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Question: The area of a triangle formed by the lines x+y-3=0, x-3y+9=0 and 3x-2y+1=0 is?...

The area of a triangle formed by the lines x+y-3=0, x-3y+9=0 and 3x-2y+1=0 is?

Explanation

Solution

This type of question is based on the concept of straight lines. We can solve this question by solving the three equations to get the 3 vertices of the triangle. After we have determined the vertices, we can find the Area can then be found by using the direct formula:
Area= 12[x1(y2y3)+x2(y3y1)+x3(y1y2)]\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]

Complete step-by-step solution:
Solving the equations, we can write,
x+y-3=0-(1)
x-3y+9=0-(2)
3x-2y+1=0-(3)

Since these 3 lines form a triangle, the point where (1) and (2) intersect is called the vertex A of the triangle. The point where (2) and (3) intersect is called the vertex B and the point where (1) and (3) intersect is called the vertex C of the triangle.
Subtracting equation (2) from equation (1) we get, y=3
Substituting the value of y=3 in (1) we get x=0
\therefore Coordinates of Vertex A = (0,3)
Similarly multiplying equation (1) with 3 and subtracting (3) from (1) we get y=2
Substituting the value of y=2 in (3) we get x=1
\therefore Coordinates of Vertex C = (1,2)
Similarly multiplying equation (3) with 3 and subtracting (3) from (2) we get x= 157\dfrac{15}{7}
Substituting the value of x in (2) we get y= 267\dfrac{26}{7}
\therefore Coordinates of Vertex B = (157\dfrac{15}{7}, 267\dfrac{26}{7})
So, we have the coordinates A, B, C. Now we have to find the area using the direct formula. Putting the values in the direct formula
AreaArea= 12[x1(y2y3)+x2(y3y1)+x3(y1y2)]\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]
Substituting the values, we get

& \Rightarrow Area= \dfrac{1}{2}\left[ 0\left( 2-\dfrac{26}{7} \right)+1\left( \dfrac{26}{7}-3 \right)+\dfrac{15}{7}\left( 3-2 \right) \right] \\\ & \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow Area=\,\,\dfrac{1}{2}\left[ 0+\dfrac{5}{7}+\dfrac{15}{7} \right] \\\ & \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow Area=\,\,\dfrac{10}{7}\,\,sq.\,units \\\ & \\\ \end{aligned}$$ **Hence, the Area of the triangle is $$\dfrac{10}{7}\,sq.\,units$$.** **Note:** While solving this question, we need to take care while subtracting the equations from one another as that could lead to a mistake and that would give an entirely different answer. We could also use another approach of solving by bringing the equation into the form of y=mx+c for all three equations and then using the coordinates so obtained to find the area using the formula given above.