A straight line T, which passes through the points A(-3, -5) and B(3, -6) is perpendicular to a line L at the point C(-2, -2). a) Find the gradient of line T. b) Find the equation of line T in the for

13. A straight line T, which passes through the points A(-3, -5) and B(3, -6) is perpendicular to a line L at the point C(-2, -2).

a) Find the gradient of line T.

Step 1: Use the formula for the gradient $m = \frac{y_2 - y_1}{x_2 - x_1}$ with points A$(-3, -5)$ and B$(3, -6)$.
Let $(x_1, y_1) = (-3, -5)$ and $(x_2, y_2) = (3, -6)$.

$$m_T = \frac{-6 - (-5)}{3 - (-3)}$$

Step 2: Simplify the expression.

$$m_T = \frac{-6 + 5}{3 + 3}$$ $$m_T = \frac{-1}{6}$$

The gradient of line T is $\boxed{-\frac{1}{6}}$.

b) Find the equation of line T in the form $y = mx + c$ where $m$ and $c$ are constants.

Step 1: Use the point-slope form $y - y_1 = m(x - x_1)$ with the gradient $m_T = -\frac{1}{6}$ and point A$(-3, -5)$.

$$y - (-5) = -\frac{1}{6}(x - (-3))$$ $$y + 5 = -\frac{1}{6}(x + 3)$$

Step 2: Distribute the gradient and simplify.

$$y + 5 = -\frac{1}{6}x - \frac{3}{6}$$ $$y + 5 = -\frac{1}{6}x - \frac{1}{2}$$

Step 3: Isolate $y$ to get the equation in the form $y = mx + c$.

$$y = -\frac{1}{6}x - \frac{1}{2} - 5$$ $$y = -\frac{1}{6}x - \frac{1}{2} - \frac{10}{2}$$ $$y = -\frac{1}{6}x - \frac{11}{2}$$

The equation of line T is $\boxed{y = -\frac{1}{6}x - \frac{11}{2}}$.

c) Given that another line Q is parallel to T and passes through (1, -3) find the equation of line Q in the form $ax + by = c$.

Step 1: Since line Q is parallel to line T, their gradients are equal.

$$m_Q = m_T = -\frac{1}{6}$$

Step 2: Use the point-slope form $y - y_1 = m(x - x_1)$ with $m_Q = -\frac{1}{6}$ and the point $(1, -3)$.

$$y - (-3) = -\frac{1}{6}(x - 1)$$ $$y + 3 = -\frac{1}{6}(x - 1)$$

Step 3: Multiply by 6 to eliminate the fraction.

$$6(y + 3) = -1(x - 1)$$ $$6y + 18 = -x + 1$$

Step 4: Rearrange the equation into the form $ax + by = c$.

$$x + 6y = 1 - 18$$ $$x + 6y = -17$$

The equation of line Q is $\boxed{x + 6y = -17}$.

d) Find the points of intersection of L and Q.

Step 1: Find the gradient of line L. Line T is perpendicular to line L, so $m_L \cdot m_T = -1$.

$$m_L \cdot \left(-\frac{1}{6}\right) = -1$$ $$m_L = 6$$

Step 2: Find the equation of line L. Line L passes through C$(-2, -2)$ with gradient $m_L = 6$.

$$y - (-2) = 6(x - (-2))$$ $$y + 2 = 6(x + 2)$$ $$y + 2 = 6x + 12$$ $$y = 6x + 10$$

Step 3: Solve the system of equations for line L and line Q simultaneously.
Equation of L: $y = 6x + 10$
Equation of Q: $x + 6y = -17$
Substitute the expression for $y$ from line L into the equation for line Q.

$$x + 6(6x + 10) = -17$$ $$x + 36x + 60 = -17$$ $$37x + 60 = -17$$ $$37x = -17 - 60$$ $$37x = -77$$ $$x = -\frac{77}{37}$$

Step 4: Substitute the value of $x$ back into the equation for line L to find $y$.

$$y = 6\left(-\frac{77}{37}\right) + 10$$ $$y = -\frac{462}{37} + \frac{10 \times 37}{37}$$ $$$$