Exams › JEE Advanced › Maths
Triangle ABC has sides BC, CA, and AB represented by the lines x - 2y + 5 = 0, x + y + 2 = 0, and 8x - y - 20 = 0 respectively. Find the area of triangle ABC.
- 41/2
- 43/2
- 45/2
- 47/2
Correct answer: 45/2
Solution
The three lines are L1: x-2y+5=0 (BC), L2: x+y+2=0 (CA), L3: 8x-y-20=0 (AB). Vertex A = L2 intersect L3: x+y+2=0 and 8x-y-20=0. Adding: 9x-18=0 => x=2, y=-4. A=(2,-4). Vertex B = L1 intersect L3: x-2y+5=0 and 8x-y-20=0. From first: x=2y-5. Sub: 8(2y-5)-y-20=0 => 16y-40-y-20=0 => 15y=60 => y=4, x=3. B=(3,4). Vertex C = L1 intersect L2: x-2y+5=0 and x+y+2=0. Subtracting: -3y+3=0 => y=1, x=-3. C=(-3,1). Area = (1/2)|x_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)| = (1/2)|2(4-1)+3(1-(-4))+(-3)(-4-4)| = (1/2)|2*3+3*5+(-3)*(-8)| = (1/2)|6+15+24| = (1/2)*45 = 45/2.
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