Question

The equations of the sides of a triangle are $x=0,y=m_{1}x+c_1$ and $y=m_{2}x+c_2$. The area of the triangle

• A. $\frac{c_1-c_2}{m_1-m_2}$
• B. $\frac{1}{2}\frac{(c_1-c_2{)}^2}{m_1-m_2}$
• C. $\frac{1}{2}\frac{(c_1-c_2)}{(m_1-m_2{)}^2}$
• D. $\frac{(c_1-c_2{)}^2}{(m_1-m_2{)}^2}$

Answer 1

The correct option is B $\frac{1}{2}\frac{(c_1-c_2{)}^2}{m_1-m_2}$

Let $x=0$ is the base of the triangle, and the length of the base of the triangle is $c_2-c_1$ from the figure.

The height of the triangle is the $x$ coordinate of the point where $y=m_{1}x+c_1$ and $y=m_{2}x+c_2$ meet i.e R.

$m_{1}x+c_1=m_{2}x+c_2⟹x=\frac{c_2-c_1}{m_1-m_2}$

Area of the triangle $=\frac{1}{2}\times base\times height=\frac{1}{2}\times (c_2-c_1)\times \frac{c_2-c_1}{m_1-m_2}$

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