If
\(\begin{vmatrix} 2a &x_{1} &y_{1} \\ 2b&x_{2} &y_{2} \\ 2c&x_{3} &y_{3} \end{vmatrix}\)
= abc / 2 ≠ 0, then the area of the triangle whose vertices are [(x1 / a), (y1 / a)], [(x2 / b), (y2 / b)] and [(x3 / c), (y3 / c)] is

1) 1/4(abc)

2) 1/8 (abc)

3) 1/4

4) 1/8

5) 1/12

Solution: (4) ⅛

\(\begin{array}{l}\begin{aligned} &\left|\begin{array}{ccc} 2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3} \end{array}\right|=\frac{a b c}{2}\\ &2\left|\begin{array}{lll} \mathrm{a} & \mathrm{x}_{1} & \mathrm{y}_{1} \\ \mathrm{~b} & \mathrm{x}_{2} & \mathrm{y}_{2} \\ \mathrm{c} & \mathrm{x}_{3} & \mathrm{y}_{3} \end{array}\right|=\frac{\mathrm{abc}}{2} ; \therefore\left|\begin{array}{lll} \mathrm{a} & \mathrm{x}_{1} & \mathrm{y}_{1} \\ \mathrm{~b} & \mathrm{x}_{2} & \mathrm{y}_{2} \\ \mathrm{c} & \mathrm{x}_{3} & \mathrm{y}_{3} \end{array}\right|=\frac{\mathrm{abc}}{4}\\ &\text { Area of triangle whose vertices are }\\ &\left(\frac{\mathrm{x}_{1}}{\mathrm{a}}, \frac{\mathrm{y}_{1}}{\mathrm{a}}\right),\left(\frac{\mathrm{x}_{2}}{\mathrm{~b}}, \frac{\mathrm{y}_{2}}{\mathrm{~b}}\right),\left(\frac{\mathrm{x}_{3}}{\mathrm{c}}, \frac{\mathrm{y}_{3}}{\mathrm{c}}\right) \text { is }\\ &A=\frac{1}{2}\left|\begin{array}{ccc} \frac{x_{1}}{a} & \frac{y_{1}}{a} & 1 \\ \frac{x_{2}}{b} & \frac{y_{2}}{b} & 1 \\ \frac{x_{3}}{c} & \frac{y_{3}}{c} & 1 \end{array}\right|=\frac{1}{2} \frac{1}{a} \frac{1}{b} \frac{1}{c}\left|\begin{array}{lll} x_{1} & y_{1} & a \\ x_{2} & y_{2} & b \\ x_{3} & y_{3} & c \end{array}\right|\\ &\frac{1}{2 \mathrm{abc}}\left|\begin{array}{lll} \mathrm{x}_{1} & \mathrm{y}_{1} & \mathrm{a} \\ \mathrm{x}_{2} & \mathrm{y}_{2} & \mathrm{~b} \\ \mathrm{x}_{3} & \mathrm{y}_{3} & \mathrm{c} \end{array}\right|\left[\because \text { changing } \mathrm{c}_{3} \text { to } \mathrm{c}_{1} \text { and } \mathrm{c}_{1} \text { to } \mathrm{c}_{2}\right]\\ &\frac{1}{2 \mathrm{abc}} \times \frac{\mathrm{abc}}{4}=\frac{1}{8}\\ &\therefore \text { Area of triangle }=\frac{1}{8} \end{aligned}\end{array} \)

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