Edexcel IAL January 2023 P4 (WMA14/01) Q2 Parametric Equations, Cartesian Form
A set of points P (x, y) is defined by the parametric equations
\( x = \frac{t-1}{2t+1} \qquad y = \frac{6}{2t+1} \qquad t \neq -\frac{1}{2} \)
(a) Show that all points P (x, y) lie on a straight line.
(4)
(4)
(a) Hence or otherwise, find the x coordinate of the point of intersection of this line and the line with equation \( y = x + 12 \)
(2)
(2)
Solution to Part a: Showing Points Lie on a Straight Line
Key Concepts Used:
- Eliminating the Parameter
- Solving for \( t \)
Step-by-Step Working:
-
Let’s rearrange the x-equation.
\( x = \frac{t-1}{2t+1} \)
-
Expand and collect like terms.
\( x(2t+1) = t-1 \)
\( 2xt + x = t-1 \)
\( x + 1 = t – 2xt \)
\( x + 1 = t(1-2x) \)
\( t = \frac{x+1}{1-2x} \)
-
Now, put it in equation of y.
\( y = \frac{6}{2t+1} \)
\( y = \frac{6}{2\left(\frac{x+1}{1-2x}\right) + 1} \)
-
Simplify to linear form.
\( y = \frac{6}{2t+1} \)
\( y = \frac{6}{2\left(\frac{x+1}{1-2x}\right) + 1} \)
\( y = \frac{6}{\frac{2(x+1)}{1-2x} + 1} \)
\( y = \frac{6}{\frac{2(x+1) + (1-2x)}{1-2x}} \)
\( y = \frac{6(1-2x)}{2(x+1) + (1-2x)} \)
\( y = \frac{6(1-2x)}{2x+2+1-2x} \)
\( y = \frac{6(1-2x)}{3} \)
\( y = 2(1-2x) \)
\( y = 2 – 4x \)
Since this is a linear equation, hence all points P (x, y) lie on a straight line.
Final Answer:
\( y = -4x + 2 \) (linear equation)
Solution to Part b: Finding the intersection
Key Concepts Used:
- Simultaneous Equations (to find intersection, solve the line from part (a) with the given line)
- Substitution Method (Substitute one equation into the other.)
Step-by-Step Working:
- To find the x-coordinate of point of intersection of lines let’s equate them and solve them simultaneously.
\( x + 12 = 2 – 4x \)
\( 5x = -10 \)
\( x = -2 \)
\( 5x = -10 \)
\( x = -2 \)
Final Answer:
\( x = -2 \)