Coordinate Geometry

Please help me!

i) Find the equation of BC

slope BA: \(\begin{array}{|rcll|} \hline m = \frac{y_A-y_B}{x_A-x_B} \\ \hline \end{array}\)

slope BC: \(\begin{array}{|rcll|} \hline m_{\perp} = -\frac{1}{m} &=& -\frac{x_A-x_B}{y_A-y_B} \\ \hline \end{array}\)

Equation of BC:

\(\begin{array}{|rcll|} \hline m_{\perp} = \frac{y -y_B}{x -x_B} &=& -\frac{x_A-x_B}{y_A-y_B} \\ & \dots & \\ y &=& -\left(\frac{x_A-x_B}{y_A-y_B}\right)\cdot (x-x_B) + y_B \\ && x_B = -2 \quad y_B = 8 \qquad x_A = 2 \quad y_A = 14 \\ &=& -\left[\frac{2-(-2)}{14-8}\right]\cdot [x-(-2)] + 8 \\ &=& -\frac{4}{6}\cdot (x+2) + 8 \\ &=& -\frac{2}{3}\cdot (x+2) + 8 \\ \mathbf{y} & \mathbf{=} & \mathbf{-\frac{2}{3}\cdot x + \frac{20}{3}} \\ \hline \end{array}\)

ii) Find the coordinates of D

coordinates of C ( \(y_C=0\) )

\(\begin{array}{|rcll|} \hline 0 & = & -\frac{2}{3}\cdot x_C + \frac{20}{3} \\ \frac{2}{3}\cdot x_C &=& \frac{20}{3} \\ x_C &=& \frac{20}{3}\cdot \frac{3}{2} \\ \mathbf{x_C} & \mathbf{=} & \mathbf{10} \\ \hline \end{array}\)

C is (10,0)

coordinates of D:

\(\begin{array}{|rcll|} \hline x_D &=& x_C + (x_A-x_B) \\ x_D &=& 10 + [2-(-2)] \\ x_D &=& 10 + 4 \\ \mathbf{x_D} & \mathbf{=} & \mathbf{14} \\\\ y_D &=& y_C + (y_A-y_B) \\ y_D &=& 0 + (14-8) \\ \mathbf{y_D} & \mathbf{=} & \mathbf{6} \\ \hline \end{array} \)

D is (14,6)

iii) Find the angle that CD makes with the positive x axis.

\(\begin{array}{|rcll|} \hline \varphi &=& \arctan(m) \quad & | \quad m = \frac{y_A-y_B}{x_A-x_B} = \frac{14-8}{2-(-2)}=\frac{6}{4}=\frac32 \\ \varphi &=& \arctan(\frac32) \\ \varphi &=& \arctan(1.5) \\ \mathbf{\varphi} & \mathbf{=} & \mathbf{56.3099324740^{\circ} } \\ \hline \end{array} \)