Question

A triangle has corners A, B, and C located at `(4 ,2 )`, `(1 ,3 )`, and `(6 ,5 )`, respectively. What are the endpoints and length of the altitude going through corner C?

Answer 1

`color(blue)("Endpoints are " (6,5), (49/10,17/10)`

`color(blue)("Length of altitude line"=(11sqrt(10))/10" units"`

Explanation:

From the diagram:

We need to find the endpoints of the altitude passing through `C`. We can see that the altitude through this vertex is perpendicular the the line `BA` produced at `P`. We need to find the equation of the line passing through points `A` and `B` first.

Using point `A` and `B`, the gradient of this line is:

`(2-3)/(4-1)=-1/3`

Using point slope form of a line and point `(4,2)`

`y-2=-1/3(x-4)`

`y=-1/3x+10/3 \ \ \ [1]`

We now need the equation of the line through point `(6,5)` and perpendicular to `[1]`

If two line are perpendicular the the product of their gradients is `bb-1`

Let the unknown gradient be `m`, then:

`m*-1/3=-1=>m=3`

Using this and point `(6,5)`

`y-5=3(x-6)`

`y=3x-13 \ \ \ \[2]`

We solve now solve `[1]` and `[2]` simultaneously to find `D(x,y)`

`-1/3x+10/3=3x-13=>x=49/10`

Substitute in `[2]`

`y=3(49/10)-13=17/10`

So the end points of the altitude are:

`C=(6,5) and D=(49/10,17/10)`

The length of the altitude line `DC` can be found using the distance formula:

`|d|=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`|DC|=sqrt((6-49/10)^2+(5-17/10)^2)`

`\ \ \ \ \ \ \ \ \ =sqrt((11/10)^2+(33/10)^2)=color(blue)((11sqrt(10))/10 " units"`