Question

Find the altitude of triangle whose vertex is a(1,-2) and the equation of the base is X+y=3.

Answer 1

Answer:

$2\sqrt {2}$

Step-by-step explanation:

Distance From a Point to a Line

Given a line with an equation :

ax + by + c = 0

And the point $(x_0,y_0)$

The distance from the line to the point is given by

$\displaystyle d=\frac {|ax_{0}+by_{0}+c|}{\sqrt {a^{2}+b^{2}}}$

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:

$\displaystyle d=\frac {|1*1+1*(-2)-3|}{\sqrt {1^{2}+1^{2}}}$

$\displaystyle d=\frac {|1-2-3|}{\sqrt {2}}$

$\displaystyle d=\frac {4}{\sqrt {2}}$

Rationalizing:

$\displaystyle d=\frac {4}{\sqrt {2}}\cdot\frac{\sqrt {2}}{\sqrt {2}}$

$\displaystyle d=2\sqrt {2}$

The altitude of the triangle is $\mathbf{4\sqrt {2}}$