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

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

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

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).

Workspace:

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

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).

Workspace:

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

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 4.

Workspace:

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

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 2.

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

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 4.

Workspace:

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

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 3.

Workspace:

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

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 4.

Workspace:

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

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 3.

Workspace:

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

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 3.

Workspace:

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

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 4.

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