If |2a&x1&y1 2b&x2&y2 2c&x3&y3|=(abc/2)≠0 , then the area of the triangle whose vertices are ((x1/a),(y1/a)),((x2/b),(y2/b)) and ((x3/a),(y3/a)) is

Question

If |2a&x1&y1 2b&x2&y2 2c&x3&y3|=(abc/2)≠0 , then the area of the triangle whose vertices are ((x1/a),(y1/a)),((x2/b),(y2/b)) and ((x3/a),(y3/a)) is (A) (1/4) abc (B) (1/8) abc (C) (1/4) (D) (1/8)

Tardigrade Admin · Accepted Answer

Q. If $∣ ∣ 2 a 2 b 2 c x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} ∣ ∣ = \frac{ab c}{2} \neq = 0,$
then the area of the triangle whose vertices are $(\frac{x _{1}}{a}, \frac{y _{1}}{a}), (\frac{x _{2}}{b}, \frac{y _{2}}{b})$ and $(\frac{x _{3}}{a}, \frac{y _{3}}{a})$ is

$\frac{1}{4} ab c$

$\frac{1}{8} ab c$

$\frac{1}{4}$

$\frac{1}{8}$

$\frac{1}{12}$

Q. If ∣∣​2a2b2c​x1​x2​x3​​y1​y2​y3​​∣∣​=2abc​=0, then the area of the triangle whose vertices are (ax1​​,ay1​​),(bx2​​,by2​​) and (ax3​​,ay3​​) is

41​abc

81​abc

41​

81​

121​

Q. If $∣ ∣ 2 a 2 b 2 c x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} ∣ ∣ = \frac{ab c}{2} \neq = 0,$
then the area of the triangle whose vertices are $(\frac{x _{1}}{a}, \frac{y _{1}}{a}), (\frac{x _{2}}{b}, \frac{y _{2}}{b})$ and $(\frac{x _{3}}{a}, \frac{y _{3}}{a})$ is

$\frac{1}{4} ab c$

$\frac{1}{8} ab c$

$\frac{1}{4}$

$\frac{1}{8}$

$\frac{1}{12}$