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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, 20:39: PM
answered-by-expert 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 = begin mathsize 12px style 1 half open vertical bar table row cell x subscript 1 end cell cell y subscript 1 end cell 1 row cell x subscript 2 end cell cell y subscript 2 end cell 1 row cell x subscript 3 end cell cell y subscript 3 end cell 1 end table close vertical bar space end style,  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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