The line L₁ has equation 4y + 3 = 2x - Edexcel - A-Level Maths Pure - Question 10 - 2012 - Paper 2

L₂ is perpendicular to L₁. First, we need the slope of L₁:

1. Rewrite L₁ in slope-intercept form: 4y = 2x - 3. Hence, y = \frac{1}{2}x - \frac{3}{4} (slope m₁ = \frac{1}{2}).
2. The slope of L₂ (m₂) will be the negative reciprocal of m₁: m₂ = -2.
3. Use the point-slope form to write the equation of L₂: y - 4 = -2(x - 2).
4. Distributing gives: y - 4 = -2x + 4 → y + 2x - 8 = 0.
5. Rearranging gives the final equation: 2x + y - 8 = 0.

Thus, L₂ is described by the equation 2x + y - 8 = 0.

To find the intersection point D of the lines L₁ and L₂:

1. Set the equations equal: 4y + 3 = 2x (from L₁) and 2x + y - 8 = 0 (from L₂).
2. Substitute y from L₂ into L₁: 4(8 - 2x) + 3 = 2x.
3. Simplifying: 32 - 8x + 3 = 2x → 35 = 10x → x = 3.5.
4. Substitute x back to find y: y = 8 - 2(3.5) = 1.

Thus, the coordinates of point D are (3.5, 1).

To find length CD where C is (2, 4) and D is (3.5, 1):

1. Use the distance formula: (CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}).
2. Substituting values: (CD = \sqrt{(3.5 - 2)^2 + (1 - 4)^2} = \sqrt{(1.5)^2 + (-3)^2} = \sqrt{2.25 + 9} = \sqrt{11.25} = \frac{3}{2} \sqrt{5}.)

This confirms that the length of CD is indeed (\frac{3}{2} \sqrt{5}).

To find the area of quadrilateral ACBE, we first find the area of triangles ABC and ABE:

1. Area of triangle ABC:

   • Vertices: A(p, 4), B (unknown), C(2, 4)
   • Using coordinates to determine the base and height.

2. Area of triangle ABE:

   • Vertices: A(p, 4), B (unknown), E (on L₂)

3. Both areas can be found using the formula:

   • Area = \frac{1}{2} \times base \times height.
   • Finally, adding areas will yield the area of ACBE.

As the calculations depend on additional specifics for points B and E, general forms can be used for area calculations.