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Problem 23

# The cartesian co-ordinates $$(x, y)$$ of a point on a curve are given by $$x: y: 1=t^{3}: t^{2}-3: t-1$$ where $$t$$ is a parameter, then the points given by $$t=a, b, c$$ are collinear, if (a) $$a b c+3(a+b+c)=a b+b c+c a$$ (b) $$3 a b c+2(a+b+c)=a b+b c+c a$$ (c) $$a b c+2(a+b+c)=3(a b+b c+c a)$$ (d) none of these

Expert verified
(c) $$abc+2(a+b+c)=3(ab+bc+ca)$$ is the right equation.
See the step by step solution

## Step 1: Understanding the Collinearity Condition

If three points are collinear, they all lie on the same straight line. Therefore there is a linear relationship between them. Suppose A, B and C are three collinear points. The slope of the line AB is equal to the slope of the line BC. Let's use this property to find the condition for a, b, c to be collinear.

## Step 2: Find Cartesian Coordinates

Given the equation: $$x: y: 1= t^{3}: t^{2}-3: t-1$$, we find the cartesian coordinates of points given by t = a, b, and c, which are ($$a^3$$, $$a^2-3$$), ($$b^3$$, $$b^2-3$$), and ($$c^3$$, $$c^2-3$$). We will now calculate the slopes of the lines AB and BC.

## Step 3: Calculate Slopes & Set Equal

The slope of line AB is: $$\frac{(a^2-3) - (b^2-3)}{a^3 - b^3} = \frac{a^2 - b^2}{a^3 - b^3} = \frac{a+b}{a^2+ab+b^2}$$. \nThe slope of the line BC is: $$\frac{(b^2-3) - (c^2-3)}{b^3 - c^3} = \frac{b^2 - c^2}{b^3 - c^3} = \frac{b+c}{b^2+bc+c^2}$$. \nSet these two equal to each other, we get: $$\frac{a+b}{a^2+ab+b^2} = \frac{b+c}{b^2+bc+c^2}$$.

## Step 4: Simplify Equation and Identify the condition

Multiplying both sides by the denominator and simplifying, we get: $$a^2c+abc+c^2a + 2(a+b+c) = a^2b+b^2c+c^2a$$, which after rearrangement results in $$abc+2(a+b+c) = 3(ab+bc+ca)$$.

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