Given:
color(white)("XXX")A=(8,3)XXXA=(8,3)
color(white)("XXX")B=(4,5)XXXB=(4,5)
color(white)("XXX")C=(6,7)XXXC=(6,7)
First let's consider the line through AA and BB
color(white)("XXX")"slope"_(AB)=(5-3)/(4-8)=-1/2XXXslopeAB=5−34−8=−12
color(white)("XXX")XXXusing the point-slope form the equation of the line through AA and BB is
color(white)("XXXXXX")(y-3)=-1/2(x-8)XXXXXX(y−3)=−12(x−8)
color(white)("XXXXXX")2y-6 =8-xXXXXXX2y−6=8−x
color(white)("XXXXXX")x+2y=14XXXXXXx+2y=14
Looking at the altitude from ABAB through CC
color(white)("XXX")XXXThe altitude line is always perpendicular to the base.
color(white)("XXX")XXXTherefore
color(white)("XXX")"slope"_"Alt"= 2XXXslopeAlt=2 ...since the base has a slope of (-1/2)(−12)
color(white)("XXX")XXXusing the slope-point form the equation of the altitude line through CC is
color(white)("XXXXXX")y-7=2(x-6)XXXXXXy−7=2(x−6)
color(white)("XXXXXX")y-7=2x-12XXXXXXy−7=2x−12
color(white)("XXXXXX")2x-y=5XXXXXX2x−y=5
Finding the point of intersection of the altitude with the base ABAB
color(white)("XXX")XXXWe have
color(white)("XXXXXX"){(x+2y=14),(2x-y=5):}
color(white)("XXX")Using standard solution techniques:
color(white)("XXXXXX")(x,y)=(24/5,23/5)
![enter image source here]()
Calculating the length of the Altitude
color(white)("XXX")Using the Pythagorean Theorem
color(white)("XXX")with endpoints (6,7) and (24/5,23/5)
color(white)("XXXXXX")abs("alt")=sqrt((6-24/5)^2+(7-23/5)^2)
color(white)("XXXXXXXX")~~2.68 (using calculator/spreadsheet)
