{y}

AC = 2AB

2^{y} = 2.x^{2}

2^{y} - 1 = x^{2}

2^{y} - 1 is a perfect square when y is odd.

Sum of any two sides of a triangle should be larger than the third side. Therefore,

2^{y} + x^{2} > 293, 293 + 2^{y} > x^{2} and 293 + x^{2} > 2^{y}

y > 8

If y = 9, AC = 512, x = 16; satisfies above equations.

If y = 11, AC = 2048, x = 32; doesn't satisfy above equations.

Therefore, AB = 256 and AC = 512

Fencing required = 256 + 512 + 293 = 1061

Answer is option D.

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**IIFT 05 Dec 2021 QA | Geometry - Coordinate Geometry omet Question**

ABCD is a rectangle in the clockwise direction. The coordinates of A are (1,3) and the coordinates of C are (5,1), the coordinates of vertices B and D satisfy the line y = 2x + c, then what will be the coordinates of the mid-point of BC.

- (a)
$\left(\frac{5}{2},\frac{7}{2}\right)$

- (b)
$\left(\frac{9}{2},\frac{5}{2}\right)$

- (c)
$\left(\frac{9}{5},\frac{7}{2}\right)$

- (d)
(3, 2)

Answer: Option B

**Explanation** :

Midpoint of AC passes through the line y = 2x + c

Midpoint of AC = (3, 2)

2 = 6 + c

c = -4

Line passing through B is y = 2x - 4

(slope of AB)(slope of BC) = -1

$\left(\frac{y-1}{x-5}\right)$$\left(\frac{y-3}{x-1}\right)$ = -1

Substituting y = 2x - 4 and solving, we get

B(2, 0) or B(4, 4)

B cannot lie on X-axis. Therefore, co-ordinates of B = (4, 4)

Midpoint of BC = $\left(\frac{9}{2},\frac{5}{2}\right)$

The answer is option B.

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**XAT 2019 QADI | Geometry - Coordinate Geometry omet Question**

Let P be the point of intersection of the lines 3x + 4y = 2a and 7x + 2y = 2018 and Q the point of intersection of the lines 3x + 4y = 2018 and 5x + 3y = 1. If the line through P and Q has slope 2, the value of a is:

- (a)
1/2

- (b)
1

- (c)
4035

- (d)
1009

- (e)
3026

Answer: Option E

**Explanation** :

P be the point of intersection of the lines 3x + 4y = 2a and 7x + 2y = 2018.

Solving these two equations we get the coordinates of P in terms of a.

∴ P ≡ $\left(\frac{4036-2a}{11},\frac{7a-3027}{11}\right)$

Q the point of intersection of the lines 3x + 4y = 2018 and 5x + 3y = 1

Solving these two equations we get the coordinates of Q.

∴ Q ≡ (-550, 917)

Now slope of line connecting P and Q is 2

∴ $\frac{{\displaystyle \frac{7a-3027}{11}}-917}{{\displaystyle \frac{4036-2a}{11}}+550}$ = $\frac{2}{1}$

⇒ $\frac{7a-13114}{10086-2a}$ = $\frac{2}{1}$

⇒ 11a = 33286

⇒ a = 3026

Hence, option (e).

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**IIFT 2019 QA | Geometry - Coordinate Geometry omet Question**

If the co-ordinates of orthocentre and the centroid of a triangle ABC are (–5, 7) and (5, 5), then the circumcentre of the triangle ABC is:

- (a)
(25, 1)

- (b)
(10, 4)

- (c)
(-5. 2)

- (d)
(0, 6)

Answer: Option B

**Explanation** :

We know that Centroid divides the line joining the Orthocenter and Circumcenter in the ratio 2 : 1.

Let the coordinates of Circumcenter be(x, y)

∴ Orthocenter ((–5, 7)) ; Centroid (5, 5); Circumcenter (x, y).

Using the section formula;

5 = (2x − 5)/3 ⇒ x = 10.

5 = (2y + 7)/3 ⇒ y = 4.

So, the coordinates of the circum-center are (10, 4).

Hence option (b).

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**IIFT 2017 QA | Geometry - Coordinate Geometry omet Question**

The coordinates of a triangle ABC are A (1, 5), B (−2, 3), and C(0, −4). What is the equation of the median AD?

- (a)
7x − 3y + 8 = 0

- (b)
5x − 4y + 15 = 0

- (c)
x + 3y − 16 = 0

- (d)
11x − 4y + 9 = 0

Answer: Option D

**Explanation** :

The co-ordinates of D are {[(−2 + 0)/2], [(3 − 4)/2]} = (−1, − 0.5)

Using the two point form, the equation of median AD is:

(y − 5) / (x − 1) = (−0.5 − 5) / (−1 − 1)

∴ (y − 5) / (x − 1) = 5.5 / 2 = 11 / 4

∴ 4y − 20 = 11x − 11

∴ 11x − 4y + 9 = 0

Hence, option (d).

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**IIFT 2016 QA | Geometry - Coordinate Geometry omet Question**

Suppose the two sides of a square are along the straight lines 6x - 8y = 15 and 4y - 3x = 2. Then the area of the square is

- (a)
2.52 sq.units

- (b)
3.61 sq.units

- (c)
4.33 sq.units

- (d)
None of the above

Answer: Option B

**Explanation** :

The two lines are 4*y* – 3*x* = 2 and 4*y* – 3*x* = −7.5

As *a*_{1} / *a*_{2} = *b*_{1 }/ *b*_{2} ≠ *c*_{1} / *c*_{2}; thetwo lines are parallel to each other.

The distance between the two parallel lines would be the side of the square

i.e. |*c*_{1} – *c*_{2}| / (√*a*^{2} + *b*^{2})

Here *a*^{2} + *b*^{2} = (−3)^{2} + (4)^{2} = 25 and |*c*_{1} – *c*_{2}| = |2 – (−15/2)| = 19/2

∴ Side of square = (19/2)/5 = 19/10

= 1.9 ∴ Area of square = (1.9)^{2}

= 3.61 sq. units

Hence, option (b).

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**XAT 2015 QA | Geometry - Coordinate Geometry omet Question**

Two diagonals of a parallelogram intersect each other at coordinates (17.5, 23.5). Two adjacent points of the parallelogram are (5.5, 7.5) and (13.5, 16). Find the lengths of the diagonals.

- (a)
15 and 30

- (b)
15 and 40

- (c)
17 and 30

- (d)
17 and 40

- (e)
Multiple solutions are possible

Answer: Option D

**Explanation** :

(13.5, 16) and (5.5, 7.5) are adjacent points of the parallelogram and (17.5, 23.5) is the point of intersection of two diagonals of the parallelogram.

Property: Diagonals of a parallelogram bisect each other.

Distance between (17.5, 23.5) and (5.5, 7.5):

$\sqrt{{(17.5-5.5)}^{2}+{(23.5-7.5)}^{2}}=20$

Length of diagonal that passes through (17.5, 23.5) and (5.5, 7.5) = 20 × 2 = 40 cm

Distance between (17.5, 23.5) and (13.5, 16):

$\sqrt{{(17.5-13.5)}^{2}+{(23.5-16)}^{2}}=8.5$

Length of diagonal that passes through (17.5, 23.5) and (13.5, 16) = 8.5 × 2 = 17 cm

Hence, option (d).

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**IIFT 2014 QA | Geometry - Coordinate Geometry omet Question**

A ladder just reaches a window that is 8 metres high above the ground on one side of the street. Keeping one end of the ladder at the same place, the ladder is moved to the other side of the street so as to reach a 12 metre high window. If the ladder is 13 metres long, what is the width of the street?

- (a)
14.6 metres

- (b)
15.8 metres

- (c)
15.2 metres

- (d)
15.5 metres

Answer: Option C

**Explanation** :

The length of the ladder is 13 metres.

Thus, for the cases given in the question, the height is given as 8 metres and 12 metres and the hypotenuse will be 13 metres for both cases.

Thus, by Pythagoras theorem, the width of the street = 5 + √105 = 15.2 metres

Hence, option (c).

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**IIFT 2013 QA | Geometry - Coordinate Geometry omet Question**

Your friend’s cap is in the shape of a right circular cone of base radius 14 cm and height 26.5 cm. The approximate area of the sheet required to make 7 such caps is

- (a)
6750 sq cm

- (b)
7280 sq cm

- (c)
8860 sq cm

- (d)
9240 sq cm

Answer: Option D

**Explanation** :

Let the base radius, height, slant height be r, h and l respectively.

r = 14 cm

h = 26.5 cm

∴ l = $\sqrt{{r}^{2}+{h}^{2}}$

∴ l = 29.97 cm ≈ 30 cm

To make a conical cap from a sheet of paper, the curved surface area of cap must be calculated.

Curved surface area = π × r × l

To calculate the area of the sheet required to make 7 caps,

total area = 7 × π × r × l = 7 × $\frac{22}{7}$ × 14 × 3

= 9240 sq.cm.

Hence, option (d).

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**IIFT 2013 QA | Geometry - Coordinate Geometry omet Question**

In an engineering college there is a rectangular garden of dimensions 34 m by 21 m. Two mutually perpendicular walking corridors of 4 m width have been made in the central part and flowers have been grown in the rest of the garden. The area under the flowers is

- (a)
320 sq.m

- (b)
400 sq.m

- (c)
510 sq.m

- (d)
630 sq.m

Answer: Option C

**Explanation** :

The following diagram displays the garden along with the paths,

The shaded area represents the flower beds,

For each rectangular flower bed

breadth = $\frac{21-4}{2}$ = 8.5 m

length = $\frac{34-4}{2}$ = 15 m

There are 4 such flower beds

Therefore, total area = 4 × 15 × 8.5 = 510 m^{2}

Hence, option (c).

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**IIFT 2012 QA | Geometry - Coordinate Geometry omet Question**

There are two buildings, one on each bank of a river, opposite to each other. From the top of one building – 60 m high, the angles of depression of the top and the foot of the other building are 30° and 60° respectively. What is the height of the other building?

- (a)
30 m

- (b)
18 m

- (c)
40 m

- (d)
20 m

Answer: Option C

**Explanation** :

Let the height of the shorter building be h.

Let the distance between the two buildings be x.

Therefore from the diagram,

tan 60 = $\frac{60}{x}$

∴ x = $\frac{60}{\sqrt{3}}$ = 20$\sqrt{3}$

∴ tan 30 = $\frac{60-h}{x}$

∴ 60 - h = 20

∴ h = 40

Hence, option (c).

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**IIFT 2009 QA | Geometry - Coordinate Geometry omet Question**

Mohan was playing with a square cardboard of side 2 metres. While playing, he sliced off the corners of the cardboard in such a manner that a figure having all its sides equal was generated. The area of this eight sided figure is:

- (a)
$\frac{4\sqrt{2}}{\left(\sqrt{2}+1\right)}$

- (b)
$\frac{4}{\left(\sqrt{2}+1\right)}$

- (c)
$\frac{2\sqrt{2}}{\left(\sqrt{2}+1\right)}$

- (d)
$\frac{8}{\left(\sqrt{2}+1\right)}$

Answer: Option D

**Explanation** :

Refer to the following figure.

As all the sides of the figure are equal,

2 - 2x = x$\sqrt{2}$

∴ x = $\frac{\sqrt{2}}{1+\sqrt{2}}$ = 2 - $\sqrt{2}$ ...(rationalising the denominator)

Area of the eight sided figure = Area of the square – 4 × Area of each small triangle

= 2 × 2 - 4 × $\frac{1}{2}$(2 - $\sqrt{2}$)^{2}

= 4 - 2(6 - 4$\sqrt{2}$)

= 8($\sqrt{2}$ - 1)

= $\frac{8}{\sqrt{2}+1}$ ...(multiplying and dividing by ($\sqrt{2}$ + 1)

Hence, option (d).

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