A triangle has corners A, B, and C located at #(2 ,7 )#, #(5 ,3 )#, and #(9 , 4 )#, respectively. What are the endpoints and length of the altitude going through corner C?

Answer 1

endpoints are:
#C=(9,4)#
#H=(5.96, 1.72)#
length of #CH=3.8#

#A=(2,7)# #B=(5,3)# #C=(9,4)# Endpoints of height passing through C are C itself and some point lying on segment (side) AB. To find that point we need to solve system of equations corresponding to the height and segment AB. To find an equation of a line we must have either 2 points (from which we can get the slope) or 1 point and the slope.

Line AB has two points, and its slope is:

#m=(rise)/(run)=(y_B-y_A)/(x_B-x_A)=(3-7)/(5-2)=-4/3#

and the following equation would result from using point B in place of point A:

#y-y_A=m_(AB)(x-x_A)# #y-7=-4/3(x-2)# #y=-4/3x+8/3+7=-4/3x+29/3#

Since height is only one point and its slope is unknown, it is inversely reciprocal of AB's slope because it is perpendicular to AB.

#m_h=3/4#

Moreover, the formula

#y-y_C=m_h(x-x_C)# #y-4=3/4(x-9)# #y=3/4x-27/4+4=3/4x-11/4#
So here we have a system of equations: #{(y=-4/3x+29/3),(y=3/4x-11/4):}#
We could solve for x: #-4/3x+29/3=3/4x-11/4#
to get #x=149/25=5 24/25=5.96#
and substitute to any equation (second one here): #y=3/4(149/25)-11/4#
to get #y=43/25=1 18/25=1.72#
Now we have point #H=(5.96, 1.72)# which is the other endpoint of height.

We can use pythagorean formula to get distance between H and C: #CH=sqrt((x_C-x_H)^2+(y_C-y_H)^2) =sqrt((9-5 24/25)^2+(4-1 18/25)^2)=sqrt((3 1/25)^2+(2 7/25)^2) =sqrt((76^2+57^2)/25^2)=sqrt(5776+3249)/25=sqrt(9025)/25=sqrt(25(90*4+1))/25=sqrt(361)/5=19/5#

Or using decimals (calculator may be handy): #CH=sqrt((9-5.96)^2+(4-1.72)^2)=sqrt((3.04)^2+(2.28)^2) =sqrt(9.2416+5.1984)=sqrt(14.44)=3.8#

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Answer 2

The altitude going through corner C has endpoints (9, 4) and the foot of the altitude, which can be found using the equation of the line containing side AB. The length of the altitude can be calculated using the distance formula between these two endpoints.

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