Question

$(3,0)$ is a point on the line joining the points $(3x^2,6x)$ and $(3y^2,6y)$. Then the value of $xy=$.

• A. -1
• B. 2
• C. 1

Answer 1

The correct option is A -1


If $(3,0)$ is a point on the line joining the points $(3x^2,6x)$ and $(3y^2,6y)$, then we must have, slope of AB = slope of BC

We know that the slope(m) of the line formed by joining the points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$m=\frac{y_2-y_1}{x_2-x_1}$

Then, we must have,

$\frac{6x-0}{3x^2-3}=\frac{6y-6x}{3y^2-3x^2}$

$⟹\frac{6x}{3(x^2-1)}=\frac{6(y-x)}{3(y+x)(y-x)}$ $[\because a^2-b^2=(a+b\left)\right(a-b\left)\right]$

$⟹\frac{x}{x^2-1}=\frac{1}{y+x}$

$⟹x^2-1=x(y+x)$

$⟹x^2-1=xy+x^2$

$⟹xy=-1$