Question

A(0;1) B(1;3) C(2;7) tìm K thuộc ox : KB=KC

Answer 1

K thuộc Ox nên K(x;0)

\(KB=\sqrt{\left(x-1\right)^2+\left(0-3\right)^2}=\sqrt{\left(x-1\right)^2+9}\)

\(KC=\sqrt{\left(2-x\right)^2+\left(7-0\right)^2}=\sqrt{\left(x-2\right)^2+49}\)

Để KB=KC thì \(\left(x-1\right)^2+9=\left(x-2\right)^2+49\)

=>\(x^2-2x+10=x^2-4x+53\)

=>-2x+10=-4x+53

=>2x=43

=>\(x=\dfrac{43}{2}\)

Vậy: \(K\left(\dfrac{43}{2};0\right)\)