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If Two Sides of a Triangle Are Unequal, Then the Angle Opposite to the Greater Side Is Greater Than Angle Opposite to the Smaller Side

Find the area...

Question

Find the area of the triangle formed by the lines $y = m_{1} x + c_{1}, y = m_{2} x + c_{2}$ and x = 0

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Solution

Let the sides AC, AB and BC of $△$ ABC be represented by the equations:

$m_{1} x - y + c_{1} = 0$ .....(i)

$m_{2} x - y + c_{2} = 0$ .....(ii)

$x = 0$ .....(iii)

On solving (i) and (ii) by cross multiplication, we have,

$\frac{x}{(- C_{2} + C_{1})} = \frac{y}{(m_{2} c_{1} - m_{1} c_{2})} = \frac{1}{(- m_{1} + m_{2})}$

$\Rightarrow x = \frac{(c_{1} - c_{2})}{(m_{2} - m_{1})} and y = \frac{(m_{2} c_{1} - m_{1} c_{2})}{(m_{2} - m_{1})}$

Thus, the lines AC and AB intersect at A $(\frac{c_{1} - c_{2}}{m_{2} - m_{1}}, \frac{m_{2} c_{1} - m_{1} c_{2}}{m_{2} - m_{1}})$

On solving (ii) and (iii), we get B $(0, c_{2}$ ).

On solving (i) and (ii), we get C(0, $c_{1}$ )

$∴$ area of $△ A B C$ is given by,

A $= \frac{1}{2} ∣ ∣ \begin{matrix} \frac{(c_{1} - c_{2})}{(m_{2} - m_{1})} . (c_{2} - c_{1}) + 0 + 0 \end{matrix} ∣ ∣$

$= \frac{1}{2} \frac{(c_{1} - c_{2})^{2}}{| m_{1} - m_{2} |}$

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Similar questions

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

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

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

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

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

y

x = 0, y = m_{1} x + c_{1}

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

y = m_{1} x + c_{1}, y = m_{2} x + c_{2}

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

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

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