Question

Assertion $\left(A\right)$ : lf two sides of a triangle are represented by $x^2-3xy+2y^2=0$ and centroid is $\left(\frac{2}{3},0\right)$ , then the third side is $2x-3y-2=0$
Reason $\left(R\right)$ : lf two sides of a triangle are represented by $a{x}^2+2hxy+b{y}^2=0$ and centroid is $(\frac{2x_1}{3},\frac{2y_1}{3})$ then the third side is $x(a{x}_1+h{y}_1)+y(h{x}_1+b{y}_1)=a{x}_1^2+2h{x}_1{y}_1+b{y}_1^2$

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true and R is not correct explanation of A
• C. A is true but R is false
• D. A is false but R is true

Answer 1

The correct option is C A is true but R is false

Given two sides of triangle is $x^2-3xy+2y^2=0$
Let the third line of triangle be $ax+by=1$
Point of intersection of given lines is $(0,0)$ and points of intersection of third line and given lines are $(\frac{1}{a+\frac{b}{2}},\frac{1}{2a+b})$,$(\frac{1}{a+b},\frac{1}{a+b})$
x-coordinate of centroid of triangle is $\frac{1}{a+b}+\frac{2}{2a+b}=2$.........(1)
y-coordinate of centroid of triangle is $\frac{1}{a+b}+\frac{1}{2a+b}=0$.........(2)
By solving above two equations we get a,b as $1,3/2$ respectively
then equation of third line will be $2x-3y-2=0$
From reason,we get
$x_1=1$ and $y_1=0$
we get third side as $x(1\times 1-\frac{3}{2}\times 0)+y(\frac{-3}{2}\times 1+2\times 0)=1\left(1^2\right)+2\left(\frac{-3}{2}\right)\left(1\right)\left(0\right)+2⟹2x-3y=6$, which is false.