Answer:
The equation of altitude AD drawn from A to BC will be:
[tex]y=x+3[/tex]
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)
[tex]\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}[/tex]
[tex]\left(x_1,\:y_1\right)=\left(2,\:2\right),\:\left(x_2,\:y_2\right)=\left(6,\:-2\right)[/tex]
[tex]m_{2} =\frac{-2-2}{6-2}[/tex]
[tex]m_{2} =-1[/tex]
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
[tex]y-y_1=m\left(x-x_1\right)[/tex]
∴ Equation of altitude AD passing through A(-2, 1) with slope 1 will be
[tex]y-1=1\cdot \left(x-\left(-2\right)\right)[/tex]
[tex]y-1=x+2[/tex]
[tex]y=x+3[/tex]
Therefore, the equation of altitude AD drawn from A to BC will be:
[tex]y=x+3[/tex]
