Very Short Answer on Points and Their Coordinates Class 11 OP Malhotra Exe-26C ISC Maths Solutions Ch-26. In this article you would learn to solve hard problems easily on Points and Their Coordinates. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.
Points and Their Coordinates Class 11 OP Malhotra Very Short Answer ISC Maths Solutions Ch-26
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-26 | Points and Their Coordinates |
| Writer | OP Malhotra |
| Exe-26(C) | Very Short Answer Type Questions. |
Very Short Answer on Points and Their Coordinates
OP Malhotra ISC Class 11 Maths Solutions
Que-1: What are the coordinates of a point lying on x-axis?
Sol: Any point on x-axis has y-coordinate = 0.
Hence, general form is (x, 0).
Que-2: In which octant do the points (a)(5, 7, 8) (b) (3, −1, −4) (c) (−6, 2, 4) (d) (3, 2, −4) lie?
Sol: Octants depend on signs of (x, y, z):
(a) (5,7,8) → (+,+,+) ⇒ 1st octant
(b) (3,−1,−4) → (+,−,−) ⇒ 8th octant
(c) (−6,2,4) → (−,+,+) ⇒ 2nd octant
(d) (3,2,−4) → (+,+,−) ⇒ 5th octant
Que-3: Find the distance between the points (−1, 3, −4) and (1, −3, 4).
Sol: Distance formula:
√[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²]
= √[(1+1)² + (−3−3)² + (4+4)²]
= √[4 + 36 + 64]
= √104 = 2√26 units.
Que-4: Find the distance of the point (3, 12, 5) from the x-axis.
Sol: Distance from x-axis = √(y² + z²)
= √(12² + 5²)
= √(144 + 25)
= √169 = 13 units.
Que-5: What is the distance of the point (3, 4, 5) from the yz-plane?
Sol: Distance from yz-plane = |x|
= |3| = 3 units.
Que-6: If A, B, C be the feet of the perpendiculars from a point (4, −3, −5) on the X, Y and Z axes, then find the coordinates of A, B and C.
Sol: Foot on X-axis ⇒ y = 0, z = 0 ⇒ A = (4, 0, 0)
Foot on Y-axis ⇒ x = 0, z = 0 ⇒ B = (0, −3, 0)
Foot on Z-axis ⇒ x = 0, y = 0 ⇒ C = (0, 0, −5)
Que-7: Show that if x² + y² = 1, then the point (x, y, √(1 − x² − y²)) is at a distance 1 unit from the origin.
Sol: Distance from origin:
= √[x² + y² + (√(1 − x² − y²))²]
= √[x² + y² + (1 − x² − y²)]
= √1 = 1 unit.
Que-8: Find the coordinates of the point which divides the line segment joining the points (−2, 3, 5) and (1, −4, 6) internally in the ratio 2 : 3.
Sol: Using section formula:
x = [2(1) + 3(−2)] / (2+3) = (2 − 6)/5 = −4/5
y = [2(−4) + 3(3)] / 5 = (−8 + 9)/5 = 1/5
z = [2(6) + 3(5)] / 5 = (12 + 15)/5 = 27/5
Required point = (−4/5, 1/5, 27/5).
Que-9: Find the coordinates of the mid-points of the line joining the points (−3, 7, 8) and (5, 11, −8).
Sol: Midpoint formula:
= [(−3+5)/2 , (7+11)/2 , (8−8)/2]
= (2/2 , 18/2 , 0)
= (1, 9, 0).
Que-10: Find the coordinates of the image of the point (3, 4, 2) in (i) xy-plane (ii) yz-plane (iii) zx-plane.
Sol: (i) xy-plane ⇒ z changes sign ⇒ (3, 4, −2)
(ii) yz-plane ⇒ x changes sign ⇒ (−3, 4, 2)
(iii) zx-plane ⇒ y changes sign ⇒ (3, −4, 2)
–: End Points and Their Coordinates Class 11 OP Malhotra Exe-26C ISC Maths Ch-26 Solutions :–
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