`a) text{Theo giả thuyết}`
`->` \(\left\{ \begin{array}{l}A (4; 0)\\B (-3; 0)\\C (6; 0)\end{array} \right.\)
`b) text{Gọi trung điểm của BA, BC, AC lần lượt là: M, N, P}`
`text{Ta có}`
\(\left\{ \begin{array}{l}x_{M} = \dfrac{x_{B} + x_{A}}{2} = \dfrac{-3 + 4}{2} = \dfrac{1}{2}\\y_{M} = 0\end{array} \right.\)
`-> M (1/2; 0)`
`text{Ta có}`
\(\left\{ \begin{array}{l}x_{N} = \dfrac{x_{B} + x_{C}}{2} = \dfrac{-3 + 6}{2} = \dfrac{3}{2}\\y_{N} = 0\end{array} \right.\)
`-> N (3/2; 0)`
`text{Ta có}`
\(\left\{ \begin{array}{l}x_{P} = \dfrac{x_{A} + x_{C}}{2} = \dfrac{4 + 6}{2} = 5\\y_{P} = 0\end{array} \right.\)
`-> P (5; 0)`
`c) text{Ta có}`
`AB = sqrt{(x_{B} - x_{A})^2 + 0} = sqrt{(-3 - 4)^2} = 7`
`AC = sqrt{(x_{C} - x_{A})^2 + 0} = sqrt{(6 - 4)^2} = 2`
`CB = sqrt{(x_{B} - x_{C})^2 + 0} = sqrt{(-3 - 6)^2} = 9`
