Find the equation of a line passing through the point ( 4 , 3 ) , which cuts a triangle of minimum area from…

The equation of the straight line passing through the point $\left(4, 3\right)$, whose slope $m$ (assume).

$y-3=m\left(x-4\right) \Rightarrow y=mx-4m+3$

So, $x-$intercept of the line is $4-\frac{3}{m}$

and $y-$intercept is $3-4m$,

So, area of the given triangle is $A=\frac{1}{2}\times \left(4-\frac{3}{m}\right)\left(3-4m\right)=\frac{1}{2}\left(24-16m-\frac{9}{m}\right)=12-8m-\frac{9}{2m}$

To find the min/max,

$\frac{dA}{dm}=0 \Rightarrow -8+\frac{9}{2m^2} =0 \Rightarrow m=\pm \frac{3}{4}$

Second derivative test :

${\overline{)\frac{d^{2}A}{d{m}^2}}}_m=-\frac{9}{m^3}\Rightarrow {\overline{)\frac{d^{2}A}{d{m}^2}}}_{m=\frac{3}{4}}<0 \mathrm{and} {\overline{)\frac{{\mathrm{d}}^2\mathrm{A}}{{\mathrm{dm}}^2}}}_{\mathrm{m}=-\frac{3}{4}}>0$

Also, the straight line makes triangle in the first quadrant, so $m=-\frac{3}{4}$

So, the equation of the line which passes though the given point is

$y-3=-\frac{3}{4}\left(x-4\right) \Rightarrow 4y-12=-3x+12 \Rightarrow 3x+4y=24$