audreybergey2501
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WILL GIVE BRAINLEST!!
Finding the area of a triangle is stralghtforward if you know the length of the base and the helght of the triangle. But is it possible to
find the area of a triangle if you know only the coordinates of its vertices? In this task, you'll find out. Consider /_\ABC, whose vertices
are A(2, 1), B(3, 3), and C(1,6); let line segment AC represent the base of the triangle.

Part A
Find the equation of the line passing through B and perpendicular to AC.

Part B
Let the point of intersection of AC with the line you found in part A be point D. Find the coordinates of point B.

Part C
Use the distance formula to find the length of the base and the height of /_\ABC.

Part D
Find the area of /_\ABC.

Sagot :

rvillalta2002

Answer:

Step-by-step explanation:

Part A

Eq of line passing through C and A

[tex]y-y_o=m(x-x_o)\\y-6=\frac{6-1}{1-2}(x-1)~=>y=-5x+11\\\\PERPEDICULAR~ LINE\\\\m_2=-\frac{1}{m_1}=-\frac{1}{-5}=\frac{1}{5}\\y-3=\frac{1}{5}(x-3)~=>y=\frac{1}{5}x+\frac{12}{5}[/tex]

Part B

Coordinates of point D

[tex]-5x+11=\frac{1}{5}x+\frac{12}{5}=>11-\frac{12}{5}=x(\frac{1}{5}+5)=>x=\frac{43}{26}\\y=-5(\frac{43}{26})+11=\frac{71}{26}\\D(\frac{43}{26}, \frac{71}{26})[/tex]

Part C

Distance DB

[tex]h=\sqrt{(\frac{43}{26}-3)^2+(\frac{71}{26}-3)^2}=\frac{7}{\sqrt{26}}[/tex]

Part D

Length of base

[tex]base=\sqrt{(2-1)^2+(1-6)^2}=\sqrt{26}[/tex]

[tex]area=\frac{1}{2}base*h=\frac{1}{2}\sqrt{26}*\frac{7}{\sqrt{26}}=\frac{7}{2}[/tex]

View image rvillalta2002
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