Tell whether the orthocenter is inside, on, or outside the triangle. Then find the coordinates of the orthocenter.

L(0,5), M(3,1), N(8,1): Is it outside, inside or on?

The orthocenter is ( , )

Answer:

see explanation

Step-by-step explanation:

The orthocentre is

• Inside all acute triangles

• Outside all obtuse triangles

• On all right triangles

To determine if the triangle is any of the above, we require the lengths of the 3 sides of the triangle.

Calculate the lengths using the distance formula

d = [tex]\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }[/tex]

L(0, 5) and M(3, 1)

LM = [tex]\sqrt{(3-0)^2+(1-5)^2}[/tex] = [tex]\sqrt{3^2+(-4)^2}[/tex] = [tex]\sqrt{9+16}[/tex] = [tex]\sqrt{25}[/tex] = 5

L(0, 5) and (N(8, 1)

LN = [tex]\sqrt{(8-0)^2+(1-5)^2}[/tex] = [tex]\sqrt{8^2+(-4)^2}[/tex] = [tex]\sqrt{64+16}[/tex] = [tex]\sqrt{80}[/tex] = 4[tex]\sqrt{5}[/tex]

M(3, 1) and (N(8, 1)

MN = [tex]\sqrt{(8-3)^2+(1-1)^2}[/tex] = [tex]\sqrt{5^2+0^2}[/tex] = [tex]\sqrt{25}[/tex] = 5

Given 3 sides of a triangle a, b, c where c is the longest side, then

• a² + b² = c² ⇒ right triangle

• a² + b² < c² ⇒ acute triangle

• a² + b² > c² ⇒ obtuse triangle

Here a = 5, b = 5, c = 4[tex]\sqrt{5}[/tex]

a² + b² = 5² + 5² = 25 + 25 = 50

c² = (4[tex]\sqrt{5}[/tex] )² = 80

Since a² + b² < c² then triangle is obtuse and the orthocentre is outside the triangle.

The orthocentre is the point of intersection of the 3 Altitudes of the triangle.

We only require to find the equation of 2 of the altitudes and solve them simultaneously to obtain the coordinates of the orthocentre.

An Altitude is a line from a vertex at right angles to the opposite side

Altitude from M to LN

Calculate the slope of LN using the slope formula

m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex] = [tex]\frac{1-5}{8-0}[/tex] = [tex]\frac{-4}{8}[/tex] = - [tex]\frac{1}{2}[/tex] ⇒ [tex]m_{perpendicular}[/tex] = 2

Equation of altitude

y - 1 = 2(x - 3)

y - 1 = 2x - 6

y = 2x - 5 → (1)

Altitude from N to LM

m = [tex]\frac{1-5}{3-0}[/tex] = - [tex]\frac{4}{3}[/tex] ⇒ [tex]m_{perpendicular }[/tex] = [tex]\frac{3}{4}[/tex]

Equation of altitude

y - 1 = [tex]\frac{3}{4}[/tex](x - 8)

y - 1 = [tex]\frac{3}{4}[/tex] x - 6

y = [tex]\frac{3}{4}[/tex] x - 5 → (2)

Equating (1) and (2)

2x - 5 = [tex]\frac{3}{4}[/tex] x - 5 ( multiply through by 4 )

8x - 20 = 3x - 20

5x - 20 = - 20 ( add 20 to both sides )

5x = 0 ⇒ x = 0

Substitute x = 0 into (1)

y = 0 - 5 = - 5

The orthocentre is (0, - 5 )

Answer:

The orthocenter is outside the triangle, at (0, -5)

Step-by-step explanation:

The orthocenter is the intersection of the altitudes. So let's find the slope-intercept equation of "LM" "MN" and "LN"  then determine it's  perpendicular:

LM: slope = m = 1-5/3-0 = -4/3, equation => y = -4/3x + 5, perpendicular => y = 3/4x + b

this line passes through the opposite point, (8, 1)...therefore we can find b:

y = 3/4x + b

1 = 3/4(8) + b, b = -5, equation => y = 3/4x - 5

______________________________________

MN: slope = m = 1-1/8-5 = 0, equation => y = 1, perpendicular => x = ?

the line passes through the opposite point (0, 5)...so we can find ? or b:

x = ?, equation => x = 0

______________________________________

LN: slope = m = 1-5/8-0  = -4/8 = -1/2, equation => y = -1/2x + 5, perpendicular => y = 2x + b

the line passes through the opposite point (3, 1)...so we can find b:

1 = 2(3) + b, b = 1-6 = -5, equation => y = 2x - 5

______________________________________

System of Equations:

[tex]\begin{bmatrix}y=\frac{3}{4}x-5\\ y=2x-5\\ x=0\end{bmatrix}[/tex]

[tex]\mathrm{Substitute\:}y=\frac{3}{4}x-5: \begin{bmatrix}x=0\\ \frac{3}{4}x-5=2x-5\end{bmatrix}[/tex]

[tex]\mathrm{Substitute\:}x=0: \begin{bmatrix}\frac{3}{4}\cdot \:0-5=2\cdot \:0-5\end{bmatrix}[/tex]

[tex]\mathrm{Simplify\:}: \begin{bmatrix}-5=-5\end{bmatrix}[/tex]

[tex]\mathrm{For\:}y=\frac{3}{4}x-5,\\\mathrm{Substitute\:}x=0,\\y=\frac{3}{4}\cdot \:0-5 = -5[/tex]

[tex]y=-5,\:x=0[/tex]

_____________________________________

The orthocenter is outside the triangle, at (0, -5)

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