Finding the area of a triangle is straightforward if you know the length of the base and the height of the triangle. But is it possible to find the area of a triangle if you know only the coordinates of its vertices? In this task, you’ll find out. Consider ΔABC, whose vertices are A(2, 1), B(3, 3), and C(1, 6); let line segment AC represent the base of the triangle.
Part A
Find the equation of the line passing through B and perpendicular to AC .

Question

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Respuesta :

profarouk profarouk · Answer 1 · 2022-05-26T12:04:23+02:00

Answer:

[tex]y : \frac{1}{5} x + \frac{12}{5}[/tex]

Step-by-step explanation:

the slope of the line AC :

[tex]\frac{6-1}{1-2} = \frac{5}{-1} =-5[/tex]

Let y : mx + p be the equation of the line D passing through B and perpendicular to AC.

Where m is the slope and p is the y value of the y intercept point.

The line D is perpendicular to AC means the product of their slopes is −1

Therefore

(−5) × m = −1

Then

[tex]m=\frac{-1}{-5} =\frac{1}{5}[/tex]

We obtain, the equation of D is :

[tex]y : \frac{1}{5} x + p[/tex]

At this point ,we still need the value of p to get the full equation of D.

B(3, 3)∈ D

[tex]3=\frac{1}{5} \left( 3\right) +p[/tex]

[tex]\Longleftrightarrow p=3-\frac{3}{5} =\frac{12}{5} =2.4[/tex]

thus , the equation of D is :

[tex]y : \frac{1}{5} x + \frac{12}{5}[/tex]

Аноним Аноним · Answer 2 · 2022-05-27T07:46:15+02:00

Yes it's possible through matrix

Let's find out the area formula for our vertices

That's

[tex]\\ \rm\Rrightarrow \left|\begin{array}{ccc}\rm 2&\rm 1&\rm 1\\ \rm 3&\rm 3&\rm 1\\ \rm 1&\rm 6&\rm 1\end{array}\right|[/tex]

The determinant values modulus is the area

Slope of AC

Perpendicular lines have slopes negative reciprocal to each other

So

Equation in point slope