(i) Find the area of the triangle bounded by the lines 4x−5y+7=0, x−2=0 and y+1=0.
(ii) Find the area of the triangle bounded by the lines x−2y−6=0, 3x−y+3=0 and 2x+y−4=0.
Background and Explanation
To find the area of a triangle formed by three intersecting lines, first determine the vertices by solving the equations pairwise to find their points of intersection. Once the three vertices (x1,y1), (x2,y2), and (x3,y3) are known, apply the determinant formula for the area of a triangle.
The given lines are:
4x−5y+7x−2y+1=0(i)=0(ii)=0(iii)
Finding Vertex A (intersection of lines (i) and (ii)):
From equation (ii), we immediately get:
x=2
Substituting x=2 into equation (i):
4(2)−5y+78−5y+715−5y5yy=0=0=0=15=3
Thus, vertex A is (2,3).
Finding Vertex B (intersection of lines (i) and (iii)):
From equation (iii), we get:
y=−1
Substituting y=−1 into equation (i):
4x−5(−1)+74x+5+74x+124xx=0=0=0=−12=−3
Thus, vertex B is (−3,−1).
Finding Vertex C (intersection of lines (ii) and (iii)):
From equation (ii): x=2
From equation (iii): y=−1
Thus, vertex C is (2,−1).
Calculating the Area:
Using the determinant formula for the area of triangle ABC:
Area=212−323−1−1111
Expanding along the first row:
=21[2−1−111−3−3211+1−32−1−1]=21[2(−1⋅1−1⋅(−1))−3(−3⋅1−1⋅2)+1((−3)⋅(−1)−(−1)⋅2)]=21[2(−1+1)−3(−3−2)+1(3+2)]=21[2(0)−3(−5)+1(5)]=21[0+15+5]=21[20]=10 sq. units
The given equations of lines are:
x−2y−63x−y+32x+y−4=0(i)=0(ii)=0(iii)
Finding Vertex A (intersection of lines (i) and (ii)):
To eliminate x, multiply equation (i) by 3 and subtract equation (ii) from it:
3x−6y−18=0−(3x−y+3=0)−5y−21=0
Solving for y:
−5yy=21=−521
Substituting y=−521 into equation (i):
x−2(−521)−6x+542−6x+542−530x+512x=0=0=0=0=−512
Thus, vertex A is (−512,−521).
Finding Vertex B (intersection of lines (i) and (iii)):
Multiply equation (i) by 2 and subtract equation (iii) from it:
2x−4y−12=0−(2x+y−4=0)−5y−8=0
Solving for y:
−5yy=8=−58
Substituting y=−58 into equation (i):
x−2(−58)−6x+516−6x+516−530x−514x=0=0=0=0=514
Thus, vertex B is (514,−58).
Finding Vertex C (intersection of lines (ii) and (iii)):
Add equations (ii) and (iii) to eliminate y:
3x−y+3=0+(2x+y−4=0)5x−1=0
Solving for x:
5xx=1=51
Substituting x=51 into equation (iii):
2(51)+y−452+y−4yyy=0=0=4−52=520−52=518
Thus, vertex C is (51,518).
Calculating the Area:
Using the determinant formula with the three vertices:
Area=21−51251451−521−58518111
Expanding along the first row:
=21[−512−5851811+5215145111+151451−58518]=21[−512(−58−518)+521(514−51)+1(514⋅518−(−58)⋅51)]=21[−512(−526)+521(513)+(25252+258)]=21[25312+25273+25260]=21[25845]=21[5169]=10169=16.9 square units
Key Formulas or Methods Used
Intersection of two lines: Solve the system of two linear equations simultaneously using substitution or elimination methods to find the point (x,y) where they meet.
Area of triangle using determinant: For vertices (x1,y1), (x2,y2), and (x3,y3):
Area=21x1x2x3y1y2y3111=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣
Summary of Steps
Identify the three lines forming the triangle and label them as equations (i), (ii), and (iii).
Find the first vertex by solving the system of equations (i) and (ii) simultaneously.
Find the second vertex by solving the system of equations (i) and (iii) simultaneously.
Find the third vertex by solving the system of equations (ii) and (iii) simultaneously.
Apply the determinant formula for the area of a triangle using the three coordinate pairs found.
Expand the determinant along any row or column (typically the first row) and evaluate the 2×2 minors.
Calculate the final value, ensuring to take the absolute value and multiply by 21 to get the area in square units.