To solve the problem of finding the area of the triangle formed by the lines through the given equations, we need to decode the intersections and geometry involved.
The first given equation:
2x2−3xy−2y2=0can be factored as:
(2x+y)(x−2y)=0This represents two intersecting lines
L1:2x+y=0 and
L2:x−2y=0.
The second equation:
2x2−3xy−2y2−x+7y−3=0can be rewritten using trial and error for factorization as:
(2x+y−3)(x−2y+1)=0yielding two more lines
L3:x−2y+1=0 and
L4:2x+y−3=0.
The line pairs are rewritten as:
L1:2x+y=0 (parallel to
L4:2x+y−3=0 )
L2:x−2y=0 (parallel to
L3:x−2y+1=0 )
Find intersection
A between lines
L1 and
L3, and
B between lines
L2 and
L4.
Intersection calculation:
For point
A, solve:
‌2x+y=0‌x−2y+1=0The solution is:
A=(−‌,‌)For point
B, solve:
‌x−2y=0‌2x+y−3=0The solution is:
B=(‌,‌)Area calculation:
Using the coordinates
A=(−‌,‌),B=(‌,‌), and the intersection point
P(1,1) of lines
L3 and
L4 :
The area of triangle
∆ABP is computed using the determinant method:
Area
=‌||Performing column operations:
Subtract the first column from the second and third:
C2⟶C2−C1,‌‌C3⟶C3−C1 Thus, we have:
‌|| −‌ | ‌ | ‌ |
| ‌ | −‌ | −‌ |
| 1 | 0 | 0 |
|Then, compute:
=‌×‌=‌×‌=‌Thus, the area of the triangle is
‌.