Question

Let a triangle be bounded by the lines L₁ : 2x + 5y = 10; L₂ : –4x + 3y = 12 and the line L₃, which passes through the point P(2, 3), intersects L₂ at A and L₁ at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to

• A. \(\frac{110}{13}\)
• B. \(\frac{132}{13}\)
• C. \(\frac{142}{13}\)
• D. \(\frac{151}{13}\)

Answer 1

The Correct Option is B

The correct answer is (B) :\(\frac{132}{13}\)
L₁ : 2x + 5y = 10
L₂ : – 4x + 3y = 12
Solving L₁ and L₂ we get
\(C≡(\frac{−15}{13},\frac{32}{13})\)
Now, Let
\(A(x_1,\frac{1}{3}(12+4x_1))\)
and
\(B(x_2,\frac{1}{5}(10−2x_2))\)
\(∴\frac{3x_1+x_2}{4}=2\)
and
\(\frac{(12+4x_1)+\frac{10−2x_2}{5}}{4}=3\)
So, 3x₁ + x₂ = 8 and 10 x₁ – x₂ = – 5
\(So, (x1,x2)=(\frac{3}{13},\frac{95}{13})\)
\(A=(\frac{3}{13},\frac{56}{13})\)
and
\(B=(\frac{95}{13},\frac{−12}{13})\)

\(=|\frac{1}{2}(\frac{3}{13}(\frac{−44}{13})\frac{−56}{13}(\frac{110}{13})+1(\frac{2860}{169}))|\)

\(=\frac{132}{13} sq. units\)