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Test Level 2: Coordinate Geometry - 2 - CAT MCQ


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20 Questions MCQ Test - Test Level 2: Coordinate Geometry - 2

Test Level 2: Coordinate Geometry - 2 for CAT 2024 is part of CAT preparation. The Test Level 2: Coordinate Geometry - 2 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 2: Coordinate Geometry - 2 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 2: Coordinate Geometry - 2 below.
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Test Level 2: Coordinate Geometry - 2 - Question 1

What kind of a quadrilateral is formed by the vertices (0, 0), (4, 3), (3, 5) and (-1, 2)?

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 1

Let the points be A(0, 0), B(4, 3), C(3, 5) and  D(-1, 2) 
Let A and B be the easy points then,

The sides are equal. So, the quadrilateral is at least a parallelogram.

and (AC)2 ≠ (AB)2 + (BC)2
Therefore, the quadrilateral is not rectangle.
So, the given quadrilateral is a parallelogram.

Test Level 2: Coordinate Geometry - 2 - Question 2

If a, b, c are in A.P. then the fixed point through which the straight line ax + 2by + c = 0 will always pass, is

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 2

Since, a, b, c are in A.P.
So, 2b = a + c.
Now, only (1, -1) satisfies the equation.

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Test Level 2: Coordinate Geometry - 2 - Question 3

In the above figure, if O(0, 0) is the centre of the circle, then which of the following points lies outside the circle?

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 3

A point will be outside the given circle if its distance from the centre, i.e. origin O, is more than 10.
Now, the distance of point (-7, -8) from O is , which is more than 10.
So, point (-7, -8) lies outside the given circle.
Hence, option (1) is correct.

Test Level 2: Coordinate Geometry - 2 - Question 4

Find the equation of a straight line which contains the point  with inclination 150o.

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 4

Slope of the line, m = tan 150o = tan (90o + 60o) = -cot 60o = -1/√3
y = mx + c 

Line contains the point

Test Level 2: Coordinate Geometry - 2 - Question 5

Find the point which divides the line joining the points A (1, 3) and B (2, 7) in the ratio of 3 : 4 externally.

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 5


Test Level 2: Coordinate Geometry - 2 - Question 6

If C is the centroid of the triangle PQR and the coordinates of points P, Q, R and C respectively are (x, 1), (0, y), (1/2, 0) and (1/2, 1/3), find the measure of angle PQR.

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 6

The centroid of ΔPQR has coordinates (1/2, 1/3).
Therefore, 1/2 = (x + 0 + 1/2)/3 and 1/3 = (1 + y + 0)/3
So, x = 1 and y = 0
Angle PQR is equal to angle between line PQ and line QR.
Slope of PQ (m1)= (0 - 1)/(0 - 1) = 1
Slope of QR (m2)= (0 - 0)/((1/2) - 0) = 0
Angle between line PQ and line QR = tan θ = (m1 - m2)/(1 + m1m2) = (1 - 0)/(1 + (0)(1)) = 1
tan θ = 1
θ = 45°

Test Level 2: Coordinate Geometry - 2 - Question 7

In the xy-plane, which of the following points is the greatest distance from the origin?

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 7

The square of the distance from the origin of (x, y) is x2 + y2.
This values for the different points given in options are 9, (4 + 9), (1 + 9) and 9, respectively. Among these, 4 + 9 is the greatest.

Test Level 2: Coordinate Geometry - 2 - Question 8

The projection (the foot of perpendicular) from (x, y) on the x-axis is

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 8

The foot of perpendicular of a point on the x-axis always has abscissa equal to the perpendicular distance of the point from the y-axis and ordinate 0.
The projection or foot of perpendicular of (x, y) on the x-axis is (x, 0).

Test Level 2: Coordinate Geometry - 2 - Question 9

If a, b and c are the sides of a triangle, and a2 + b2 + c2 = ab + bc + ca, then the triangle is

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 9

a2 + b2 + c2 = ab + bc + ac
(a - b)2 + (b - c)2 + (c - a)2 = 0  ...(1)
Since the sum of squares is zero, each term should be zero. 
⇒ (a - b)2 = 0, (b - c)2 = 0, (c - a)2 = 0   
⇒ a = b = c
Hence, the triangle is equilateral.

Test Level 2: Coordinate Geometry - 2 - Question 10

In the given figure, if PQ is parallel to OR, then what is the area of quadrilateral PQRO?

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 10

Since ∠POR = 90° and ∠QOR = 45°, POQ is also equal to 45°.
So, ∠PQO = 45° and ∠QPO = 90°
Therefore, ΔQPO is an isosceles right-angled triangle.
So, OP = PQ = 4 units and OR = 5 units.
Now, OPQR is a trapezium with PQ and OR as its parallel sides.
So, area = ½(PQ + OR) × OP = ½ × (4 + 5) × 4 = 18 sq. units

Test Level 2: Coordinate Geometry - 2 - Question 11

The point that divides the line joining the points (1, 2) and (3, 4) internally in the ratio of 1 : 1 lies in the  

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 11

Since the coordinates of both the points are positive, the line joining them lies in the first quadrant. The division of the line in the ratio of 1 : 1 means that the mid-point of this line also lies in the first quadrant, with both coordinates positive.

Test Level 2: Coordinate Geometry - 2 - Question 12

The line (1 + K)X + (3 – K)Y = 2(1 + 3K) passes through a fixed point P for any value of K. Find the co-ordinates of P.

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 12

(1 + K)X + (3 – K)Y = 2(1 + 3K)            
Check the options.
Option (2): L.H.S. = (1 + K)(5) + (3 – K)(–1)
= 5 + 5K – 3 + K
= 2 + 6K
= 2(1 + 3K) = R.H.S.

Test Level 2: Coordinate Geometry - 2 - Question 13

The vertices of a square S have coordinates (-1, -2), (-1, 1), (2, 1) and (2, -2). What are the coordinates of the point where the diagonals of S intersect?

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 13


The diagonals of a square bisect each other.
So, the point at which the diagonals meet will be the midpoint of a line joining a pair of opposite vertices.
As can be seen from the figure, (-1, -2) and (2, 1) are opposite vertices.
Their midpoint has coordinates ((-1 + 2)/2, (-2 + 1)/2), or (1/2, -1/2).

Test Level 2: Coordinate Geometry - 2 - Question 14

In the rectangular coordinate system shown below, the area of triangle RST is _______.

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 14

x–coordinates of R and T are 1 and c, respectively. So, the length of RT = (c – 1).
Length of the altitude from S = Its distance from the x–axis = Its y–coordinate = b 
So, the area of ΔRST = b(c – 1)/2 

Test Level 2: Coordinate Geometry - 2 - Question 15

The incentre of a triangle formed by the lines y = 0, 3x + 4y = 48 and 3y - 4x = 36 is

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 15


Point C(x3 ,y3) is the intersection of 3x + 4y = 48 with y = 0.  
So, the coordinates of C are (16, 0).
Point A(x1 ,y1) is the intersection of 3y - 4x = 36 with y = 0. 
So, the coordinates of A are (-9, 0).
Point B(x2 ,y2) is the intersection of 3x + 4y = 48 and 3y - 4x = 36.
So, the coordinates of B are (0, 12).

Coordinates of the incentre 


= (1, 5)

Test Level 2: Coordinate Geometry - 2 - Question 16

Two points (a, 0) and (0, b) are joined by a straight line. Another point on this line is

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 16

The equation of the line passing through (a, 0) and (0, b) is: 

On putting the options, we get to know that option (1) is correct.

Test Level 2: Coordinate Geometry - 2 - Question 17

If the coordinates of the centroid of a triangle are (2/3, 5/3) and the coordinates of two vertices of the triangle are (-1, 0) and (3, 0), then what are the coordinates of the third vertex of the triangle?

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 17

Coordinates of centroid (G)


y3 = 5
∴ (x3, y3) = (0, 5)

Test Level 2: Coordinate Geometry - 2 - Question 18

A line segment with end points A(-6, 10) and B(3, -8) is divided in the ratio 2 : 7 by a point P. Find the coordinates of point P.

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 18

If a line segment with end points A(x1, y1) and B(x2, y2) is divided in the ratio p : q by a point P(x, y).

For given condition

Test Level 2: Coordinate Geometry - 2 - Question 19

If point (t, 1) is located inside the circle x2 + y2 = 10, then t must lie between 

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 19

As (t, 1) is located inside the given circle, its distance from the centre, i.e. origin, should be less than the radius, i.e. √10 units.
∴ Distance between (t, 1) and origin 

so, 

Test Level 2: Coordinate Geometry - 2 - Question 20

Find the inclination and perpendicular distance of a line represented by (x/2) + (√3y/2) = 12√5 from the origin.

Detailed Solution for Test Level 2: Coordinate Geometry - 2 - Question 20

Equation of the line (x/2) + (√3y/2) = 12√5 can also be written as: 
x cos 60° + y sin 60° = 12√5 
Comparing this with the normal form of equation, x cos θ + y sin θ = p, we get the inclination as 60° and the perpendicular distance from the origin as 12√5 units.

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