Question

show that the points (-1, -2),(5,4),(-3,0) are the vertices of a right triangle, and find its area.

Answer 1

Answer on Question #82973 – Math – Analytic Geometry

Question

Show that the points (-1, -2), (5,4), (-3,0) are the vertices of a right triangle, and find its area.

Solution

Let points be A(-1, -2), B(5, 4) and C(-3,0). Now let’s check whether the points are on the one straight line. Find equation of the straight line AB:

$$\frac {x - x _ {B}}{x _ {A} - x _ {B}} = \frac {y - y _ {B}}{y _ {A} - y _ {B}}$$ $$\frac {x - 5}{- 1 - 5} = \frac {y - 4}{- 2 - 4}$$ $$\frac {x - 5}{- 6} = \frac {y - 4}{- 6}$$ $$x - 5 = y - 4$$ $$y = x - 1$$

If point C belongs to the line AB, its coordinates satisfy the equation: $y = x - 1$ :

$$0 = - 3 - 1$$ $$0 \neq - 4$$

We make sure that the point C does not belong to the line AB. It also means that these points form the triangle ABC. Now let’s check if it is right.

Find the coordinates of vectors $\overrightarrow{AB}$, $\overrightarrow{BC}$ and $\overrightarrow{AC}$:

$$\overrightarrow {A B} = \left\{x _ {B} - x _ {A}, y _ {B} - y _ {A} \right\} = \left\{5 - (- 1), 4 - (- 2) \right\} = \{6, 6 \}$$ $$\overrightarrow {B C} = \left\{x _ {C} - x _ {B}, y _ {C} - y _ {B} \right\} = \{- 3 - 5, 0 - 4 \} = \{- 8, - 4 \}$$ $$\overrightarrow {A C} = \left\{x _ {C} - x _ {A}, y _ {C} - y _ {A} \right\} = \{- 3 - (- 1), 0 - (- 2) \} = \{- 2, 2 \}$$

Now find angles between vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ (named $\alpha$), $\overrightarrow{BC}$ and $\overrightarrow{AC}$ (named $\beta$), $\overrightarrow{AC}$ and $\overrightarrow{AB}$ (named $\gamma$):

$$\alpha = \arccos \left(\frac {x _ {A B} * x _ {B C} + y _ {A B} * y _ {B C}}{| A B | * | B C |}\right) = \arccos \left(\frac {6 * (- 8) + 6 * (- 4)}{\sqrt {6 ^ {2} + 6 ^ {2}} * \sqrt {(- 8) ^ {2} + (- 4) ^ {2}}}\right) = 0.55$$ $$= \arccos \left(- \frac {7 2}{6 * \sqrt {2} * 4 * \sqrt {5}}\right) = \arccos \left(- \frac {3}{\sqrt {1 0}}\right);$$ $$\begin{array}{l} \beta = \arccos \left(\frac{x_{AC} \cdot x_{BC} + y_{AC} \cdot y_{BC}}{|AC| \cdot |BC|} \right) = \arccos \left(\frac{(-2) \cdot (-8) + 2 \cdot (-4)}{\sqrt{(-2)^2 + 2^2} \cdot \sqrt{(-8)^2 + (-4)^2}}\right) = \\ = \arccos \left(\frac{8}{2 \cdot \sqrt{2} \cdot 4 \cdot \sqrt{5}}\right) = \arccos \left(\frac{1}{\sqrt{10}}\right); \end{array}$$ $$\begin{array}{l} \gamma = \arccos \left(\frac{x_{AB} \cdot x_{AC} + y_{AB} \cdot y_{AC}}{|AB| \cdot |AC|}\right) = \arccos \left(\frac{6 \cdot (-2) + 6 \cdot 2}{\sqrt{6^2 + 6^2} \cdot \sqrt{(-2)^2 + 2^2}}\right) = \arccos \left(\frac{0}{6 \cdot \sqrt{2} \cdot 2 \cdot \sqrt{2}}\right) = \\ = \arccos(0) = 90{}^\circ; \end{array}$$

We proved that the triangle ABC is right with $\gamma = 90{}^\circ$. It means that segments AB and AC are legs of the triangle ABC.

Thus, the area of the triangle ABC is

$$\begin{array}{l} S_{\Delta ABC} = \frac{1}{2} \cdot |AB| \cdot |AC| = \frac{1}{2} \cdot \sqrt{6^2 + 6^2} \cdot \sqrt{(-2)^2 + 2^2} = \frac{1}{2} \cdot 6\sqrt{2} \cdot 2\sqrt{2} = \\ = 6 \cdot 2 = 12 \text{ square units}. \end{array}$$

Answer: the points (-1, -2), (5,4), (-3,0) are the vertices of a right triangle, and its area is 12 square units.

Answer provided by https://www.AssignmentExpert.com