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Ksivusya [100]
3 years ago
12

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

Mathematics
1 answer:
Sophie [7]3 years ago
6 0

Answer:

2\sqrt {2}

Step-by-step explanation:

<u>Distance From a Point to a Line</u>

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}}

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