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

Answer: (B)

$\frac{132}{13}$

Answer 2

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
$B=(\frac{95}{13},\frac{−12}{13})$

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

$=\frac{132}{13} sq. units$