Name: Angle Bisectors of Lines
Uploaded: 2019-07-19
Duration: 8 min 21 s
Description: How do you find the equation of an angle bisector of two lines? Well, watch this video to get all the answers related to this! Have fun learning!

Question

A vertex of a square is at the point

(1, - 1)

and its one side lies along the line

3 x - 4 y - 17 = 0

, then find the area of the square.

Accepted Answer

Let $A B C D$ be a square.

Let the coordinates of the given vertex be $A (1, - 1)$ .

Now, it is given that one of its side lies along $3 x - 4 y - 17 = 0$ .

Now, $3 (1) - 4 (- 1) - 17 = 3 + 4 - 17 = - 10$ , clearly $(1, - 1)$ does not satisfy the $3 x - 4 y - 17 = 0$ , so $A B$ and $A D$ are not along the line $3 x - 4 y - 17 = 0$ .

We know that each angle of a square is right angle.

Side of square $=$ Length of perpendicular from $(1, - 1)$ to the line $3 x - 4 y - 17 = 0$ .

$\Rightarrow$ Side of the square $= \frac{| 3 (1) - 4 (- 1) - 17 |}{\sqrt{(3)^{2} + (4)^{2}}} [∵ d = \frac{|a x_{1} + b y_{1} + c|}{\sqrt{a^{2} + b^{2}}}]$

$\Rightarrow$ Side of square $= \frac{| 3 + 4 - 17 |}{\sqrt{9 + 16}}$

$\Rightarrow$ Side of square $= \frac{|10|}{5}$

$\Rightarrow$ Side of square $= \pm \frac{10}{5} [∵ | x | = a \Rightarrow x = \pm a]$

$\Rightarrow$ Side of square $= \pm 2$

$\Rightarrow$ Side of square $= 2$ (Length is always is positive)

Hence, the area of the square $= {(side)}^{2} = 2^{2} = 4$ sq.units.

A vertex of a square is at the point $(1, - 1)$ and its one side lies along the line $3 x - 4 y - 17 = 0$ , then find the area of the square.

Important Questions on Straight Lines

Learn and Prepare for any exam you want !

A vertex of a square is at the point 1,-1 and its one side lies along the line 3x-4y-17=0, then find the area of the square.

Important Questions on Straight Lines

Learn and Prepare for any exam you want !

A vertex of a square is at the point $(1, - 1)$ and its one side lies along the line $3 x - 4 y - 17 = 0$ , then find the area of the square.