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:-

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

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

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

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

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

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

$\dashrightarrow \sf d = \sqrt{9 +36 }$

$\dashrightarrow \sf d = \sqrt{45}$

$\dashrightarrow \sf d = \sqrt{3 \times 3 \times 5}$

$\dashrightarrow \sf d =3 \sqrt{5}$

$\dashrightarrow \sf d =3 \times 2.23$

$\dashrightarrow \bf d =6.70$

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