Question

What is the orthocenter of a triangle with corners at `(4 ,1 )`, `(7 ,4 )`, and (3 ,6 )#?

Answer 1

The trick to this little problem is to find the slope between two points from there find the slope of perpendicular line which simply given by:
1) `m_(perp) = -1/m_("original")` then
2) find the equation of line that passes through the angle opposite the original line for you case give: A(4,1), B(7, 4) and C(3,6)
step1:
Find the slope of `bar(AB) => m_(bar(AB))`
`m_(bar(AB)) = (4-1)/(7-4) = 3 :. m_(perp) = m_(bar(CD)) = -1/1 = -1`
To get the equation of line write:
`y = m_bar(CD)x + b_bar(CD);`use point C(3, 6) to determine `barB`
`6 = -3 + b_bar(CD); b_bar(CD) = 9 :.`
`y_bar(CD) = color(red)(-x + 9)` `color(red)" Eq. (1)"`

step2
Find the slope of `bar(CB) => m_(bar(CB))`
`m_(bar(AB)) = (6-4)/(3-7) = -1/2 :. m_(perp) = m_(bar(AE)) = 2`
To get the equation of line write:
`y = m_bar(AE)x + b_bar(AE);`use point A(4, 1) to determine `barB`
`1 = 8 + b_bar(AE); b_bar(CD) = -7 :.`
`y_bar(AE) = color(blue)(2x - 7)` `color(blue)" Eq. (2)"`
Now equate `color(red)" Eq. (1)"` = `color(blue)" Eq. (2)"`
Solve for => `x = 16/3`
Insert `x=2/3` into `color(red)" Eq. (1)"`
`y = -2/3 + 9 = 11/3`

Answer 2

The trick to this little problem is to find the slope between two points from there find the slope of perpendicular line which simply given by:
1) `m_(perp) = -1/m_("original")` then
2) find the equation of line that passes through the angle opposite the original line for you case give: A(4,1), B(7, 4) and C(3,6)
step1:
Find the slope of `bar(AB) => m_(bar(AB))`
`m_(bar(AB)) = (4-1)/(7-4) = 3 :. m_(perp) = m_(bar(CD)) = -1/1 = -1`
To get the equation of line write:
`y = m_bar(CD)x + b_bar(CD);`use point C(3, 6) to determine `barB`
`6 = -3 + b_bar(CD); b_bar(CD) = 9 :.`
`y_bar(CD) = color(red)(-x + 9)` `color(red)" Eq. (1)"`
step2
Find the slope of `bar(CB) => m_(bar(CB))`
`m_(bar(AB)) = (6-4)/(3-7) = -1/2 :. m_(perp) = m_(bar(AE)) = 2`
To get the equation of line write:
`y = m_bar(AE)x + b_bar(AE);`use point A(4, 1) to determine `barB`
`1 = 8 + b_bar(AE); b_bar(CD) = -7 :.`
`y_bar(AE) = color(blue)(2x - 7)` `color(blue)" Eq. (2)"`
Now equate `color(red)" Eq. (1)"` = `color(blue)" Eq. (2)"`
Solve for => `x = 16/3`
Insert `x=2/3` into `color(red)" Eq. (1)"`
`y = -2/3 + 9 = 11/3`

Answer 3

Orthocenter(16/2, 11/3)

Explanation:

The trick to this little problem is to find the slope between two points from there find the slope of perpendicular line which simply given by:
1) `m_(perp) = -1/m_("original")` then
2) find the equation of line that passes through the angle opposite the original line for you case give: A(4,1), B(7, 4) and C(3,6)
step1:
Find the slope of `bar(AB) => m_(bar(AB))`
`m_(bar(AB)) = (4-1)/(7-4) = 3 :. m_(perp) = m_(bar(CD)) = -1/1 = -1`
To get the equation of line write:
`y = m_bar(CD)x + b_bar(CD);`use point C(3, 6) to determine `barB`
`6 = -3 + b_bar(CD); b_bar(CD) = 9 :.`
`y_bar(CD) = color(red)(-x + 9)` `color(red)" Eq. (1)"`
step2
Find the slope of `bar(CB) => m_(bar(CB))`
`m_(bar(AB)) = (6-4)/(3-7) = -1/2 :. m_(perp) = m_(bar(AE)) = 2`
To get the equation of line write:
`y = m_bar(AE)x + b_bar(AE);`use point A(4, 1) to determine `barB`
`1 = 8 + b_bar(AE); b_bar(CD) = -7 :.`
`y_bar(AE) = color(blue)(2x - 7)` `color(blue)" Eq. (2)"`
Now equate `color(red)" Eq. (1)"` = `color(blue)" Eq. (2)"`
Solve for => `x = 16/3`
Insert `x=2/3` into `color(red)" Eq. (1)"`
`y = -2/3 + 9 = 11/3`