Question

A straight line is drawn through a given point $P(1,4)$. Determine the length of the larger intercept from the triangle formed by the line and co-ordinate axes having least area.

• A. $4$
• B. $6$
• C. $8$
• D. $10$

Answer 1

The correct option is C $8$

Any line through (1,4) is $y-4=m(x-1)$ or mx-y=m-4 where $m=\tan \theta =-ive.$

Its intercepts on the exes are $\frac{m-4}{m},-(m-4)$

Area of triangle formed $A=\frac{1}{2}\times \frac{m-4}{m}\times -(m-4)$

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

For minimum area, $\frac{dA}{dm}=0$

$\frac{dA}{dx}=-\frac{1}{2}\times \frac{2(m-4)m-(m-4{)}^2}{m^2}$

It is zero when $m^2=16$

Hence, $m=\pm 4$

But, $m<0$

Hence, $m=4$

Intercepts, are $(2,0)$ and $(0,8)$

Larger intercept is $8$.

Hence, option C.