The area of the triangle formed by the lines | vedclass

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(C) To find the vertices of the triangle,we solve the equations of the lines pairwise: $1$. Intersection of $y = m_1x + c_1$ and $x = 0$ gives the poi

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) To find the vertices of the triangle,we solve the equations of the lines pairwise: $1$. Intersection of $y = m_1x + c_1$ and $x = 0$ gives the point $(0, c_1)$. $2$. Intersection of $y = m_2x + c_2$ and $x = 0$ gives the point $(0, c_2)$. $3$. Intersection of $y = m_1x + c_1$ and $y = m_2x + c_2$ gives: $m_1x + c_1 = m_2x + c_2 \implies x(m_1 - m_2) = c_2 - c_1 \implies x = \frac{c_2 - c_1}{m_1 - m_2}$. Substituting $x$ into $y = m_1x + c_1$,we get $y = m_1(\frac{c_2 - c_1}{m_1 - m_2}) + c_1 = \frac{m_1c_2 - m_1c_1 + m_1c_1 - m_2c_1}{m_1 - m_2} = \frac{m_1c_2 - m_2c_1}{m_1 - m_2}$. The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Substituting the vertices $(0, c_1), (0, c_2), (\frac{c_2 - c_1}{m_1 - m_2}, \frac{m_1c_2 - m_2c_1}{m_1 - m_2})$: Area $= \frac{1}{2} |0(c_2 - y_3) + 0(y_3 - c_1) + (\frac{c_2 - c_1}{m_1 - m_2})(c_1 - c_2)|$ Area $= \frac{1}{2} |\frac{-(c_1 - c_2)^2}{m_1 - m_2}| = \frac{1}{2} \frac{(c_1 - c_2)^2}{|m_1 - m_2|}$.

The area of the triangle formed by the lines $y = m_1x + c_1$,$y = m_2x + c_2$ and $x = 0$ is:

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