If the vertices of a triangle be $(a m_{1}^{2}, 2 a m_{1}), (a m_{2}^{2}, 2 a m_{2})$ and $(a m_{3}^{2}, 2 a m_{3})$ , then the area of the triangle is

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Solution

Distance, $d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Here\quad $x_{1} = a, x_{2} = - a, y_{1} = b, y_{2} = - b$

$d = \sqrt{{(- a - a)}^{2} + {(- b - b)}^{2}}$

$d = \sqrt{{4 a}^{2} + 4 b^{2}}$

$d = 2 \sqrt{a^{2} + b^{2}}$

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If the vertices of a triangle be $(a m_{1}^{2}, 2 a m_{1}), (a m_{2}^{2}, 2 a m_{2}) a n d (a m_{3}^{2}, 2 a m_{3})$ , then the area of the triangle is

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If the vertices of a triangle be (am21,2am1),(am22,2am2) and (am23,2am3), then the area of the triangle is

Distance,d=√(x2−x1)2+(y2−y1)2Here\quad x1=a,x2=−a,y1=b,y2=−bd=√(−a−a)2+(−b−b)2d=√4a2+4b2d=2√a2+b2

If the vertices of a triangle be $(a m_{1}^{2}, 2 a m_{1}), (a m_{2}^{2}, 2 a m_{2})$ and $(a m_{3}^{2}, 2 a m_{3})$ , then the area of the triangle is

Distance, $d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Here\quad $x_{1} = a, x_{2} = - a, y_{1} = b, y_{2} = - b$

$d = \sqrt{{(- a - a)}^{2} + {(- b - b)}^{2}}$

$d = \sqrt{{4 a}^{2} + 4 b^{2}}$

$d = 2 \sqrt{a^{2} + b^{2}}$