Answer:
The distance between AC is 15 units.
Step-by-step explanation:
Here's the required formula to find distance between points A(2, 10) and C(14, 1) :
[tex]{\implies{\small{\pmb{\sf{Distance = \sqrt{\Big(x_{2} - x_{1} \Big)^{2} + \Big(y_{2} - y_{1} \Big)^{2}}}}}}}[/tex]
As per given question we have provided that :
[tex]\begin{gathered}\begin{gathered} \footnotesize\rm {\underline{\underline{Where}}}\begin{cases}& \sf x_2 = 14\\ & \sf x_1 = 2\\ & \sf y_2 = 1\\& \sf y_1 = 10\end{cases} \end{gathered}\end{gathered}[/tex]
Substituting all the given values in the formula to find the distance between points A(2, 10) and C(14, 1) :
[tex]\begin{gathered} \quad{\implies{\small{\sf{AC = \sqrt{\Big(x_{2} - x_{1} \Big)^{2} + \Big(y_{2} - y_{1} \Big)^{2}}}}}}\\\\\quad{\implies{\small{\sf{AC = \sqrt{\Big(14 - 2\Big)^{2} + \Big(1 - 10\Big)^{2}}}}}}\\\\\quad{\implies{\small{\sf{AC = \sqrt{\Big(\: 12 \:\Big)^{2} + \Big( - 9\Big)^{2}}}}}}\\\\ \quad{\implies{\small{\sf{AC = \sqrt{\Big(12 \times 12\Big)+ \Big( - 9 \times - 9\Big)}}}}}\\\\\quad{\implies{\small{\sf{AC = \sqrt{\Big( \: 144 \: \Big)+ \Big( \: 81 \: \Big)}}}}}\\\\ \quad{\implies{\small{\sf{AC = \sqrt{\Big(144 + 81\Big)}}}}}\\\\\quad{\implies{\small{\sf{AC = \sqrt{\Big(225\Big)}}}}}\\\\\quad{\implies{\small{\sf{\underline{\underline{\red{AC = 15}}}}}}} \end{gathered}[/tex]
Hence, the distance between AC is 15 units.
[tex]\rule{300}{2.5}[/tex]
