Question

A line through the origin intersects $x=1,y=2,$ and $x+y=4$, at $A,B$ and $C$respectively, such that $OA.OB.OC=8\sqrt{2}$. Find the equation of the line is $y=mx$.Find the value of $m$

Answer 1

Let equation of line $y=mx$

At $x=1,y=m$

At $y=2,$ $x=\frac{2}{m}$

$x+y=4$

$x(m+1)=4$

$x=\frac{4}{m+1}$

$y=\frac{4m}{m+1}$

$A(1,m)$

$B(\frac{2}{m},2)$

$C(\frac{4}{m+1},\frac{4m}{m+1})$

$OA=\sqrt{1^2+m^2}$

$OB=\sqrt{\frac{4}{m^2}+4}$

$OC=\sqrt{{\left(\frac{4}{m+1}\right)}^2+{\left(\frac{4m}{m+1}\right)}^2}$

Acc to question

$\sqrt{1+m^2}\ast \frac{2}{m}\sqrt{1+m^2}\ast \frac{4}{m+1}\sqrt{1+m^2}=8\sqrt{2}$

$\frac{8}{m(m+1)}(1+m^2{)}^{\frac{3}{2}}=8\sqrt{2}$

Squaring both

$\frac{(1+m^2{)}^3}{m^2(m+1{)}^2}=2$

$(1+m^2{)}^3=2m^2(m+1{)}^2$

$(1+m^6)+3m^2+3m^4=2m^2(m^2+1+2m)$

$1+m^6+3m^2+3m^4=2m^4+2m^2+4m^3$

Using hit and trial method

$m=1$ Ans