Question

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.

Answer 1

We need distance first

$\\ \sf\longmapsto AB=\sqrt{(12-2)^2+(5+10)^2}$

$\\ \sf\longmapsto AB=\sqrt{14^2+15^2}$

$\\ \sf\longmapsto AB=\sqrt{196+225}$

$\\ \sf\longmapsto AB=\sqrt{321}$

$\\ \sf\longmapsto AB\approx 18$

Now

2/5 of AB

$\\ \sf\longmapsto \dfrac{2}{5}(18)$

$\\ \sf\longmapsto \dfrac{36}{5}$

$\\ \sf\longmapsto 7.2$

Answer 2

Answer: $\large \boldsymbol {\sf 2\sqrt{13} \approx7,2}$

Step-by-step explanation:

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

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

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

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

• $\sf \large \boldsymbol {} \dfrac{2}{5} \cdot D=5\sqrt{13} \cdot \dfrac{2}{5} =\boxed{\sf 2\sqrt{13}}$

• 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=$\boldsymbol {\sf \sqrt{15^2+10^2}=\sqrt{325}=5\sqrt{13} }$

• Then 2/5* AB=2$\sqrt{13}$