Question

find the distance between the point (-7\5,3)and (3/5,3)​

Answer 1

Answer:

2 units

Step-by-step explanation:

Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by:

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

Let $(x_1,y_1)$ be $(\frac{3}{5},3)$ and $(x_2,y_2)$ be $(-\frac{7}{5},3)$

Applying the formula:

$\sqrt{(\frac{3}{5}-(-\frac{7}{5})^2+(3-3)^2$

= $\sqrt{(\frac{3}{5}+\frac{7}{5})^2+0^2$

= $\sqrt{\frac{10}{5}^2$

= $\sqrt{2^2}$

= $2 \; \text{units}$                      Answer

Answer 2

Answer:

d ( P, Q ) = √( -7/5 - 3/5 )² + ( 3 - 3 )²

= √( -7 - 3 /5)²

= √( -10/5 )²

= √( -2 )²

= -2

Thus, the distance between the point

( -7\5, 3 )and ( 3/5, 3 ) is -2 units.