The diagram shows part of the curve $y = -\sqrt{10 - 6x}$ meeting the x-axis at the point A. The normal to the curve at B, where $x = -1$, meets the x-axis at the point C.

1. Find the $y$-coordinate of point B

When $x = -1$,
$y = -\sqrt{10 - 6(-1)}$
$y = -\sqrt{16} = -4$
So, B is $(-1, -4)$.

2. Differentiate the equation of the curve to find $\frac{dy}{dx}$

$y = -(10 - 6x)^{\frac{1}{2}}$
$\frac{dy}{dx} = -\frac{1}{2}(10 - 6x)^{-\frac{1}{2}} \cdot (-6)$
$\frac{dy}{dx} = \frac{3}{\sqrt{10 - 6x}}$

3. Find the gradient of the tangent at B

At $x = -1$,
$\frac{dy}{dx} = \frac{3}{\sqrt{10 - 6(-1)}} = \frac{3}{\sqrt{16}} = \frac{3}{4}$

4. Find the gradient of the normal at B

Gradient of normal $m_n = -\frac{1}{\frac{dy}{dx}}$
$m_n = -\frac{1}{\frac{3}{4}} = -\frac{4}{3}$

5. Find the equation of the normal at B

Using $y - y_1 = m(x - x_1)$ with $(-1, -4)$ and $m = -\frac{4}{3}$:
$y - (-4) = -\frac{4}{3}(x - (-1))$
$y + 4 = -\frac{4}{3}(x + 1)$
$y = -\frac{4}{3}x - \frac{4}{3} - 4$
$y = -\frac{4}{3}x - \frac{16}{3}$

1. Find the equation of the normal at B and the coordinates of C

From part (a), $y = -(10-6x)^{1/2}$.
$\frac{dy}{dx} = -\frac{1}{2}(10-6x)^{-1/2}(-6) = \frac{3}{\sqrt{10-6x}}$.
At $x = -1$, $\frac{dy}{dx} = \frac{3}{\sqrt{16}} = \frac{3}{4}$.
Gradient of normal $= -\frac{4}{3}$.
Equation of normal at B$(-1, -4)$:
$y - (-4) = -\frac{4}{3}(x - (-1)) \Rightarrow y = -\frac{4}{3}x - \frac{16}{3}$.
At C, $y = 0 \Rightarrow -\frac{4}{3}x - \frac{16}{3} = 0 \Rightarrow x = -4$.
So C is $(-4, 0)$.

2. Find the area of the triangle bounded by the normal, x-axis, and the vertical line $x=-1$

The region from $x = -4$ to $x = -1$ is a triangle with vertices $(-4,0)$, $(-1,0)$, and $(-1,-4)$.
Base $= -1 - (-4) = 3$.
Height $= 4$.
Area of triangle $= \frac{1}{2} \times 3 \times 4 = 6$.

3. Set up the definite integral for the area under the curve from $x=-1$ to $x=5/3$

The curve meets the x-axis at A where $10-6x=0 \Rightarrow x = \frac{5}{3}$.
The area between the curve and the x-axis from $x=-1$ to $x=\frac{5}{3}$ is:
$\int_{-1}^{5/3} |-\sqrt{10-6x}| \, dx = \int_{-1}^{5/3} (10-6x)^{1/2} \, dx$

4. Evaluate the definite integral

$\int_{-1}^{5/3} (10-6x)^{1/2} \, dx = \left[ \frac{(10-6x)^{3/2}}{\frac{3}{2} \times (-6)} \right]_{-1}^{5/3}$
$= \left[ -\frac{1}{9}(10-6x)^{3/2} \right]_{-1}^{5/3}$
$= -\frac{1}{9}(0)^{3/2} - \left( -\frac{1}{9}(10-6(-1))^{3/2} \right)$
$= 0 + \frac{1}{9}(16)^{3/2}$
$= \frac{1}{9}(64) = \frac{64}{9} = 7\frac{1}{9}$

5. Calculate the total shaded area

Total area $= \text{Area of triangle} + \text{Area under curve}$
$= 6 + 7\frac{1}{9} = 13\frac{1}{9}$