Question

If sides of a triangle are $y=mx+a,y=nx+b$ and $x=0$, then its area is-

• A. $\frac{1{(a-b)}^2}{2(m-n)}$
• B. $\frac{1{(a-b)}^2}{2(m+n)}$
• C. $\frac{1{(a+b)}^2}{2(m-n)}$
• D. $noneofthese$

Answer 1

The correct option is A $\frac{1{(a-b)}^2}{2(m-n)}$

Given lies are

$L_1:y=mx+a$

$L_2:y=nx+b$

Line $L_1$ intersects $y-axis$ at $A=(0,a)$

Line $L_2$ intersects $y-axis$ at $B=(0,b)$

Comparing the given two equations,

$mx+a=nx+b$

$\Rightarrow$ $nx-mx=a-b$

$\Rightarrow$ $x=\frac{a-b}{n-m}$

Substituting above value in $L_1$ we get,

$y=m\frac{a-b}{n-m}+a$

$\Rightarrow$ $y=\frac{ma-mb+na-ma}{n-m}$

$\Rightarrow$ $y=\frac{na-mb}{n-m}$

$\therefore$ The co-ordinates of intersection of the lines is $C=(\frac{a-b}{n-m},\frac{na-mb}{n-m})$

Distance between $A$ and $B$

$d=\sqrt{(0-0{)}^2+(a-b{)}^2}$

$\Rightarrow$ $d=\sqrt{(a-b{)}^2}$

$\Rightarrow$ $d=a-b$

Length of perpendicular from $C$ on the $y-$ axis.

$p=\frac{a-b}{n-m}$

$\therefore$ Area of the triangle fromed by the lines $L_1,L_2$ and $y-axis$

$A=\left|\frac{1}{2}dp\right|$

$\Rightarrow$ $A=\frac{1}{2}\times (a-b)\times \frac{a-b}{n-m}$

$\Rightarrow$ $A=\frac{(a-b{)}^2}{2(m-n)}$