Geometry - Coordinate Geometry - Previous Year CAT/MBA Questions

Answer: Option C

Explanation :

Let (x, y) be the coordinates of 4^th vertex.
In a parallelogram, diagonals bisect each other.
∴ Intersection point of diagonals is same as the mid-point of two opposite vertices.

Sum of coordinates of two pair of opposite vertices is same.

A and C are opposite vertices while B and D are opposite vertices.

(1, 1) is opposite (-2, 8) while (3, 4) is opposite (x, y)
⇒ 1 - 2 = 3 + x ⇒ x = -4
⇒ 1 + 8 = 4 + y ⇒ y = 5
∴ (x, y) = (-4, 5)

Of the three possibilities only (-4, 5) is mentioned in options.

Hence, option (c).

Answer: Option D

Explanation :

One of the diagonals of the parallelogram connects the points (2, 1) and (-3, -4).

Line x + 9y + c = 0, represents the other diagonal of the parallelogram.

We know, diagonals of a parallelogram bisect each other. 

∴ x + 9y + c = 0 should pass through the midpoint of the line joining (2, 1) and (-3, -4).

The midpoint of the line joining (2, 1) and (-3, -4) = $\left(\frac{2- 3}{2},\frac{1 - 4}{2}\right)$ = (-0.5, -1.5)

⇒ Point (-0.5 , -1.5) lies on x + 9y + c = 0.

⇒ -0.5 + 9 × (-1.5) + c = 0

⇒ -0.5 - 13.5 + c = 0

⇒ c = 14.

Hence, option (d).

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).

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).

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.

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.
[In a right triangle, hypotenuese is the diameter of circumcircle.]

Diameter = √(15² + 9²) = √306.

Radius = (√306)/2 = 8.74 ≈ 9 units.

Hence, 9.

Answer: Option B

Explanation :

The area of the triangle is 32 sq units.

Therefore, 32 = ½ × BC × height

Given, 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 A should lie on the x-axis as shown.

Hence, point A is at a distance of 8 units from x = 4 and lies on x-axis.

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

Hence, option (b).

Answer: Option C

Explanation :

The closed region bounded by |ax| + |by| = c in the two-dimensional plane has x-intercepts of ± $\frac{\mathrm{c}}{\left|\mathrm{a}\right|}$ and y-intercepts of ± $\frac{\mathrm{c}}{\left|\mathrm{b}\right|}$. 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 (c).

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 [Diagonal of a rectangle bisect each other.]

Hence co-ordinates of O will satisfy the equation

y = 3x + c
∴ 4 = 3 × 4 + c
⇒ c = – 8

Hence, option (d).

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)

∴ Total solutions = 1 + 2 + 3 + ... + 39 = 39 × 40/2 = 780 integer solutions to x + y < 41.
 
The number of points with integer coordinates lying inside the circle = 780

Hence, option (a).

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 x₁ = y₁ = a, x₂ = a + 1, y₂ = a, x₃ = a and y₃ = 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) - a²]} = $\frac{1}{2}$ {- 2a - a + 3a + 2} = 1

Hence, option (b).

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 (b).

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 (d).

Answer: Option C

Explanation :

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

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)

Answer: Option A

Explanation :

The distance between two points (x₁, y₁) and (x₂, y₂) 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).