(a 3 - 2a + 5) - (4a 3 - 5a 2 + a - 2)
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The parametric equations for the line passes through the points (X₁ , Y₁) and (X₂ , Y₂) are :
X = X₁ + t (X₂ - X₁) ......................... (1)
Y = Y₁ + t (Y₂ - Y₁) .......................... (2)
For eliminating the parameter (t) , first we will solve any one equation for 't' and then substitute that into another equation.
From the equation (1),
⇒ t =
Substituting this t = into the equation (2), we will get :
Y = Y₁ + (Y₂ - Y₁)
Y = Y₁ +
So, this is the standard form of rectangular equation.
For two given points (0, 0) and (4, -4)
X₁ = 0 , Y₁ = 0, X₂ = 4 and Y₂ = -4
For finding the parametric equations, we will plug these values into the given parametric equations.
X = X₁ + t (X₂ - X₁)
⇒ X = 0+ t (4 - 0)
⇒ X = 4t
and Y = Y₁ + t (Y₂ - Y₁)
⇒ Y = 0+ t (-4 - 0)
⇒ Y = - 4t
So, the parametric equations for the line passing through (0,0) and (4, -4) are: X = 4t and Y = -4t