Question

Find the distance between (-2, 3) & (-7, -7). Round to the nearest tenth.

Answer 1

`{ \sf{find \: the \: distance \: between \: ( - 2,3)and \: ( - 7, - 7).}} \\ { \sf{round \: to \: the \: nearest \: tenth}}`

`{ \sf{ \red{distance = \sqrt{( {x2 - x1})^(2) + ( {y2 - y1})^(2) } }}} \\ \\ { \sf{x1 = - 2}} \\ { \sf{x2 = - 7}} \\ { \sf{y1 = 3}} \\ { \sf{ y2 = - 7}}`

`{ \sf{ \green{ distance = \sqrt{ {( - 7 - ( - 2)})^(2) + {( - 7 - 3)}^(2)}}} } \\ \\ { : {\implies{ \green{ \sf{distance = \sqrt{ {( - 5)}^(2) + {( - 10)}^(2) } }}}}} \\ \\ { : { \implies{ \green{ \sf{distance = √(25 + 100)}}}}}`

`{ : { \implies{ \sf{ \green{distance = √(125)}}}}} \\ \\ { : { \implies{ \green{ \sf{distance = 5 √(5)}}}}} \\ \\ { : { \implies{ \green{ \sf{distance = 5 * 2.236}}}}}`

`{ : { \implies{ \underline{ \green { \sf{distance = 11.18= 11.2}}}}}}`

Answer 2

11.2 units

Explanation:

From (-7, -7) to (-2, 3) is 5 units horizontally and 10 units vertically. Thus we have a right triangle with sides 5 and 10 respectively. The length of the hypotenuse of this triangle is the distance between (-2, 3) & (-7, -7):

d = √(5² + 10²) = √(25 + 100) = √125 = √25√5, or 5√5.

This is approximately 11.2 units