a) D nằm trên trục Ox nên D(x; 0) $\Rightarrow$ $\vec{AD}$ = (x - 1; -3); $\vec{BD}$ = (x - 4; -2)
Ta có: DA = DB $\Rightarrow$ $(x - 1)^{2} + (-3)^{2}$ = $(x - 4)^{2} + (-2)^{2}$
$\Leftrightarrow$ $x^{2}$ - 2x + 1 + 9 = $x^{2}$ - 8x + 16 + 4 $\Leftrightarrow$ 6x = 10 $\Leftrightarrow$ x = $\frac{5}{3}$
Vậy D($\frac{5}{3}$; 0)
b) Ta có: $\vec{OA}$ = (1; 3); $\vec{OB}$ = (4; 2); $\vec{AB}$ = (3; -1)
Suy ra: OA = |$\vec{OA}$| = $\sqrt{1^{2} + 3^{2}}$ = $\sqrt{10}$
OB = |$\vec{OB}$| = $\sqrt{4^{2} + 2^{2}}$ = $2\sqrt{5}$
AB = |$\vec{AB}$| = $\sqrt{3^{2} + (-1)^{2}}$ = $\sqrt{10}$
$\Rightarrow$ Chu vi tam giác OAB là: OA + OB + AB = $\sqrt{10}$ + $2\sqrt{5}$ + $\sqrt{10}$ = $2\sqrt{10}$ + $2\sqrt{5}$
c) Ta có: $\vec{OA}$.$\vec{AB}$ = 1. 3 + 3. (-1) = 0
$\Rightarrow$ $\vec{OA}$ $\perp$ $\vec{AB}$
$\Rightarrow$ $S_{OAB}$ = $\frac{1}{2}$OA. AB = $\frac{1}{2}$. $\sqrt{10}$. $\sqrt{10}$ = 5