The distance between the points (-9, -5) and (4, -3) is 13.153 units approximately
Solution:
Given, two points are (-9, -5) and (4, -3).
We need to find the distance between above two points.
We know that, distance between two points [tex]P\left(x_{1}, y_{1}\right) \text { and } Q\left(x_{2}, y_{2}\right)[/tex] is given by:
[tex]d(P, Q)=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}[/tex]
[tex]\text { here, } x_{1}=-9, y_{1}=-5, x_{2}=4 \text { and } y_{2}=-3[/tex]
[tex]\begin{array}{l}{\text { distance }=\sqrt{(4-(-9))^{2}+(-3-(-5))^{2}}=\sqrt{(4+9)^{2}+(5-3)^{2}}} \\\\ {=\sqrt{(13)^{2}+(2)^{2}}=\sqrt{169+4}=\sqrt{173}=13.152946}\end{array}[/tex]
Hence the distance between points is 13.153 units approximately
