Following quiz provides Multiple Choice Questions (MCQs) related to Co-ordinate Geometry. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
The point (-3,-5) lies in 3rd quadrant
Q 2 - The separation of the point A (7, 4) and B (0, a) is:
AB2 = (7-3)2+ (4-1)2 = (42+32) = (16+9) =25 ⇒ AB= √25 = 5 unit.
Q 3 - A will be a point on y-pivot at a separation of 4unit from y-hub lying underneath x-hub. The co-ordinates of an are:
Clearly, The co- ordinates Of A are A (0,-4).
Q 4 - The vertices of a quadrilateral ABCD are A(0,0) ,B(4,4) ,C(4,8) and D(0,4).Then ABCD is a
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.
Q 5 - The vertices of a triangle are A (3, 8), B (- 4, 2) and C (5, - 1). The region of ∆ ABC is:
Here x₁=3, x₂= -4, Xᴈ = 5, y₁ =8, y₂= 2 and Yᴈ = -1 ∆ = 1/2 [x₁(y₂-Yᴈ) +x₂(Yᴈ-y₁) +Xᴈ (y₁-y₂)] =1/2 [3(2+1)-4(-1-8) +5(8-2)] = 1/2 (9+36+30) = 75/2 sq.unit.
Q 6 - A point C partitions the join Of A (1, 3) and B (2, 7) in the proportion 3:4. The co-ordinates of C are:
Co-ordinates of C are [(5*2+4*1/3+4), (3*7+4*3/ 3+4) i.e. (10/7, 33/7)
Q 7 - In what proportion is the line portion joining the focuses A (6, 3) and B (- 2, - 5) partitioned by the x-hub?
Let x-axis cut the join of A (6, 3) and B (-2,-5) in the ratio ℏ:1 at the point. Then, co-ordinates of P are P [(-2ℏ+6/ℏ+1), (-5ℏ+3/ℏ+1) But, P lies on x-axis. So, its ordinate is 0. ∴ -5ℏ+3/ℏ+1 =0 ⇒ -5ℏ+3 =0 ⇒ ℏ= 3/5 Required ratio is 3/5:1 i.e., 1:2.
Q 8 - On the off chance that for a line m= tan ℴ>0, then
m tan ℴ>0 ⇒ℴ is acute.
Q 9 - The slop of a line going through the focuses A (3, - 1) and B (3, 2) is:
Slop = (y₁ ?y₂)/( x₁-x₂) = (2+1)/ (3-3) = 3/0, which is not defined
Q 10 - The point made by the line x+√3y-6 =0 with positive course of x-pivot is
x+√3y-6= 0 ⇒ √3y= -x+6 ⇒ y = -1x/ √3+ 6/√3 ∴ m = tan ℴ =-1/√3 = -tan 30⁰ = tan (180⁰-30⁰) = tan150⁰ ∴ ℴ = 150⁰