Question

Consider a family of lines $\left(4a+3\right)x-\left(a+1\right)y-\left(2a+1\right)=0$ where $a\in R$
minimum area of the triangle which a member of this family with negative gradient can make with the positive semi axes, is

• A. $8$
• B. $6$
• C. $4$
• D. $2$

Answer 1

The correct option is C $4$

Given family of lines

$(4a+3)x-(a+1)y-(2a+1)=0,$ where $a\in R$

Rearranging we get

$4ax+3x-ay-y-2a-1=0$

or $(4x-y-2)a+(3x-y-1)=0$

Obviously the all lines of the family must pass through the point of intersection of two lines represented by the following two equations

$4x-y-2=0$ .......$\left(1\right)$

and $3x-y-1=0$ .......$\left(2\right)$

Subtracting $\left(2\right)$ from $\left(1\right)$ we get

$\Rightarrow 4x-y-2-3x+y+1=0$

$\Rightarrow x=1$

Substituting $x=1$ in $\left(1\right)$ we get

$y=4x-2=4-2=2$

So the coordinates of common point of intersection of all lines of the family will be $(1,2)$

Any line passing through this common point $(1,2)$ may be represented as

$y-2=m(x-1)$ ......$\left(3\right)$

where $m$ is the gradient of the line , a variable one.

Now rearranging $\left(3\right)$ in intercept form we get

$y-2=m(x-1)$

$\Rightarrow mx-y=m-2$

$\Rightarrow \frac{x}{\frac{m-2}{m}}+\frac{y}{2-m}=1$

So area of the triangle made by this line with the positive semi axes will be given by

$A=\frac{1}{2}\times \frac{m-2}{m}\times (2-m)=\frac{-m^2+4m-4}{2m}$

$\Rightarrow A=-\frac{m}{2}+2-\frac{2}{m}$

Differentiating w.r.t to $m$ we get

$\frac{dA}{dm}=-\frac{1}{2}+0+\frac{2}{m^2}$

Imposing the condition of minimization of $A$

$\Rightarrow \frac{dA}{dm}=0$ we get

$-\frac{1}{2}+\frac{2}{m^2}=0$

$\frac{2}{m^2}=\frac{1}{2}$

$\Rightarrow m^2=4$

$\Rightarrow m=-2$ as per given condition the sraight line which forms minimum area with the positive semi axes must have negative gradient.

Hence minimum area of the the triangle should be for $m=-2$

$\therefore A_{min}=\frac{-{(-2)}^2+4(-2)-4}{2\times -2}=\frac{-4-8-4}{-4}=\frac{-16}{-4}=4$