Question

(1) Using vectors prove that the perpendicular bisectors of the sides
of a triangle are concurrent​

Answer 1

Answer:

hope these will help you

Answer 2

Answer:

We have proved that, perpendicular bisectors of the sides of a triangle are concurrent

Step-by-step explanation:

Step 1: Let $\mathrm{ABC}$ be the given triangle and $\mathrm{O}$ be the point of intersection of perpendicular bisectors $\mathrm{OD}$ and $\mathrm{OE}$ of sides $\mathrm{BC}$ and CA respectively. Let $\mathrm{F}$ be mid-point of $\mathrm{AB}$. Join $\mathrm{O}$ to $\mathrm{F}$.

Take $\mathrm{O}$ as origin.

Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of A, B, C respectively.

$\therefore \mathrm{OA}=\overrightarrow{\mathrm{a}}, \mathrm{OB}=\overrightarrow{\mathrm{b}}, \mathrm{OC}=\overrightarrow{\mathrm{c}}$

Step 2:The position vectors of D, E, F are $\frac{\vec{b}+\vec{c}}{2}, \frac{\vec{c}+\vec{a}}{2}, \frac{\vec{a}+\vec{b}}{2}$ respectively.

Since, $\mathrm{OD} \perp \mathrm{BC}$

$\therefore \overrightarrow{\mathrm{OD}} \cdot \overrightarrow{\mathrm{BC}}=0$

$\Rightarrow\left(\frac{\vec{b}+\vec{c}}{2}\right) \cdot(\vec{c}-\vec{b})=0$

$\Rightarrow \frac{1}{2}(\vec{c}+\vec{b}) \cdot(\vec{c}-\vec{b})=0$

$\Rightarrow \frac{1}{2}\left[(\vec{c})^2-(\vec{b})^2\right]=0$..........(i)

Again, $\mathrm{OE} \perp \mathrm{CA}$

$\therefore \overrightarrow{\mathrm{OE}} \cdot \overrightarrow{\mathrm{CA}}=0$

$\begin{aligned}& \Rightarrow\left(\frac{\vec{c}+\vec{a}}{2}\right) \cdot(\vec{a}-\vec{c})=0 \\& \Rightarrow \frac{1}{2}(\vec{a}+\vec{c}) \cdot(\vec{a}-\vec{c})=0\end{aligned}$

$\Rightarrow \frac{1}{2}\left[(\vec{a})^2-(\vec{c})^2\right]=0$............(ii)

Step 3:Adding (1) and (2), we get

$\Rightarrow \frac{1}{2}\left[(\vec{a})^2-(\vec{b})^2\right]=0$

$\begin{aligned}& \Rightarrow \frac{1}{2}\left[(\vec{a})^2-(\vec{b})^2\right]=0 \\& \Rightarrow \frac{1}{2}(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0 \\& \Rightarrow\left(\frac{\vec{a}+\vec{b}}{2}\right) \cdot(\vec{b}-\vec{a})=0 \\& \Rightarrow \overrightarrow{O F} \cdot \overrightarrow{A B}=0 \\& \Rightarrow \mathrm{OF} \perp \mathrm{AB}\end{aligned}$

Hence we have proved that, perpendicular bisectors of the sides of a triangle are concurrent.

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