Given triangle ABC with vertices A(−3, 0), B (0, 6), and C(4, 6). Find the equations of the three altitudes of the same triangle

Answer 1

Graphing the points and connecting them to form a triangle helps a lot.

You can divide up the areas with more triangles and a square to calculate the area

left side of the y-axis' part of the triangle has area 9/7. right side has 48/7, for a total

57/7 or just over 8.

You get a very flat triangle, longest side is the square root of 36+49 or sqr(85)

A= bh/2

57/7 = sqr85/2 times the height or altitude

h= 114/7sqr85 = about 1.8

But when you do that to the other 2 sides, you seem to get a height that's less than how high the triangle actual goes. So maybe you mean something different by "altitude" than "height"?

If you use the top of the triangle as given, as the base, that's b= 4. Area =bh/2. h=57/28 = about 2

while you can see vertically, the triangle extends a full 6. so maybe you mean 6 as the altitude? which is not the height?

same with using the other side as the base = sqr45

divide 57/7 by sqr45 = 57/7sqr45 = 57(sqr45)/7(45) = about 1.2, yet the the "altitude" is about 3.5

The calculations are tedious and easy to make mistakes

You could also calculate the area of the triangle using Heron's formula

Answer 2

I will not do the arithmetic for you, but I will tell you how to solve this.

There are 3 tasks involved for each altitude..

First

Find the slope of each side of the triangle using the co-ordinates given, e.g.

(6.4) and (0,6) slope is (6-4)/(0-6).

Second

The slope of the line perpendicular to each of the sides is the negative reciprocal of the slope of the side. Remember the altitude will be perpendicular to the side.

Third

The altitude will be perpendicular to the side and though the vertex opposite. Use the point slope form to get the equation of the altitude.

Answer 3

The given question provides us with a theoretical triangle ABC, such that the vertices A,B, and C make it up. We need to find all three altitudes or heights, that is when any given side is considered to the base, what is the height of the triangle?

To do this, we first need to find the lengths of the sides which we can do using the distance formula that we know. Set up the sides AB, BC, and AC

Let AB = c, BC = a, and AC = b

We know the distance between two vertices follows the formula √( (x₂-x₁)² + (y₂-y₁)² )

So, set up each side according to the formula.

a = √( (4-0)² + (6-6)² ) = 4

b = √( (4-(-3))² + (6-0)² ) = √( 7² + 6² ) = √85 Unfortunate to see ugly numbers but it will be okay

c = √( (0-(-3))² + (6-0)² ) = √( 3² + 6² ) = √45 = 3√5

Now for the point of the question. There exists a formula called Heron's Formula that states the following:

A = √( s(s-a)(s-b)(s-c) ) where s is the semi-perimeter of the triangle such that s = (a+b+c)/2

Plugging into the formulae we will get an Area of the triangle which will not have pretty numbers due to the radicals.

Then, we know that A = 1/2 of base * height so we isolate for height giving us h = 2A/base. Then we plug in the Area that we found of the triangle using Heron's Formula and we can then plug in our three different sides for 3 different equations of height that will get us our 3 altitudes required for the question and we are done.