Prove that a point can be found which is at the same distance from each of the four points
`(am_(1),(a)/(m_(1))),(am_(2),(a)/(m_(2))),(am_(2),(a)/(m_(3)))and(am_(1)m_(2)m_(3),(a)/(m_(1)m_(2)m_(3)))`

Prove that a point can be found which is at the same distance from each of the four points :(a m_1, a/(m_1))*(a m_2, a/(m_2))*(a m_3, a/(m_3))"and"(a/(m_1m_2 m_3), a m_1m_2m_3)

The lines y=m_(1)x,y=m_(2)xandy=m_(3)x make equal intercepts on the line x+y=1 Then (a) 2(1+m_(1))(1+m_(3))=(1+m_(2))(2+m_(1)+m_(3))(1+m_(1))(1+m_(3))=(1+m_(2))(1+m_(1)+m_(3))(1+m_(1))(1+m_(2))=(1+m_(3))(2+m_(1)+m_(3))(1+m_(1))(1+m_(3))=(1+m_(2))(1+m_(1)+m_(3))2(1+m_(1))(1+m_(3))=(1+m_(2))(1+m_(1)+m_(3))

Factorise : a^(3)(1+m)^(3)-((al)/(3)+(2am)/(3))^(3)-((2al)/(3)+(am) )/(3))^(3)

Three sided of a triangle have equations L_(1)-=y-m_(i)x=o;i=1,2 and 3. Then L_(1)L_(2)+lambda L_(2)L_(3)+mu L_(3)L_(1)=0 where lambda!=0,mu!=0, is the equation of the circumcircle of the triangle if 1+lambda+mu=m_(1)m_(2)+lambda m_(2)m_(3)+lambda m_(3)m_(1)m_(1)(1+mu)+m_(2)(1+lambda)+m_(3)(mu+lambda)=0(1)/(m_(3))+(1)/(m_(1))+(1)/(m_(1))=1+lambda+mu none of these