Question

A triangle has corners A, B, and C located at `(7 ,3 )`, `(4 ,8 )`, and `(3 , 7 )`, respectively. What are the endpoints and length of the altitude going through corner C?

Answer 1

The end points are `=(71/17,131/17)` and the length is `=1.37`

Explanation:

The corners of the triangle are

`A=(7,3)`

`B=(4,8)`

`C=(3,7)`

The slope of the line `AB` is `m=(8-3)/(4-7)=-5/3`

The equation of line `AB` is

`y-8=-5/3(x-4)`

`y-8=-5/3x+20/3`

`y+5/3x=44/3`...........................`(1)`

`mm'=-1`

The slope of the line perpendicular to `AB` is `m'=3/5`

The equation of the altitude through `C` is

`y-7=3/5(x-3)`

`y-7=3/5x-9/5`

`y=3/5x+26/5`................................`(2)`

Solving for `x` and `y` in equations `(1)` and `(2)`, we get

`-5/3x+44/3=3/5x+26/5`

`3/5x+5/3x=44/3-26/5`

`34/15x=132/15`

`x=142/34=71/17`

`y=3/5*71/17+26/5=655/85=131/17`

The end points of the altitude is `=(71/17,131/17)`

The length of the altitude is

`=sqrt((3-71/17)^2+(7-131/17)^2)`

`=sqrt((-20/17)^2+(-12/17)^2)`

`=sqrt(544)/17`

`=1.37`

Answer 2

End points are `(3,7) and (4.18, 7.71)`
Length of altitude is `1.38` unit

Explanation:

Let `CD` be the altitude going through `C` touches `D` on line

`AB`. `C` and `D` are the endpoints of altitude `CD. CD` is

perpendicular on `AB`. Slope of `AB= m_1= (y_2-y_1)/(x_2-x_1)=(8-3)/(4-7) = -5/3 :.` Slope of `CD=m_2= -1/m_1= 3/5`

Equation of line `AB` is `y - y_1 = m_1(x-x_1) or y- 3 = -5/3(x-7)` or
`3y-9 = -5x + 35 or 5x+3y= 35+9 or 5x+3y = 44 (1)`

Equation of line `CD` is `y - y_3 = m_2(x-x_3) or y- 7 = 3/5(x-3)` or
`5y-35 = 3x -9 or 3x-5y= -35+9 or 3x-5y = -26 (2)`

Sollving equation (1) and (2) we get the co-ordinates of D.

`15x+9y = 132 (3)`
`15x - 25y = -130 (4)` Subtracting we get,
`34y=262 or y =131/17=7.71 ; x = (44-3*131/17)/5= (44-393/17)/5 =(748-393)/85 =355/85=4.18`
So end points are `(3,7) and (4.18, 7.71)`
Length= `sqrt((3-4.18)^2+ (7-7.71)^2)) 1.38` unit [Ans]