JEE Class main Answered
Find area oF triangle formed by joining origin to the point of intersection of the line x√5+2y=3√5 and the circle x2+y2 =10
Asked by harshitj291 | 03 Apr, 2019, 08:39: PM
Expert Answer
Let us get points of intersection of line √5x +2y = 3√5 and x2 + y2 = 10
√5x +2y = 3√5 or y = (3/2)√5 - (√5/2) x ......................(1)
If we substitute y from eqn.(1) in circle eqn., we get , x2 + [ (3/2)√5 - (√5/2) x ]2 = 10 .................(2)
after simplification, we write eqn.(2) as 9x2 - 30x +5 = 0 .................(3)
Quadratic eqn.(3) has roots , x = (5±2√5)/3 .......................(4)
corresponding y-coord for x values given in eqn.(4) is obtained from line eqn √5x +2y = 3√5, y = (2√5±5)/3 ...................(5)
hence vertices of triangle are A(0,0), B( (5+2√5)/3, (2√5-5)/3 ) and C ( (5-2√5)/3, (2√5+5)/3 )
Area of triangle = 
, we consider (x1 , y1 ) as ( 0, 0 )
, we consider (x1 , y1 ) as ( 0, 0 ) Then area of triangle = (1/2) {[(5+2√5)/3]2 + [(5-2√5)/3]2} = 5
Answered by Sneha shidid | 04 Apr, 2019, 10:04: AM
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