Question

find the distance between the following pairs of points
1.(3,4) and (-7,2)
2.(a,b) and (c,b)
3.(3.- 9) and (- 2,3)

Answer 1

1. ( 3, 4) and ( -7, 2)

We know that :

$\bf \: distance \: = \: \sqrt{( x_{1 } - x _{2})^{2} + {( y_{1} - y_{2}) }^{2} }$

Here,
$x_{1} =3 \\ x_{ 2} = - 7 \\ y_{1} = 4 \\ y _{2} = 2$

Putting the values :

$= \sqrt{ {(3 - ( - 7))}^{2} + {(4 - 2)}^{2} } \\ \\ = \sqrt{ ({10)}^{2} + {(2)}^{2} } \\ \\ = \sqrt{100 + 4} \\ \\ = \sqrt{104} \\ \\ = 2 \sqrt{26}$

2. (a, b) and (c, b)

Here,

$x_{1} =a \\ x_{2} = c \\ y _{1} = b\\ y _{2} = b$

Putting the values,

$= \sqrt{ {(a - c)}^{2} + {(b - b)}^{2} } \\ \\ = \sqrt{ {a}^{2} + {c}^{2} - 2ac + {(0)}^{2} } \\ \\ = \sqrt{ {a}^{2} + {c}^{2} - 2 ac}$

3. ( 3, -9) and ( -2, 3)

Here,

$x _{1} =3 \\ x _{2} = - 2\\ y _{1} = - 9\\ y_{2} = 3$

Putting the values we get,

$= \sqrt{ {(3 - ( - 2))}^{2} + {( - 9 - 3)}^{2} } \\ \\ = \sqrt{ {(5)}^{2} + {( - 12)}^{2} } \\ \\ = \sqrt{25 + 144} \\ \\ = \sqrt{169} \\ \\ = 13$