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The endpoint coordinates of AB are A(2, -10) & B(12, 5). Find the coordinates of the point that is 2/5 from A to B.

Sagot :

kamilmatematik100504

Answer: [tex]\large \boldsymbol {\sf 2\sqrt{13} \approx7,2}[/tex]

Step-by-step explanation:

  • Find the distance between points A and B by the formula:

  • [tex]\large \boldsymbol {\sf D=\sqrt{(x_1-x_2)^2+(y_1+y_2)^2} }[/tex]

  • [tex]\large \boldsymbol{} \sf D=\sqrt{(2-12)^2+(-10-5)^2} =\sqrt{100+225} =\boldsymbol {\sqrt{325}=5\sqrt{13} }[/tex]  

  • By the condition we are told to find 2/5 the distance between points A and B

  • [tex]\sf \large \boldsymbol {} \dfrac{2}{5} \cdot D=5\sqrt{13} \cdot \dfrac{2}{5} =\boxed{\sf 2\sqrt{13}}[/tex]

  • We can also find the distance between them through the  Pythagorean theorem we will complete a right triangle
  • AB²=AC²+BC²
  • AB=√AC²+BC²
  • Where AC=10  ; BC=15  
  • AB=[tex]\boldsymbol {\sf \sqrt{15^2+10^2}=\sqrt{325}=5\sqrt{13} }[/tex]
  • Then 2/5* AB=2[tex]\sqrt{13}[/tex]

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