Question

A triangle has corners A, B, and C located at `(2 ,3 )`, `(8 ,1 )`, and `(4 , 2 )`, respectively. What are the endpoints and length of the altitude going through corner C?

Answer 1

Please read the explanation.

Explanation:

A triangle ABC with the three vertices of the triangle ABC:

`A(2,3), B(8,1) and color(red)(C(4,2)`

An altitude through the vertex `color(red)(C(4,2)` is on the side AB.

We can find the magnitude of the altitude CD using the distance between the points formula:

Distance`color(blue)(=sqrt((x_2-x_1)^2+(y_2-y_1)^2" "`

We must find the coordinates of the point at `color(red)(D`.

This is the point of intersection of the altitude CD and `bar (AB)`.

Find two important equations to solve this problem:

Equation-1 is the equation of the (side) segment AB.

Equation-2 is the equation of the altitude CD.

For Equation-1, we can use the formula `color(red)(y = mx+b`

Use the points `A(2,3), B(8,1)`

Slope `color(purple)(m=(y_2-y_1)/(x_2-x_1)`

`rArr m=(1-3)/(8-2)=-2/6=-1/3`

Work on the Equation-1 next.

`y=-1/3x+b rArr y = -8/3+b`

`1=-8/3+b`

Add `8/3` to both sides.

`1+8/3=-cancel(8/3)+b+cancel(8/3)`

`b=8/3+1 rArr b=8/3+3/3 rArr b = 11/3`

Hence,

`color(blue)(y=-1/3x+11/3" "`Equation 1.

Work on Equation 2 to obtain the equation of the altitude CD.

To find the slope of the Altitude CD: Altitude CD is perpendicular at point C.. `CD` is perpendicular to `AB`. Slopes of perpendicular lines are negative reciprocals of one another.

Slope of altitude CD is: `color(red)(-(-3/1)=3`

Slope-intercept form `y=mx+b`

Find `b` using the point `C(4,2)`

`2=3(4)+b rArr 2=12+b rArr b=-10`

`color(blue)(y=3x-10" "`Equation 2.

Solve the the system of linear equations:

`color(blue)(y=-1/3x+11/3" "`Equation 1.

`color(blue)(y=3x-10" "`Equation 2.

We obtain `x=41/10=4.1 and y=23/10=2.3`

Hence, `D(4.1, 2.3)`

`bar (CD) = sqrt((4.1-4)^2+(2.3-2)^2)=sqrt(0.0.01+0.09`

`rArr sqrt(0.1)~~0.316228`

`bar (CD) ~~ "0.32 units."`