Pyramid’s volume is ##56/3##

Let they be

##A(8;5);B(6;7);C(5;1);h=8##

First you would calculate the area of the triangle ABC by getting the lenght of a side:

##AC=sqrt((x_C-x_A)^2+(y_C-y_A)^2)=sqrt((5-8)^2+(1-5)^2)=sqrt(9+16)=5##

and then the height relative to AC by finding the distance between the point B and the line AC:

##m_(AC)=(y_C-y_A)/(x_C-x_A)=(1-5)/(5-8)=4/3##

Then the equation of the line AC is

##y-y_A=m_(AC)(x-x_A)##

that’s

##y-5=4/3(x-8)##

##y-5=4/3x-32/3##

##3y-15=4x-32##

##4x-3y-17=0## in the form ##ax+by+c=0##

Then let’s calculate the distance:

##d=|ax_B+by_B+c|/sqrt(a^2+b^2)##

##d=|4*6-3*7-17|/sqrt(4^2+3^2)=|24-21-17|/sqrt25=14/5##

Now let’s calculate the area of the triangle:

##Area=(side AC)*(d)*1/2=cancel5*cancel14^7/cancel5*1/cancel2=7##

The pyramid’s volume is obtained by:

##V=1/3Area_(base)*h=1/3*7*8=56/3##