The vertices of a triangle are A(-2, 1), B(2, 2) and C(6, - 2). Find the equation of the altitude of AD drawn from A to BC.

The equation of altitude AD drawn from A to BC will be:

Step-by-step explanation:

Let m₁ and m₂ be the slope of line AD and BC respectively.

Now, AD⊥BC

∴ m₁ × m₂ = -1

be the slope of line AD and BC respectively.

⇒ m₁ = -1/m₂    → (A)

Finding the slope of BC using points

B(2, 2)

C(6, - 2)

On substituting the value of m₂ in equation (A)

m₁ = -1/m₂

    = -1/(-1)

    = 1

We know that the point-slope form of the line equation is

∴ Equation of altitude AD passing through A(-2, 1) with slope 1 will be

Therefore, the equation of altitude AD drawn from A to BC will be:

g(x) is the negative form of f(x)? because whenever its the negative form it reflects its self over the x axis. idk if this is right though.

step-by-step explanation:

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.