The area of the triangle formed by the lines y = m_1 x + c_1 , , … - Vidyadip

MCQ

The area of the triangle formed by the lines $y = {m_1}x + {c_1},\,$ $y = {m_2}x + {c_2}$ and $x = 0$is

A
$\frac{1}{2}\frac{{{{({c_1} + {c_2})}^2}}}{{({m_1} - {m_2})}}$
B
$\frac{1}{2}\frac{{{{({c_1} - {c_2})}^2}}}{{({m_1} + {m_2})}}$
✓
$\frac{1}{2}\frac{{{{({c_1} - {c_2})}^2}}}{{({m_1} - {m_2})}}$
D
$\frac{{{{({c_1} - {c_2})}^2}}}{{({m_1} - {m_2})}}$

✓

Answer

Correct option: C.

$\frac{1}{2}\frac{{{{({c_1} - {c_2})}^2}}}{{({m_1} - {m_2})}}$

c
(c)On solving the equation of lines, we get the vertices of triangle $(0,\,\,{c_1}),\,\,(0,\,\,{c_2})$,
and $\left( {\frac{{{c_2} - {c_1}}}{{{m_1} - {m_2}}},\,\,\frac{{{m_1}{c_2} - {m_2}{c_1}}}{{{m_1} - {m_2}}}} \right)$
Hence, the area $ = \frac{1}{2}\,\left| {\,\begin{array}{*{20}{c}}0&{{c_1}}&1\\0&{{c_2}}&1\\{\frac{{{c_2} - {c_1}}}{{{m_1} - {m_2}}}}&{\frac{{{m_1}{c_2} - {m_2}{c_1}}}{{{m_1} - {m_2}}}}&1\end{array}\,} \right|$
$ = \frac{1}{2}\left[ {0 + {c_1}\left( {\frac{{{c_2} - {c_1}}}{{{m_1} - {m_2}}}} \right) - {c_{2\,}}\left( {\frac{{{c_2} - {c_1}}}{{{m_1} - {m_2}}}} \right)} \right]$
$ = \frac{1}{2}\frac{{({c_2} - {c_1})\,\,({c_1} - {c_2})}}{{{m_1} - {m_2}}} = \frac{1}{2}\frac{{{{({c_1} - {c_2})}^2}}}{{({m_1} - {m_2})}}$.
(sign is not considered).

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