Question

А = (1,1) B=(-2,7) C=(3,-3)
find 1/AB+1/BC+1/CA=?​

Answer 1

The value of $\dfrac{1}{AB}+\dfrac{1}{BC}+\dfrac{1}{CA}$ is $\frac{31\sqrt{5}}{150}$.

Step-by-step explanation:

The given points are А = (1,1) B=(-2,7) C=(3,-3).

Distance formula:

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula we get

$AB=\sqrt{\left(-2-1\right)^2+\left(7-1\right)^2}=3\sqrt{5}$

$BC=\sqrt{\left(3-\left(-2\right)\right)^2+\left(-3-7\right)^2}=5\sqrt{5}$

$CA=\sqrt{\left(1-3\right)^2+\left(1-\left(-3\right)\right)^2}=2\sqrt{5}$

We need to find the value of $\dfrac{1}{AB}+\dfrac{1}{BC}+\dfrac{1}{CA}$.

$\dfrac{1}{3\sqrt{5}}+\dfrac{1}{5\sqrt{5}}+\dfrac{1}{2\sqrt{5}}$

$\frac{10}{30\sqrt{5}}+\frac{6}{30\sqrt{5}}+\frac{15}{30\sqrt{5}}$

$\frac{10+6+15}{30\sqrt{5}}$

$\frac{31}{30\sqrt{5}}$

After rationalization we get

$\frac{31\sqrt{5}}{150}$

Therefore, the value of $\dfrac{1}{AB}+\dfrac{1}{BC}+\dfrac{1}{CA}$ is $\frac{31\sqrt{5}}{150}$.

#Learn more

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