Question

If G be the centroid and I be the incentre of the triangle with vertices A(-36,7),B(20,7),C(0,-8) and $GI=\frac{25}{2}\sqrt{205\lambda }$ ,then find the value of $\lambda$:

• A. $\frac{1}{25}$
• B. $\frac{1}{28}$
• C. $\frac{1}{22}$
• D. $\frac{1}{13}$
• E. None

Answer 1

The correct option is E None

According to the problem

$A\left(-36,7\right),B\left(20,7\right)andC\left(0,-8\right)$

Hence the sides be

$AB=c=\sqrt{{\left(-36-20\right)}^2+{\left(7-7\right)}^2}$

$=\sqrt{{56}^2}=56$

$BC=a=\sqrt{{\left(20-0\right)}^2+{\left(7+8\right)}^2}$

$=\sqrt{400+225}=\sqrt{625}$

$=25$

$CA=b=\sqrt{{\left(-36-0\right)}^2+{\left(7+8\right)}^2}=\sqrt{1296+225}$

$=\sqrt{1521}=39$

therefore the coordinates are

$=\left(\frac{a{x}_A+b{x}_B+c{x}_C}{a+b+c},\frac{a{y}_A+b{y}_B+c{y}_C}{a+b+c}\right)$

$=\left(\frac{25\times -36+39\times 20+56\times 0}{25+39+56},\frac{25\times 7+39\times 7+56\times -8}{25+39+56}\right)$

$=\left(-\frac{120}{120},\frac{0}{120}\right)$

$=\left(-1,0\right)$

Therefore point of G

$=\left(\frac{x_A+x_B+x_C}{3},\frac{y_A+y_B+y_C}{3}\right)$

$=\left(\frac{-36+20+0}{3},\frac{7+7-8}{3}\right)$

$\frac{-16}{3},2$

$G.I=\sqrt{{\left(\frac{-16}{3}+1\right)}^2+{\left(2-0\right)}^2}$

$=\sqrt{{\left(\frac{-13}{3}\right)}^2+4}$

$=\sqrt{\frac{169}{9}+4}$

$=\sqrt{\frac{169+36}{9}}$

$=\sqrt{\frac{205}{9}}$

Hence the value of $\lambda$ is none of these.