Show that the area of the triangle formed by the lines `y=m_1x+c_1,""""y=m_2x+c_2` and `x=0` is `((c_1-c_2)^2)/(2|m_1-m_2|)`

Explanation

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

`m_1x−y+c_1=0`........(i)
`m_2x−y+c_2=0`........(ii)
x=0........(iii) On solving (i) and (ii) by cross multiplication, we have

`x/(−c_2+c_1)=y/(m_2c_1−m_1c_2)=1/(−m_1+m_2) `

⇒`x=(c_1−c_2)/(m_2−m_1)`and `y=(m_2c_1−m_1c_2)1/(m_2−m_1)`

Thus, the lines AC and AB intersect at

`A((c_1−c_2)/(m_2−m_1),(m_2c_1−m_1c_2)/(m_2−m_1))`

On solving (ii) and (iii), we get B(0,c2) On solving (i) and (iii), we get C(0,c1)

∴area of △ABCis given by `△=1/2.|x_1(y_2−y_3)+x_2(y_3−y_1)+x_3(y_1−y_2)| `

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