In an isosceles triangle ABC,the coordinates of the points B and C on the base BC are respectively (1,2) and (2,1).If the equation of line AB is y=2x,then equation of line AC is?

Dear Piyush
Use
|AB| = |AC|
And you will simply get your answer.
Regards
Arun (askIITians forum expert)

Hi, As we know that the perpendicular bisected of base of an isosceles triangle will always pass through the third vertex.Let the midpoint of BC be D=(3/2,3/2)As AD is perpendicular to BC so m1xm2=-1(m1=slope of the line BC, m2=slope of the line AD), => -1 x m2 = -1 => m2 = 1Now the equation of line AD will be=> (y-y1)=m(x-x1)=> (y-3/2)=(1)(x-3/2)=> y=xOn solving the equation of AB and AD we will get the vertices of the point A which is (0,0). Hence the equation of the line AC=(y-0)=(1-0/2-0)(x-0) =>2y=xHence 2y=x is the required equation.

Approved