Coordinate Geometry Aptitude Questions

Q1 What is the equation of a circle of radius 6 units centered at (3, 2)?

Solution

x² + y² - 6x - 4y = 23

Q2 Find k, if the line 2x - 3y = 11 is perpendicular to the line 3x + ky = -4?

Solution

None

Q3 In which quadrant does the point (4, - 6) lies?

Solution

The point (4,-6) lies in 4th quadrant

Q4 The separation of the point p (8, - 6) from the beginning is:

Solution

Op = √ (8-0)2+ (-6-0)2 = √64+ 36 = √100 =10 unit

Q5 P is a point on x-hub at a separation of 3 units from y-pivot on its right side. The co-ordinates of P are:

Solution

the co-ordinates of P are P (3, 0).

Q6 The vertices of a quadrilateral ABCD are A(0,0) ,B(4,4) ,C(4,8) and D(0,4).Then ABCD is a

Solution

AB2= (4-0) 2+ (4-0) 2= 32
 BC2= (4-4) 2+ (8-4) 2= 0+16= 16
 CD2= (0-4) 2+ (4-8) 2= (16+16) = 32
 AB= CD= √32 = 4√2, BC= AD =√16 = 4
 AC2= (4-0) 2+ (8-0) 2= (16+64) = 80
 BD2= (0-4) 2+ (4-4) 2= 16+0= 16
 ∴ Diag = AC≠Diag BD.
 ∴ ABCD is a parallelogram.

Q7 The end purposes of the distance across of a circle are A (4,- 1) and B(- 2,- 5). The co-ordinates of its middle are:

Solution

Co-ordinates of the center are [4+ (-2)/2, -1+ (-5)/2], i.e. (1,-3)

Q8 The lines x+2y-9 =0 and 3x+6y+8 =0 are commonly.

Solution

x+2y-9 =0 ⇒ 2y = -x+9 ⇒ y= -x/2+9/2
3x+6y+8 =0 ⇒ 6y= -3x-8 ⇒ y=-x/2 -4/3
∴ m₁: m₂ = -1/2
Hence, the given lines are parallel.

Q9 On the off chance that the slant of a line joining the focuses A(x,- 3) and B(2,5) is 135⁰ then x=?

Solution

Slop of AB = (5+3)/(2-x) =8/2-x
∴ 8/2-x = tan135⁰ = tan (180⁰- 45⁰) = -tan 45⁰ =-1
8/2-x = -1 ⇒8 = x-2 ⇒ x= 10

Q10 A line goes through the focuses A (- 2, 3) and B (- 6, 5). The slop of line AB is

Solution

  Slop = (y₁ ?y₂)/( x₁-x₂) = (5-3)/(-6+2) = 2/-4 =-1/2