Problem and Solution
We are given a line $l$ with gradient $$ -\frac{1}{12} $$ that passes through the points $$ A(p,10) \quad \text{and} \quad B(123,0) $$
Part (a): Show that $p = 3$
The gradient of the line passing through two points $$ (x_1,y_1) \quad \text{and} \quad (x_2,y_2) $$ is given by
$$ \text{Gradient} = \frac{y_2 - y_1}{x_2 - x_1} $$
For the points $$ A(p,10) \quad \text{and} \quad B(123,0) $$ the gradient is
$$ \text{Gradient of line} = \frac{0 - 10}{123 - p} $$
We are given that the gradient is $$ -\frac{1}{12} $$ so
$$ \frac{-10}{123 - p} = -\frac{1}{12} $$
Multiplying both sides by $123 - p$:
$$ -10 = \frac{-1}{12}(123 - p) $$
Now multiply both sides by $12$:
$$ -120 = -(123 - p) $$
This simplifies to
$$ -120 = -123 + p $$
Adding $123$ to both sides:
$$ p = 3 $$
Thus, $$ p = 3 $$
Part (b): Find an equation for $l$ in the form $rx + sy + t = 0$
We know the gradient of line $l$ is $$ -\frac{1}{12} $$ and it passes through the point $$ A(3,10) $$
The point-slope form of a line is
$$ y - y_1 = m(x - x_1) $$
Substitute $$ m = -\frac{1}{12}, \quad x_1 = 3, \quad y_1 = 10 $$
$$ y - 10 = -\frac{1}{12}(x - 3) $$
Simplify:
$$ y - 10 = -\frac{1}{12}x + \frac{3}{12} $$
$$ y - 10 = -\frac{1}{12}x + \frac{1}{4} $$
Multiply through by $12$:
$$ 12(y - 10) = -x + 3 $$
$$ 12y - 120 = -x + 3 $$
Rearranging into the form $$ rx + sy + t = 0 $$
$$ x + 12y - 123 = 0 $$
Thus, the equation of line $l$ is
$$ x + 12y - 123 = 0 $$
Part (c): Find an equation for $k$ in the form $y = mx + c$
The line $k$ is perpendicular to $l$ and passes through $$ A(3,10) $$
The gradient of line $l$ is $$ -\frac{1}{12} $$
Therefore, the gradient of line $k$ is the negative reciprocal:
$$ m_k = 12 $$
Using point-slope form:
$$ y - 10 = 12(x - 3) $$
Simplify:
$$ y - 10 = 12x - 36 $$
$$ y = 12x - 26 $$
Thus, the equation of line $k$ is
$$ y = 12x - 26 $$
Part (d): Find the exact area of triangle $ABC$
We are given the points:
- $A(3,10)$
- $B(123,0)$
- $C$ is the x-intercept of line $k$
To find point $C$, set $$ y = 0 $$ in $$ y = 12x - 26 $$
$$ 0 = 12x - 26 $$
Solving for $x$:
$$ 12x = 26 \quad \Rightarrow \quad x = \frac{26}{12} = \frac{13}{6} $$
Thus, $$ C\left(\frac{13}{6},0\right) $$
Use the area formula for a triangle:
$$ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| $$
Substitute $$ A(3,10), \quad B(123,0), \quad C\left(\frac{13}{6},0\right) $$
$$ \text{Area} = \frac{1}{2} \left| 3(0 - 0) + 123(0 - 10) + \frac{13}{6}(10 - 0) \right| $$
$$ = \frac{1}{2} \left| 0 + 123(-10) + \frac{130}{6} \right| $$
$$ = \frac{1}{2} \left| -1230 + \frac{65}{3} \right| $$
$$ = \frac{1}{2} \times 1208\frac{1}{3} $$
$$ = 604\frac{1}{6} $$
Thus, the exact area of triangle $ABC$ is
$$ \boxed{604\frac{1}{6}} $$
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