Find the equation of line passing through the points A(-1,3) and B(0,2). Hence show that the points A, B and C (1,-1) are collinear.

A line passes through A( - 1 , 3) & B(0 , 2).

A , B , C(1 , 1) are collinear.

We know that,

Slope of a line passing through the co - ordinates (x₁, y₁) and (x₂ , y₂) is :

m = (y₂ - y₁)/(x₂ - x₁)

Let,

Hence,

→ m = (2 - 3)/( 0 - ( - 1))

→ m = - 1/1

→ m = - 1

Now,

Equation of a line ⟹ y - y₁ = m (x - x₁)

Putting the values we get,

→ y - 3 = - 1(x - ( - 1))

→ y - 3 = - x - 1

→ y + x = - 1 + 3

→ x + y = 2

Hence, the equation of the line is x + y = 2.

Now,

we have to prove that,

A , B , C are collinear.

i.e.,

Slope of AB should be equal to Slope of BC.

We have;

Slope of AB = - 1.

→ - 1 = ( 1 - 2)/(1 - 0)

→ - 1 = - 1/1

→ - 1 = - 1

Hence, A , B , C are collinear points.

I hope it will help you.

Regards.

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)

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step-by-step explanation: