A(-11,8)B(2,-6) and C(-19,-8) are the vertices of Triangle ABC. N(X,Y) is a point on AC such the BN is Perpendicular to AC. Find Area of Triangle ABC. PLEASE HELP!!!

The second sentence is not needed.

Draw the triangle. Compute the side lengths

s1=AB= sqrt ((bx-ax)^2+(by-ay)^2 )
= sqrt ( 13^2 + (-14)^2 )= sqrt (169+256)=20.6

and do that for all sides. Then knowing the sides, calculate the perimeter, divide by 2. Let that be p.

Area=sqrt (p(p-a)(p-b)(p-c))

Or you can do it here
http://www.analyzemath.com/Geometry_calculators/perimeter_area_tri_verti.html

To find the area of triangle ABC, we can use the formula for the area of a triangle:

Area = 1/2 * base * height

In this case, the base of the triangle is AC, and the height is the length of the perpendicular line BN.

To calculate the length of AC, we need to find the distance between points A and C. We can use the distance formula:

Distance AC = √((x2 - x1)^2 + (y2 - y1)^2)

Substituting the coordinates of A(-11,8) and C(-19,-8) into the formula:

Distance AC = √((-19 - (-11))^2 + (-8 - 8)^2)
= √((-19 + 11)^2 + (-16)^2)
= √((-8)^2 + (-16)^2)
= √(64 + 256)
= √320
= 4√20

Now, to find the length of BN, we can use the fact that BN is perpendicular to AC, so the slope of BN will be the negative reciprocal of the slope of AC.

The slope of AC can be calculated using the formula:

Slope AC = (y2 - y1) / (x2 - x1)

Substituting the coordinates of A(-11,8) and C(-19,-8) into the formula:

Slope AC = (-8 - 8) / (-19 - (-11))
= (-16) / (-8)
= 2

The slope of BN will be the negative reciprocal of 2, which is -1/2.

Now we can use the coordinates of B(2,-6) and the slope of BN to find the equation of line BN:

y - y1 = m(x - x1)

Substituting the coordinates of B(2,-6) and the slope -1/2 into the equation:

y - (-6) = -1/2(x - 2)
y + 6 = -1/2(x - 2)
y + 6 = -1/2x + 1
y = -1/2x - 5

The equation of line BN is y = -1/2x - 5.

Next, we need to find the coordinates of the point N where line BN intersects line AC.

To find the intersection point, we can solve the system of equations formed by the equations of lines AC and BN:

y = -1/2x - 5 (from equation of BN)
y = mx + b (from equation of AC)

Substituting the coordinates of A(-11,8) into the equation of AC, we have:

8 = m(-11) + b

Substituting the coordinates of C(-19,-8) into the equation of AC, we have:

-8 = m(-19) + b

Now we have a system of equations:

8 = m(-11) + b
-8 = m(-19) + b

Solving this system of equations will give us the coordinates of point N.

Once we have the coordinates of point N, we can calculate the length of BN using the distance formula:

Distance BN = √((x2 - x1)^2 + (y2 - y1)^2)

With the length of AC (4√20) and the length of BN, we can calculate the area of triangle ABC using the formula:

Area = 1/2 * base * height

Plugging in the values, we can find the area of triangle ABC.