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Chapter

Coordinate Geometry

EXERCISE

Type - 1
Question 1 :
Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243
(ii) 256
(iii) 72
(iv) 675
Formulae used :

Distance Formula to find distance between two points (x1,y1) and (x2,y2) is :                     D = √[(x2 – x1)2 + (y2 – y1)2 ]. 

Solution :
  (i) We have 243 = 3 × 3 × 3 × 3 × 3
                  
      The prime factor 3 is not a group of three.
      ∴ 243 is not a perfect cube.
       Now, [243] × 3 = [3 × 3 × 3 × 3 × 3] × 3
            or 729 =3 × 3 × 3 × 3 × 3 × 3
        Now, 729 becomes a perfect cube.
        Thus, the smallest required number to multiply 243 to make it a perfect cube is 3.
Class Work
Q.1.   Find the distance between the  following pairs of points:
      (i)   (2, 3), (4, 1)
     (ii)   (–5, 7), (–1, 3)
     (iii)  (a, b), (–a, –b)
Q.1.   Find the distance between the  following pairs of points:
      (i)   (2, 3), (4, 1)
     (ii)   (–5, 7), (–1, 3)
     (iii)  (a, b), (–a, –b)
Q.1.   Find the distance between the  following pairs of points:
      (i)   (2, 3), (4, 1)
     (ii)   (–5, 7), (–1, 3)
     (iii)  (a, b), (–a, –b)
Answer
(i) 5 m
(i) 5 m
(i) 5 m
Assignment
Q.1.   Find the distance between the  following pairs of points:
      (i)   (2, 3), (4, 1)
     (ii)   (–5, 7), (–1, 3)
     (iii)  (a, b), (–a, –b)
Answer
(i) 5 m
(i) 5 m
(i) 5 m
NCERT Questions
Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients:
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 – x – 4
Answer
(i) 5 m
(i) 5 m
(i) 5 m
CBSE Questions
Q.1.   Find the distance between the  following pairs of points:
      (i)   (2, 3), (4, 1)
     (ii)   (–5, 7), (–1, 3)
     (iii)  (a, b), (–a, –b)
Answer
(i) 5 m
(i) 5 m
(i) 5 m
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