Figure 3 shows a sketch of part of the curve C with equation y = \frac{1}{2} x + 27 - 12, \quad x > 0 The point A lies on C and has coordinates \left( 3, -\frac{3}{2} \right) - Edexcel - A-Level Maths Pure - Question 10 - 2018 - Paper 1

To find the equation of the normal at point A, we first determine the derivative of the curve:

$y = \frac{1}{2}x + 27 - 12$

Calculating the derivative, we get:

$\frac{dy}{dx} = \frac{1}{2}$

At point A ( \left( 3, -\frac{3}{2} \right) ), substituting ( x = 3 ):

$\frac{dy}{dx} = \frac{1}{2}$

The gradient of the normal ( m_n ) is the negative reciprocal of the tangent gradient:

$m_n = -\frac{1}{m_t} = -\frac{1}{\frac{1}{2}} = -2$

Using the point-slope form of the equation of a line:

$y - y_1 = m_n (x - x_1) \Rightarrow y + \frac{3}{2} = -2\left(x - 3\right)$

Rearranging this, we get:

$y + \frac{3}{2} = -2x + 6\Rightarrow y = -2x + \frac{9}{2}$

To write this in the form 10y = 4x - 27, we can multiply through by 10:

$10y = -20x + 45 \Rightarrow 10y + 20x = 45 \Rightarrow 10y = 4x - 27.$

Thus, we have shown the required equation.