Name: Family of Straight Lines: Example
Uploaded: 2019-07-19
Duration: 3 min 23 s
Description: Do you know how to find the equation of a straight line passing through the point of intersection of a family of two lines and also through a point? Let's find out.

Question

Prove that a median of an equilateral triangle is perpendicular to the corresponding side.

Accepted Answer

Let $△ A B C$ be the equilateral triangle with $B C = 2 k$ and middle point of $B C$ as the origin.

Let the coordinates of $B$ and $C$ are $(- k, 0)$ and $(k, 0)$ respectively.

Now, $\tan 60 ° = \frac{O A}{O C} = \frac{O A}{k}$

$\Rightarrow \sqrt{3} = \frac{O A}{k}$

$\Rightarrow O A = \sqrt{3} k$

$∴$ Coordinates of $A$ is $(0, \sqrt{3} k)$

Now, coordinates of midpoint of $A C$ is $D \equiv (\frac{k}{2}, \frac{\sqrt{3} k}{2})$

And coordinates of midpoint of $A B$ is $E \equiv (- \frac{k}{2}, \frac{\sqrt{3} k}{2})$

And coordinates of midpoint of $B C$ is $O \equiv (0, 0)$

Now, slope of median $B D$ is $m_{B D} = \frac{\frac{\sqrt{3} k}{2} - 0}{\frac{k}{2} - (- k)} = \frac{1}{\sqrt{3}}$

Slope of line $A C$ is $m_{A C} = \frac{0 - \sqrt{3} k}{k - 0} = - \sqrt{3}$

Since, $m_{A C} \times m_{B D} = - 1$

$∴$ $A C ⊥ B D$

Now, slope of median $C E$ is $m_{C E} = \frac{\frac{\sqrt{3} k}{2} - 0}{- \frac{k}{2} - k} = - \frac{1}{\sqrt{3}}$

And slope of line $A B$ is $m_{A B} = \frac{0 - \sqrt{3} k}{- k - 0} = \sqrt{3}$

Since, $m_{A B} \times m_{C E} = - 1$

$∴ A B ⊥ C E$

Now, slope of median $A O$ is $m_{A O} = \frac{0 - \sqrt{3} k}{0 - 0} = \infty$ i.e $y -$ axis.

And slope of line $B C$ is $m_{B C} = 0$ i.e $x -$ axis.

$∴ A O ⊥ B C$

Prove that a median of an equilateral triangle is perpendicular to the corresponding side.

Important Questions on Straight Lines

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