Q 11 Share with your friends Share 1 Rahul Raj answered this Let ABC be the ∆.Let the equation of AB is 3x + 4y = 10Let the equation of BC is 7x + y = -10Let the equation of CA is 4x - 3y = 5The given equations are :3x + 4y = 10 ....17x + y = -10 .....24x - 3y = 5 ....3Point of intersection of 1 and 2 :Multiply 2 by 4, we get28x + 4y = -40 ....4Subtracting 1 from 4, we get25x = -50 ⇒ x = -2Put x = -2 in 2, we gety = 4So, coordinates of B are -2,4Point of intersection of 2 and 3 :Multiply 2 by 3, we get21x + 3y = -30 ....5Adding 3 and 5, we get25x = -25⇒x = -1Put x = -1 in 2, we gety = -3So, coordinates of C are -1, -3Points of intersection of 1 and 3 :Solving 1 and 3 simultaneously, we getcoordinates of point A as 2, 1Put x = 2; y = 1 in equation of BC, we get7×2+1+10 = 25 > 0Put x = 0; y = 0 in equation of BC, we get7×0+0+10 = 10 > 0So, A and origin lies on same side of BC.Put x = -2; y = 4 in equation of CA, we get4×-2-3×4-5 = -8-12-5 = -25 < 0Put x = 0; y = 0 in equation of BC, we get4×0-3×0-5 =-5 <0So, B and origin lies on same side of CA.Put x = -1; y = -3 in the equation of AB, we get3×-1 + 4×-3 - 10 = -3-12-10 = -25 < 0Put x = 0; y = 0 in the equation of AB, we get3×0 + 4×0 - 10 = -10 < 0So, C and origin lies on same side of ABHence, origin lies inside the ∆ABC. 1 View Full Answer
