Question

WILL GIVE BRAINLEST!!

Answer 1

Answer:

Step-by-step explanation:

Part A

Eq of line passing through C and A

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

Part B

Coordinates of point D

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

Part C

Distance DB

$h=\sqrt{(\frac{43}{26}-3)^2+(\frac{71}{26}-3)^2}=\frac{7}{\sqrt{26}}$

Part D

Length of base

$base=\sqrt{(2-1)^2+(1-6)^2}=\sqrt{26}$

$area=\frac{1}{2}base*h=\frac{1}{2}\sqrt{26}*\frac{7}{\sqrt{26}}=\frac{7}{2}$