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What is the distance between (4, 3) and (9, 15) on the coordinate plane? Select two that apply. 13 units V 169 units V144 units 12 units 5 units

Sagot :

[tex]\begin{gathered} \sqrt[]{169} \\ \text{and} \\ 13 \end{gathered}[/tex]

Explanation

the distance between 2 points P1 and P2 is given by:

[tex]\begin{gathered} \text{distance}=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \text{where} \\ P1(x_1,y_1) \\ P2(x_2,y_2) \end{gathered}[/tex]

Step 1

Let

P1=(4,3)

P2=(9,15)

replace

[tex]\begin{gathered} \text{distance}=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \text{distance}=\sqrt[]{(9-4)^2+(15-3)^2} \\ \text{distance}=\sqrt[]{(5)^2+(12)^2} \\ \text{distance}=\sqrt[]{25+144^{}} \\ \text{distance}=\sqrt[]{169} \\ \text{also} \\ \text{distance}=13 \end{gathered}[/tex]

I hope this help you