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# Find the area of the triangle bounded by lines $4x + 3y + 8 = 0$, $x - y + 2 = 0$ and $9x - 2y - 17 = 0$.

Last updated date: 26th Feb 2024
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Hint: Here, we are required to find the area of the triangle bounded by the given equations. We will solve these equations to find all the three vertices of the triangle. We will then use the formula of area of a triangle and substitute the values of vertices in that formula to find the required area. By this, we will get the required answer.

Formula Used:
Area of a triangle $= \left| {\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]} \right|$

The given equations of the sides of a triangle are:
$4x + 3y + 8 = 0$……………………$\left( 1 \right)$
$x - y + 2 = 0$……………………...$\left( 2 \right)$
And $9x - 2y - 17 = 0$………….$\left( 3 \right)$
Now, we will solve the first two equations using elimination.
Hence, multiplying the equation $\left( 2 \right)$ by 4 and then subtracting it from equation $\left( 1 \right)$, we get
$4x + 3y + 8 - 4\left( {x - y + 2} \right) = 0$
$\Rightarrow 4x + 3y + 8 - 4x + 4y - 8 = 0$
Solving further, we get
$\Rightarrow 7y = 0$
Dividing both sides by 7,
$\Rightarrow y = 0$
Substituting this in equation $\left( 2 \right)$, we get
$\Rightarrow x - 0 + 2 = 0$
$\Rightarrow x + 2 = 0$
Subtracting 2 from both the sides, we get
$\Rightarrow x = - 2$
Hence, the common vertex of the equations $\left( 1 \right)$ and $\left( 2 \right)$ is $\left( { - 2,0} \right)$.
Similarly, we will solve the equations $\left( 2 \right)$ and $\left( 3 \right)$ using elimination.
Hence, multiplying the equation $\left( 2 \right)$ by 9 and then subtracting the equation $\left( 3 \right)$ from it, we get
$9\left( {x - y + 2} \right) - \left( {9x - 2y - 17} \right) = 0$
$\Rightarrow 9x - 9y + 18 - 9x + 2y + 17 = 0$
Solving the above equation further, we get
$\Rightarrow - 7y + 35 = 0$
$\Rightarrow 7y = 35$
Dividing both sides by 7, we get
$\Rightarrow y = 5$
Substituting this in equation $\left( 2 \right)$, we get
$\Rightarrow x - 5 + 2 = 0$
$\Rightarrow x - 3 = 0$
Adding 3 on both the sides, we get
$\Rightarrow x = 3$
Hence, the common vertex of the equations $\left( 2 \right)$ and $\left( 3 \right)$ is $\left( {3,5} \right)$.
Also, we will solve the equations $\left( 1 \right)$ and $\left( 3 \right)$ using elimination.
Hence, multiplying the equation $\left( 1 \right)$ by 2 and then equation $\left( 2 \right)$ by 3 and adding both of them, we get
$2\left( {4x + 3y + 8} \right) + 3\left( {9x - 2y - 17} \right) = 0$
$\Rightarrow 8x + 6y + 16 + 27x - 6y - 51 = 0$
Solving the above equation further, we get
$\Rightarrow 35x - 35 = 0$
$\Rightarrow 35\left( {x - 1} \right) = 0$
Dividing both sides by 35, we get
$\Rightarrow x - 1 = 0$
Adding 1 on both sides, we get
$\Rightarrow x = 1$
Substituting this in equation $\left( 1 \right)$, we get
$4\left( 1 \right) + 3y + 8 = 0$
$\Rightarrow 3y + 12 = 0$
Dividing both sides by 3, we get
$\Rightarrow y + 4 = 0$
Subtracting 4 from both the sides, we get
$\Rightarrow y = - 4$
Hence, the common vertex of the equations $\left( 1 \right)$ and $\left( 3 \right)$ is $\left( {1, - 4} \right)$.
Now, we are getting a triangle with the vertices $A\left( { - 2,0} \right)$, $B\left( {3,5} \right)$ and $C\left( {1, - 4} \right)$.
Substituting $\left( {{x_1},{y_1}} \right) = \left( { - 2,0} \right)$, $\left( {{x_2},{y_2}} \right) = \left( {3,5} \right)$ and $\left( {{x_3},{y_3}} \right) = \left( {1, - 4} \right)$ in the formula Area of triangle $ABC$ $= \left| {\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]} \right|$, we get
Area of triangle $ABC = \left| {\dfrac{1}{2}\left[ { - 2\left( {5 - \left( { - 4} \right)} \right) + 3\left( { - 4 - 0} \right) + 1\left( {0 - 5} \right)} \right]} \right|$
$\Rightarrow$ Area of triangle $ABC = \left| {\dfrac{1}{2}\left[ { - 2\left( 9 \right) + 3\left( { - 4} \right) + 1\left( { - 5} \right)} \right]} \right|$
Solving the above equation further, we get
$\Rightarrow$ Area of triangle $ABC = \left| {\dfrac{1}{2}\left[ { - 18 - 12 - 5} \right]} \right| = \left| {\dfrac{{ - 35}}{2}} \right| = \left| { - 17.5} \right|$
Hence, area of triangle $ABC = 17.5$ square units.

Therefore, the area of the triangle bounded by lines $4x + 3y + 8 = 0$, $x - y + 2 = 0$ and $9x - 2y - 17 = 0$ is $17.5$ square units.

Note:
Here, we have used the ‘modulus sign’ while finding the area of the triangle because it means that we have to take the absolute value of the terms present inside it, i.e. we will only take the non-negative values of the terms present inside the modulus when we will remove it. Therefore, we should keep in mind that area of a triangle can never be negative.