Answer to Question #293931 in Analytic Geometry for bookaddict

Question

a. The points 𝐴,𝐵 and 𝐶 have co-ordinates (−1,2),(1,1) and (2,3) respectively. Sketch the triangle 𝐴𝐵𝐶 .By calculating the lengths of the sides of this triangle, determine if it is scalene, isosceles, or equilateral.

b. Find the distance 𝑀𝐵, where 𝑀 is the midpoint of 𝐴𝐶, and hence find the area of the triangle 𝐴𝐵𝐶.

Answer 1

a.

$$AB=\sqrt{(1-(-1))^2+(1-2)^2}=\sqrt{5}$$

$$BC=\sqrt{(2-1)^2+(3-1)^2}=\sqrt{5}$$

$$AC=\sqrt{(2-(-1))^2+(3-2)^2}=\sqrt{10}$$

The triangle ABC is isosceles.

b.

$$M(\dfrac{x_A+x_C}{2}, \dfrac{y_A+y_C}{2})$$

$$M(\dfrac{-1+2}{2}, \dfrac{2+3}{2})$$

$$M(\dfrac{1}{2}, \dfrac{5}{2})$$

$$BM=\sqrt{(x_M-x_B)^2+(y_M-y_B)^2}$$

$$=\sqrt{(\dfrac{1}{2}-1)^2+(\dfrac{5}{2}-1)^2}=\dfrac{\sqrt{10}}{2}(units)$$

$$AC=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}$$

$$=\sqrt{(2+1)^2+(3-2)^2}=\sqrt{10}(units)$$

$$S_{ABC}=\dfrac{1}{2}BM\cdot AC=\dfrac{1}{2}(\dfrac{\sqrt{10}}{2})(\sqrt{10})$$

$$=\dfrac{5}{2}({units}^2)$$