Question

Find the area of the triangle formed by sides x + 4y − 9 = 0, 9x + 10y + 23 = 0, 7x + 2y −11 = 0

Answer 1

Given:

The sides of a triangle are given as:-

AB ⇔ x + 4y - 9 = 0

BC ⇔ 9x +10y +23 = 0

AC ⇔ 7x + 2y - 11 =0

To Find:

We have to find the area of the triangle using the given sides.

Solution:

First step

We have to find the coordinates of the triangle using the given line equations.

Coordinate of A can be obtained by solving the equations of line AB and AC using the elimination method.

∵  AC ⇔ 7x + 2y - 11 =0 .....Eq.1  [ Multiply line AC by 2 on both sides ]    

⇒ AC ⇔ 14x + 4y - 22 =0

Subtract the two equations, we get

 AC ⇔ 14x + 4y - 22 = 0

 AB ⇔    x + 4y -  9 = 0

             13x + 0y-  13 =0

⇒ 13x = 13

⇒ x = 1

Substitute the value of x in Eq.1, then we get

⇒  7 + 2y = 11

⇒ y = 2

∴ Coordinate of A is ( 1,2)

Similarly using the elimination method, we get the coordinates of B and C.

∴ Coordinate of B is ( -7,4)

∴ Coordinate of C is ( 3,-5)

Second Step

We can find the area of the triangle using the equation,

 $\[A = \frac{1}{2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|\]$

Let ( x₁ ,y₁ ) = (1,2)

     ( x₂ ,y₂) = (-7,4)

     ( x₃ , y₃) = (3,-5)

By substituting the values obtained in the above equation, we get

  $\[\begin{array}{l}A = \frac{1}{2}\left| {1\left( {4 - - 5} \right) + - 7\left( { - 5 - 2} \right) + 3\left( {2 - 4} \right)} \right|\\\\\,\,\,\,\, = \frac{1}{2}\left| {52} \right|\\\\\,\,\,\,\, = 26\,sq.units\end{array}\]$

∴ The area of the triangle is equal to 26 sq.units