Answer:
[tex]2\sqrt {2}[/tex]
Step-by-step explanation:
Distance From a Point to a Line
Given a line with an equation :
ax + by + c = 0
And the point [tex](x_0,y_0)[/tex]
The distance from the line to the point is given by
[tex]\displaystyle d=\frac {|ax_{0}+by_{0}+c|}{\sqrt {a^{2}+b^{2}}}[/tex]
The triangle has a vertex at (1,-2) and the base lies on the equation
x + y = 3
Rearranging:
x + y - 3 = 0
The altitude of the triangle is the distance from the point to the line.
The values to use in the formula of the distance are: a=1, b=1, c=-3, xo=1, yo=-2:
[tex]\displaystyle d=\frac {|1*1+1*(-2)-3|}{\sqrt {1^{2}+1^{2}}}[/tex]
[tex]\displaystyle d=\frac {|1-2-3|}{\sqrt {2}}[/tex]
[tex]\displaystyle d=\frac {4}{\sqrt {2}}[/tex]
Rationalizing:
[tex]\displaystyle d=\frac {4}{\sqrt {2}}\cdot\frac{\sqrt {2}}{\sqrt {2}}[/tex]
[tex]\displaystyle d=2\sqrt {2}[/tex]
The altitude of the triangle is [tex]\mathbf{4\sqrt {2}}[/tex]
