The orthocenter is the point where the three altitudes meet.
sketch the graph and you will see that AB is a horizontal line, the altitude is a vertical line through the point (1,3), so the equation of this altitude is x=1
next, find another altitude. I'll use the altitude of BC.
the slope of BC is (6-3)/(4-1)=1, so the slope of the altitude, which is perpendicular to BC going through the point A (0,6), is -1, the equation of the altitude of BC is y=-x+6
the system of equation : x=1
y=-x+6
has a solution (1, 5)
the solution is where the two lines meet, the meeting point is the orthocenter.
Double check by find the equation for other altitude:
slope of AC: (3-6)/(1-0)=-3
slope of altitude of AC: 1/3
equation of altitude of AC: y=(1/3)x+b
the altitude of AC goes through point B (4,6), so we can find out b by plug x=4, y=6 in the equation: 6=(1/3)*4+b, b=14/3
y=(1/3)x+14/3
Is (1,5) also a solution to this equation? Plug x=1 in the equation, we get y=5, so yes, (1,5) is a point on the third altitude.
Anything raised do 0 is always 1
I hope that’s correct
Plug in the values of p and q where you see them in the equation:
-(2+4)2 / (-6) - Distribute the -1
(-2-4)2 / (-6) - Distribute the 2
(-4-8) / (-6) - Subtract what's inside the parenthesis
(-12) / (-6) - Divide
The answer is 2
It would be 31.80 because if you do 1.59 x 20 that will be the answer
