# find the altitiude of AD BE and CF in traingle ABC whose vertices are A(-3,4) B(3,6) C(2,9)..also show that AD BE and CF are concurrent?u may change the verices value..jus need to kno the method

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### 2 Answers

You need to evaluate the area of triangle `Delta ABC` using the following formula, such that:

`A = (1/2)*|[(-3,4,1),(3,6,1),(2,9,1)]|`

`A = (1/2)*|(-18 + 27 + 8 - 12 + 27 - 12)| => A = 10`

You also may evaluate the area of triangle using the following formula, such that:

`A = (b*h)/2`

b represents the length of one base of triangle

h represents the height of triangle

`A = (BC*AD)/2 = (AC*BE)/2 = (AB*CF)/2`

You need to evaluate the lengths BC,AC,AB, using distance formula such that:

`AB = sqrt((x_B - x_A)^2 + (y_B - y_A)^2)`

`AB =sqrt((3+3)^2 + (6-4)^2) => AB =sqrt40 = 2sqrt10`

`BC = sqrt((x_C - x_B)^2 + (y_C - y_B)^2)`

`BC = sqrt((2-3)^2 + (9-6)^2) =>BC = sqrt10`

`AC = sqrt((x_C - x_A)^2 + (y_C - y_A)^2)`

`AC = sqrt((2+3)^2 + (9-4)^2) => AC = sqrt50 = 5sqrt2`

You need to evaluate AD such that:

`A = (BC*AD)/2 => 20 = sqrt10*AD => AD = 20/sqrt10 = 2sqrt10`

You need to evaluate BE such that:

`A = (AC*BE)/2 => 20 = 5sqrt2*BE => BE = 2sqrt2`

You need to evaluate CF such that:

`A = (AB*CF)/2 => 20 = 2sqrt10*CF => CF = 10/sqrt10 = sqrt10`

**Hence, evaluating the lengths of the heights of triangle `Delta ABC` , yields `AD = 2sqrt10, BE = 2sqrt2, CF = sqrt10.` **

plz post fig also..andi how the area of traingle ABC is equal to that of A=bc.ad /2 ..xplain plz.or if any other method...