# Coordinate Geometry-18

### From Homeworkwiki

**Find the distance between the pair of points**

(a) (2, 5), (1, 6); (b) (-9, -2), (4, -3); (c) (a, b), (-a, -b); (d) (a cos Θ, a sin Θ), (b cos Θ, b sin Θ)

**Solution:** Distance between the points (x_{1}, y_{1}) and (x_{2}, y_{2}) is given by

√[ (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} ]

(a) distance between (2, 5) and (1, 6)

= √[ (2 - 1)^{2} + (5 - 6)^{2} ] = √(1^{2} + 1^{2}) = √2 units

(b) Distance between (-9, -2) and (4, -3)

= √[ (-9 - 4)^{2} + (-2 + 3)^{2} ] = √( 169 + 1) = √170 units

(c) Distance between (a, b) and (-a, -b)

= √[ (a + a)^{2} + (b + b)^{2} ] = √( 4a^{2} + 4a^{2} )

= 2 √(a^{2} + b^{2})

(d) Distance between (a cos Θ, a sinΘ) and (b cos Θ, b sin Θ)

= √[(a cos Θ - b cos Θ)^{2} + (a sinΘ - b sinΘ)^{2}]

= √[ cos^{2} Θ (a - b)^{2} + sin^{2} Θ (a - b)^{2} ]

√ [ (a - b)^{2} (cos^{2} Θ + sin^{2} Θ)] = √ (a - b)^{2} units

= (a - b) units