If ∣∣∣∣2a x1 y12b x2 y22c x3 y3∣∣∣∣=abc2≠0, then the area of the triangle whose vertices are(x1a,y1a),(x2b,y2b),(x3c,y3c) is

If ∣∣∣∣2a x1 y12b x2 y22c x3 y3∣∣∣∣=abc2≠0, then the

1.	If ∣∣∣∣2a x1 y12b x2 y22c x3 y3∣∣∣∣=abc2≠0, then the area of the triangle whose vertices are(x1a,y1a),(x2b,y2b),(x3c,y3c) is
Answer» If $∣ ∣∣ ∣ \begin{matrix} 2 a x_{1} y_{1} 2 b x_{2} y_{2} 2 c x_{3} y_{3} \end{matrix} ∣ ∣∣ ∣ = \frac{a b c}{2} \neq 0$ , then the area of the triangle whose vertices are $(\frac{x_{1}}{a}, \frac{y_{1}}{a}), (\frac{x_{2}}{b}, \frac{y_{2}}{b}), (\frac{x_{3}}{c}, \frac{y_{3}}{c})$ is