Fill in the table using this function rule.
1 answer:
You might be interested in
<u>Given</u>:
Given that the graph OACE.
The coordinates of the vertices OACE are O(0,0), A(2m, 2n), C(2p, 2r) and E(2t, 0)
We need to determine the midpoint of EC.
<u>Midpoint of EC:</u>
The midpoint of EC can be determined using the formula,
Substituting the coordinates E(2t,0) and C(2p, 2r), we get;
Simplifying, we get;
Dividing, we get;
Thus, the midpoint of EC is (t + p, r)
Hence, Option A is the correct answer.
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