Question Statement
Find the equations of the sides, altitudes, and medians of a triangle with vertices at A(β3,2), B(5,4), and C(3,β8).
Background and Explanation
To solve this, we need to use the following concepts:
- Slope of a Line: The slope between two points (x1β,y1β) and (x2β,y2β) is given by:
m=x2ββx1βy2ββy1ββ
- Midpoint Formula: The midpoint of two points (x1β,y1β) and (x2β,y2β) is:
(2x1β+x2ββ,2y1β+y2ββ)
- Point-Slope Form of a Line: The equation of a line passing through a point (x1β,y1β) with slope m is:
yβy1β=m(xβx1β)
- Perpendicular Line: The slope of a line perpendicular to another with slope m is βm1β.
Solution
Step 1: Finding the Equation of the Sides
1. Side AB
- Slope of AB:
m1β=5β(β3)4β2β=82β=41β
- Using the point-slope form for point A(β3,2):
yβ2=41β(xβ(β3))
Simplifying:
4(yβ2)=x+3β4yβ8=x+3βxβ4y+11=0
So, the equation of side AB is:
xβ4y+11=0
2. Side BC
- Slope of BC:
m2β=3β5β8β4β=β2β12β=6
- Using the point-slope form for point B(5,4):
yβ4=6(xβ5)
Simplifying:
yβ4=6xβ30β6xβyβ26=0
So, the equation of side BC is:
6xβyβ26=0
3. Side AC
- Slope of AC:
m3β=3β(β3)β8β2β=6β10β=3β5β
- Using the point-slope form for point A(β3,2):
yβ2=β35β(xβ3)
Simplifying:
3(yβ2)=β5(x+3)β3yβ6=β5xβ15β5x+3y+9=0
So, the equation of side AC is:
5x+3y+9=0
Step 2: Finding the Equations of the Altitudes
1. Altitude through Vertex A
- The slope of the perpendicular line to AB is β4 (since the slope of AB is 41β).
- Using point A(β3,2) and slope β4:
yβ2=β4(xβ(β3))
Simplifying:
6(yβ2)=β1(x+3)βx+6yβ9=0
So, the equation of the altitude through vertex A is:
x+6yβ9=0
2. Altitude through Vertex B
- The slope of the perpendicular line to BC is β61β (since the slope of BC is 6).
- Using point B(5,4) and slope β61β:
yβ4=β61β(xβ5)
Simplifying:
6(yβ4)=β1(xβ5)β3xβ5y+5=0
So, the equation of the altitude through vertex B is:
3xβ5y+5=0
3. Altitude through Vertex C
- The slope of the perpendicular line to AC is β53β (since the slope of AC is 3β5β).
- Using point C(3,β8) and slope β53β:
yβ(β8)=β53β(xβ3)
Simplifying:
5(y+8)=β3(xβ3)β4x+yβ4=0
So, the equation of the altitude through vertex C is:
4x+yβ4=0
- The midpoint of BC is (1,3).
- The slope of the median through A is the slope between A(β3,2) and (1,3):
m=1β(β3)3β2β=41β
- Using point A(β3,2) and slope 7β4β:
yβ2=β74β(xβ(β3))
Simplifying:
4x+7yβ2=0
- The midpoint of AC is (0,β3).
- The slope of the median through vertex B is
0β5β3β4β=57β
The equation of the median:
yβ4=7/5(xβ5)wherearefoundtoo....