A triangle has vertices at (1,10), (-5,2), and (7,2). What is its orthocenter. Show your work.

Question: What is the orthocenter of a triangle with the vertices (-1,2) (5,2) and (2,1)?

The coordinates of point A are (-1,2), point B are (5,2), and point C are (2,1).

The orthocent is the intersection of the three altitudes. An altitude goes from a vertex and is perpendicular to the line containing the opposite side.

In the coordinate plane the equations of the altitudes can be found and then a system of equations can be solved.

Altitude 1. From point C perpendicular to the line containing side AB.

Slope of line AB is 0 (horizontal line), a vertical line is perpendicular to a horizontal line. Thus, the equation of altitude 1 is  x=2 .

Altitude 2. From point B perpendicular to the line containing side AC.

Slope of line AC is  −13 , the slope of a line perpendicular to line AC is 3. The equation of altitude 2 is  y=3x−13  

Altitude 3. From point A perpendicular to the line containing side BC.

Slope of line BC is  13 , the slope of a line perpendicular to line BC is  −3 . The equation of altitude 3 is  y=−3x−1  

The orthocenter is the point where all three altitudes intersect.

x=2  

y=3x−13  

y=−3x−1  

Use substitution to solve the first two equations  y=3(2)−13=−7  

The orthocenter is the point  (2,−7)  

we did not need the third equation, but we can use it as a check, plug the coordinates into the third equation:

−7=−3(2)−1  

−7=−6−1  

−7=−7  it works.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2