Question

2. Triangle ABC is drawn in the coordinate plane.1) Calculate the coordinates of the midpoints of the sides of the triangle.2) Calculate the lengths of the triangle's bisectors.3) Calculate the length of the segments into which the bisectors of the triangle are divided by their point of intersection.

Answer 1

To find:

1. The coordinates of the midpoints of the sides of the triangle.

2. The lengths of the triangle's bisectors.

3. The length of the segments into which the bisectors of the triangle are divided by their point of intersection.

Solution:

The coordinates of the point A is (-7, -2), point B is (-3, 5) and point C is (10, -6)

1. The midpoint formula is given by:

`(x,y)=((x_1+x_2)/(2),(y_1+y_2)/(2))`

So, the midpoint of line AB is:

`D=((-7-3)/(2),(-2+5)/(2))=(-5,(3)/(2))`

The midpoint of the side BC is:

`E=((-3+10)/(2),(-6+5)/(2))=((7)/(2),-(1)/(2))`

The midpoint of the side AC is:

`F=((-7+10)/(2),(-2-6)/(2))=((3)/(2),-4)`

2)

The distance formula is given by:

`D=√((x_1-x_2)^2+(y_1-y_2)^2)`

So, the length of AE is:

`\begin{gathered} AE=\sqrt{(-7-(7)/(2))^2+(-2+(1)/(2))^2} \\ =√(110.25+2.25) \\ =10.61 \end{gathered}`

The length of BF is:

`\begin{gathered} BF=\sqrt{(-3-(3)/(2))^2+(5+4)^2} \\ =√(20.25+81) \\ =10.06 \end{gathered}`

The length of CD is:

`\begin{gathered} CD=\sqrt{(10+5)^2+(-6-(3)/(2))^2} \\ =√(225+56.25) \\ =16.77 \end{gathered}`

3)

The centroid of the circle is the point of intersection of the medians of the triangle and given by :

`((x_1+x_2+x_3)/(3),(y_1+y_2+y_3)/(3))`

So, the centroid of the triangle is:

`((-7-3+10)/(3),(-2+5-6)/(3))=(0,-1)`

Thus, the centroid is O(0, -1).

Now, the distance of each vertex from the centroid is:

`\begin{gathered} AO=√((-7-0)^2+(-2+1)^2) \\ =√(49+1) \\ =7.07 \end{gathered}`
`\begin{gathered} BO=√((-3-0)^2+(5+1)^2) \\ =√(9+36) \\ =6.71 \end{gathered}`
`\begin{gathered} CO=√((10-0)^2+(-6+1)^2) \\ =√(100+25) \\ =11.18 \end{gathered}`