Area of triangle with following verticies (0,-9), (3,-9), and (0,-2)

Mathematics

2 answers:

Neporo4naja [7]3 years ago

8 0

Answer:

Area = 21/2 = 10.5 area units

Step-by-step explanation:

Two of the points are on the same line (the y-axis) as they both have x-coordinates of 0. This can thus be the base of your triangle.

Two more are on the same line (y= -9) as they both have the same y-coordinate of -9. This can be the height of your triangle.

So you have a base which is between -2 and -9 (7 units) and a height which is between 0 and 3 (3 units) so now just use the equation for area of a triangle $A=0.5*b*h$

mylen [45]3 years ago

4 0

Answer:

Area = 21/2 = 10.5 area units

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7Ba%5E%7B2%7D-1%7D%7B2-5a%7D%20times%20%5Cfrac%7B15a-6%7D%7Ba%5E%7B2%7D%2B5a-6%7D" id=

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$\bf \cfrac{a^2-1}{2-5a}\times \cfrac{15-6}{a^2+5a-6}\\\\
-----------------------------\\\\
recall\quad \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
thus\quad a^2-1\iff a^2-1^2\implies (a-1)(a+1)
\\\\\\
now\quad a^2+5a-6\implies (a+6)(a-1)\\\\
-----------------------------\\\\
thus
\\\\\\
\cfrac{a^2-1}{2-5a}\times \cfrac{15-6}{a^2+5a-6}\implies \cfrac{(a-1)(a+1)}{2-5a}\times \cfrac{3(5a-2)}{(a+6)(a-1)}\\\\
-----------------------------\\\\
$

$\bf now\quad 3(5a-2) \iff -3(2-5a)\\\\
-----------------------------\\\\
thus
\\\\\\
\cfrac{\underline{(a-1)}(a+1)}{\underline{2-5a}}\times \cfrac{-3\underline{(2-5a)}}{(a+6)\underline{(a-1)}}\implies \cfrac{-3(a+1)}{a+6}$

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