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use the determinant method to find the area of a triangle ABC with vertices at A (1,6) B (4,2) and C (-3,3)​

Sagot :

absor201

Answer:

Area of triangle is 12.5

Step-by-step explanation:

We need to use the determinant method to find the area of a triangle ABC with vertices at A (1,6) B (4,2) and C (-3,3)​

The formula used is: [tex]Area=\pm \frac{1}{2}\left|\begin{array}{ccc}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{array}\right|[/tex]

We are given:

[tex]x_1=1,y_1=6,x_2=4,y_2=2,x_3=-3,y_3=3[/tex]

Putting values and finding area

[tex]Area=\frac{1}{2}\left|\begin{array}{ccc}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{array}\right|\\Area=\frac{1}{2}\left|\begin{array}{ccc}1&6&1\\4&2&1\\-3&3&1\end{array}\right|\\Area=\frac{1}{2}(1\left|\begin{array}{cc}2&1\\3&1\end{array}\right|-6\left|\begin{array}{cc}4&1\\-3&1\end{array}\right|+1\left|\begin{array}{cc}4&2\\-3&3\end{array}\right| )\\Area=\frac{1}{2}(1(2-3)-6(4-(-3))+1(12-(-6)))\\Area= \frac{1}{2}(1(-1)-6(4+3)+1(12+6))\\Area= \frac{1}{2}(1(-1)-6(7)+1(18))\\Area= \frac{1}{2}(-1-42+18)\\[/tex]

[tex]Area=\frac{1}{2}(-25)\\Area=-12.5\\[/tex]

We take mode of -12.5 i.e. |-12.5| because area can't be negative.

[tex]Now, \\Area=|-12.5|\\Area=12.5[/tex]

So, Area of triangle is 12.5

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