Question
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?
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= 21[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
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
∴ 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
∴ Coordinates of Vertex C = (1,2)
Similarly multiplying equation (3) with 3 and subtracting (3) from (2) we get x= 715
Substituting the value of x in (2) we get y= 726
∴ Coordinates of Vertex B = (715, 726)
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
Area= 21[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
Substituting the values, we get