Find the area of the triangle formed by the straight lines whose equations are
$y = m_{1} x + c_{1}$ , $y = m_{2} x + c_{2}$ , and $y = m_{3} x + c_{3}$ .

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Solution

$y = m_{1} x + c_{1} . . . . . . (i) y = m_{2} x + c_{2} . . . . . (i i) y = m_{3} x + c_{3} . . . . . . (i i)$
Consider the first two line and $y$ axis
$y = m_{1} x + c_{1}, y = m_{2} x + c_{2}, x = 0$
On solving the equations the verticies of triangle fomed by these lines are
$(\frac{c_{2} - c_{1}}{m_{1} - m_{2}}, \frac{m_{1} c_{2} - m_{2} c_{1}}{m_{1} - m_{2}}), (0, c_{1}), (0, c_{2})$
Area of triangle formed by these lines is
$Δ = \frac{1}{2} \times \frac{c_{2} - c_{1}}{m_{1} - m_{2}} \times (c_{2} - c_{1}) Δ = \frac{1}{2} \frac{{(c_{2} - c_{1})}^{2}}{m_{1} - m_{2}}$
If we take consecutive two lines and $y$ axis and find area of all the three triangles , then area of triangle formed by given three lines will be the sum of areas of the three triangles formed.

$Δ^{'} = \frac{1}{2} (\frac{{(c_{2} - c_{1})}^{2}}{m_{1} - m_{2}} + \frac{{(c_{3} - c_{2})}^{2}}{m_{2} - m_{3}} + \frac{{(c_{3} - c_{1})}^{2}}{m_{3} - m_{1}})$

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Similar questions

y = m_{1} x + c_{1}

y = m_{2} x + c_{2}

y

y = m_{1} x + \frac{a}{m_{1}}

y = m_{2} x + \frac{a}{m_{2}}

y = m_{3} x + \frac{a}{m_{3}}

y = m_{1} x + c_{1}

y = m_{2} x + c_{2}

y = m_{3} x + c_{3}

m_{1} (c_{2} - c_{3}) + m_{2} (c_{3} - c_{1}) + m_{3} (c_{1} - c_{2}) = 0

Find the conditions that the straight lines $y = m_{1} x + c_{1}, y = m_{2} x + c_{2}$ and $y = m_{3} x + c_{3}$ may meet in a point.

If the three lines $y = m_{1} x + c_{1}, y = m_{2} x + c_{2}$ and $y = m_{3} x + c_{3}$ are concurrent then show that,

$m_{1} (c_{2} - c_{3}) + m_{2} (c_{3} - c_{1}) + m_{3} (c_{1} - c_{2}) = 0$

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