Activity 14: Altitudes, Medians, Angle & Perp. Bisectors

What if I gave you the slope & a random point. What is the equation of the line?

m = 5/4 through (-4, -3)

• y = 5/4x +2

• y = -5/4x + 2

• y = 2x - 5/4

• y = 3/4x + 2

Write an equation 
m = 1/3 through (4, 4)

• y = -1/3x + 8/3
• y = 8/3x - 1/3
• y = 8/3x + 1/3
• y = 1/3x + 8/3

Great...the fractions tomorrow are going to be wild. Sometimes it is nice to GET RID OF FRACTIONS when you can. Solve the following system.

• (-4, 2)

• (4, 2)

• (2, -4)

• (2, 4)

Let's talk altitudes. What is the defining feature of any altitude?

• 90 degree angle

• Midpoint

• 90 degree angle & midpoint

• Angle Bisectors

The vertices of a triangle are A(4,3) B(0, 5) C(3, -6). Find the slope of each ALTITUDE.

• ⊥AB: 2 ⊥BC: 3/11 ⊥AC: -1/9

• ⊥AB: -1/2 ⊥BC: -11/3 ⊥AC: 9

• ⊥AB: -2 ⊥BC: 3/11 ⊥AC: -1/9

• ⊥AB: -2 ⊥BC: -3/11 ⊥AC:1/9

A(4,3) B(0, 5) C(3, -6) & the slopes are ⊥AB: 2 ⊥BC: 3/11 ⊥AC: -1/9. Find the equation of the ALTITUDE to AB. (Hint: What point will you use?)

• y=2x-12

• y=2x

• y=2x-5

• y=2x+5

A(4,3) B(0, 5) C(3, -6) & the slopes are ⊥AB: 2 ⊥BC: 3/11 ⊥AC: -1/9. Find the equation of the ALTITUDE to BC. (Hint: What point will you use?)

• $y=\frac{3}{11}x+\frac{21}{11}$

• $y=\frac{3}{11}x-\frac{9}{11}$

• $y=\frac{3}{11}x+5$

• $y=\frac{3}{11}x+\frac{57}{11}$

A(4,3) B(0, 5) C(3, -6) & the slopes are ⊥AB: 2 ⊥BC: 3/11 ⊥AC: -1/9. Find the equation of the ALTITUDE to AC. (Hint: What point will you use?)

•  $y=-\frac{1}{9}x+5$ 

•  $y=-\frac{1}{9}x-5$ 

•  $y=-\frac{1}{9}x+\frac{31}{9}$ 

•  $y=-\frac{1}{9}x-\frac{17}{3}$ 

A(4,3) B(0, 5) C(3, -6) 

Using the equations for your ALTITUDES, calculate the ORTHOCENTER. (HINT: Go to desmos calculator to graph your 3 lines to check your answer)

• (8.1, 4.1)

• (19, 26)

• (-33,  -78)

• (9, 6)

What is the defining feature of any perpendicular bisector?

• 90 degree angle

• Midpoint

• 90 degree angle & midpoint

• Angle Bisectors