Find the distance between the following pair of points:
(i) (-6,7) and (-1, -5)
(ii) (a + b, b + c) and (a - b, c - b)
(iii) (a sin a, - b cos a) and (-a cos a, b sin a)
(iv) (a, 0) and (0, b)
Answers
(i) (-6,7) and (-1, -5)
By using distance formula : √(x2 - x1)² + (y2 - y1)²
Distance = √(-1 - (- 6))² + (- 5 - 7)²
= √(-1 + 6)² + (-12)²
= √5² + 144
= √25 + 144
= √169
= 13 units
Distance = 13 units
Hence, the distance between the (-6,7) and (-1, -5) pair of points is 13 units.
(ii) (a + b, b + c) and (a - b, c - b)
By using distance formula : √(x2 - x1)² + (y2 - y1)²
Distance = √{(a - b) - (a + b) }² + {(c - b) - (b + c)²}
= √{a - b - a - b}² + {c - b - b - c)²}
= √(-2b)² + (-2b)²
= √4b² + 4b²
= √8b²
= 2√2b units
Hence, the distance between the (a + b, b + c) and (a - b, c - b) pair of points is 2√2b units.
(iii) (a sin a, - b cos a) and (-a cos a, b sin a)
By using distance formula : √(x2 - x1)² + (y2 - y1)²
Distance = √{(-acos a - asin a }² + {( bsin a) - ( - bcos a)²}
= √{(a²cos² a + a²sin²a + 2acos asin a } + {( b²sin² a + b²cos² a + 2 bsina × bcosa)}
=√ {(a²cos² a + a²sin²a + 2a²sina cosa } + {(b²sin² a + b²cos² a + 2b²sina cosa)}
= √a² (cos² a + sin²a ) b²(sin² a + cos² a) + 2sina cosa (a² + b²)
= √a² + b² + sin2a(a² + b²)
[cos² a + sin²a = 1 , 2sinacosa = sin2a]
= √a² + b² [1 + sin2a]
Hence, the distance between the ((a sin a, - b cos a) and (-a cos a, b sin a) pair of points is √a² + b² [1 + sin2a] units.
(iv) (a, 0) and (0, b)
By using distance formula : √(x2 - x1)² + (y2 - y1)²
Distance = =√(0 - a)² + (b - 0)²
= √a² + b
Hence, the distance between the (a, 0) and (0, b) pair of points is √a² + b units.
HOPE THIS ANSWER WILL HELP YOU……
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