Question

Find the distance, c, between (–3, –1) and (3, 2) on the coordinate plane. Round to the nearest tenth if necessary ??

Answer 1

Point distance formula:

d=√((x_2-x_1)²+(y_2-y_1)²)

Substitute

d=√((3-(-3))²+(2-(-1)²)

Simplify

d=√((6)²+(3)²)

d=√(36+9)

d=√(45)

d=3√(5)

-Hunter

Answer 2

Given :-

• (x 1 , y1) = (-3 , -1)

• (x2 , y2) = (3 , 2)

`\\ \\`

To find:

• Distance between two points

We know :-

`\bigstar \boxed{ \rm d = \sqrt{(y_2 - y_1 {)}^(2) + (x_2 - x_1 {)}^(2) } }`

So:-

`\dashrightarrow \sf d = \sqrt{(y_2 - y_1 {)}^(2) + (x_2 - x_1 {)}^(2) } \\`

`\\ \\`

`\dashrightarrow \sf d = \sqrt{(2 - ( - 1) {)}^(2) + (3 - ( - 3) {)}^(2) } \\`

`\\ \\`

`\dashrightarrow \sf d = \sqrt{(2 + 1 {)}^(2) + (3 + 3{)}^(2) } \\`

`\\ \\`

`\dashrightarrow \sf d = \sqrt{(3 {)}^(2) + (3 + 3{)}^(2) } \\`

`\\ \\`

`\dashrightarrow \sf d = \sqrt{(3 {)}^(2) + (6{)}^(2) } \\`

`\\ \\`

`\dashrightarrow \sf d = \sqrt{9 + (6{)}^(2) } \\`

`\\ \\`

`\dashrightarrow \sf d = √(9 +36 ) \\`

`\\ \\`

`\dashrightarrow \sf d = √(45) \\`

`\\ \\`

`\dashrightarrow \sf d = √(3 * 3 * 5) \\`

`\\ \\`

`\dashrightarrow \sf d =3 √(5)`

`\\ \\`

`\dashrightarrow \sf d =3 * 2.23`

`\\ \\`

`\dashrightarrow \bf d =6.70`

`\\ \\`

`\therefore \underline{ \textsf{ \textbf{distance \: between \: two \: points \: is \: equal \: to \red{6.70 \{approx\}}}}}`