Question 2
The point A has coordinates (−5, 3), the point B has coordinates (4, 0) and the point C has coordinates (−1, 5).
The line l passes through C and is perpendicular to AB.
(a) Find an equation of l.
Give your answer in the form $ax + by + c = 0$ where $a$, $b$ and $c$ are integers.
(4)
The line l intersects AB at the point D.
(b) Show that the coordinates of D are (−2, 2).
(3)
(c) Show that l is not the perpendicular bisector of AB.
(2)
(d) Find the value of $\tan \angle ABC$.
Give your answer in its simplest form.
(4)
Problem Solution
Part (a): Find an equation of line $ l $.
The line $ l $ is perpendicular to $ AB $ and passes through point $ C(-1, 5) $. To find the equation of $ l $, we first calculate the slope of line $ AB $.
1. Find the slope of $ AB $:
The slope of a line through two points $ (x_1, y_1) $ and $ (x_2, y_2) $ is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]For points $ A(-5, 3) $ and $ B(4, 0) $:
\[ m_{AB} = \frac{0 - 3}{4 - (-5)} = \frac{-3}{9} = -\frac{1}{3} \]2. Find the slope of line $ l $:
Since line $ l $ is perpendicular to line $ AB $, the slope of line $ l $, denoted $ m_l $, is the negative reciprocal of $ m_{AB} $:
\[ m_l = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{1}{3}} = 3 \]3. Find the equation of line $ l $:
Using the point-slope form of the equation of a line, $ y - y_1 = m(x - x_1) $, and the point $ C(-1, 5) $, we have:
\[ y - 5 = 3(x + 1) \]Simplifying:
\[ y - 5 = 3x + 3 \quad \Rightarrow \quad y = 3x + 8 \]Rearranging to the form $ ax + by + c = 0 $:
\[ 3x - y + 8 = 0 \]Thus, the equation of line $ l $ is:
\[ 3x - y + 8 = 0 \]Part (b): Show that the coordinates of point $ D $ are $ (-2, 2) $.
The point $ D $ is the intersection of lines $ AB $ and $ l $. We already know the equation of line $ l $ is $ 3x - y + 8 = 0 $. Now, we find the equation of line $ AB $.
1. Find the equation of line $ AB $:
The slope of line $ AB $ is $ m_{AB} = -\frac{1}{3} $. Using the point-slope form of the equation of a line, we get:
\[ y - 3 = -\frac{1}{3}(x + 5) \]Simplifying:
\[ y - 3 = -\frac{1}{3}x - \frac{5}{3} \quad \Rightarrow \quad y = -\frac{1}{3}x + \frac{4}{3} \]Thus, the equation of line $ AB $ is:
\[ y = -\frac{1}{3}x + \frac{4}{3} \]2. Solve the system of equations:
We now solve the system of equations:
\[ y = 3x + 8 \quad \text{(line $ l $)} \] \[ y = -\frac{1}{3}x + \frac{4}{3} \quad \text{(line $ AB $)} \]Set the two expressions for $ y $ equal to each other:
\[ 3x + 8 = -\frac{1}{3}x + \frac{4}{3} \]Multiply through by 3 to eliminate fractions:
\[ 9x + 24 = -x + 4 \]Solve for $ x $:
\[ 9x + x = 4 - 24 \quad \Rightarrow \quad 10x = -20 \quad \Rightarrow \quad x = -2 \]Substitute $ x = -2 $ into the equation of line $ AB $ to find $ y $:
\[ y = -\frac{1}{3}(-2) + \frac{4}{3} = \frac{2}{3} + \frac{4}{3} = 2 \]Thus, the coordinates of $ D $ are $ (-2, 2) $.
Part (c): Show that line $ l $ is not the perpendicular bisector of $ AB $.
To be the perpendicular bisector, line $ l $ must pass through the midpoint of $ AB $ and be perpendicular to $ AB $. We already know that line $ l $ is perpendicular to $ AB $, but we need to check if it passes through the midpoint of $ AB $.
1. Find the midpoint of $ AB $:
The midpoint $ M $ of a line segment with endpoints $ A(x_1, y_1) $ and $ B(x_2, y_2) $ is given by:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]For $ A(-5, 3) $ and $ B(4, 0) $, the midpoint is:
\[ M = \left( \frac{-5 + 4}{2}, \frac{3 + 0}{2} \right) = \left( \frac{-1}{2}, \frac{3}{2} \right) \]2. Check if the midpoint lies on line $ l $:
The equation of line $ l $ is $ y = 3x + 8 $. Substituting $ x = -\frac{1}{2} $ into the equation:
\[ y = 3\left(-\frac{1}{2}\right) + 8 = -\frac{3}{2} + 8 = \frac{13}{2} \]Since the $ y $-coordinate of the midpoint is $ \frac{3}{2} $, which is not equal to $ \frac{13}{2} $, point $ M $ does not lie on line $ l $.
Therefore, line $ l $ is not the perpendicular bisector of $ AB $.
Part (d): Find the value of $ \tan \angle ABC $.
To find $ \tan \angle ABC $, we use the lengths of sides $ AB $, $ BC $, and $ AC $.
1. Find the length of $ AB $:
\[ AB = \sqrt{(4 - (-5))^2 + (0 - 3)^2} = \sqrt{(4 + 5)^2 + (-3)^2} = \sqrt{9^2 + (-3)^2} = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10} \]2. Find the length of $ BC $:
\[ BC = \sqrt{(-1 - 4)^2 + (5 - 0)^2} = \sqrt{(-5)^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \]3. Find the length of $ AC $:
\[ AC = \sqrt{(-1 - (-5))^2 + (5 - 3)^2} = \sqrt{(4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \]4. Use the law of cosines to find $ \cos \angle ABC $:
\[ \cos \angle ABC = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC} \] \[ \cos \angle ABC = \frac{(3\sqrt{10})^2 + (5\sqrt{2})^2 - (2\sqrt{5})^2} {2 \times 3\sqrt{10} \times 5\sqrt{2}} = \frac{120}{30\sqrt{20}} = \frac{4}{2\sqrt{5}} = \frac{2}{\sqrt{5}} \]5. Find $ \tan \angle ABC $:
\[ \tan \angle ABC = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} = \frac{1}{2} \]
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