The vertices of a triangle are at A(-3,-2), B(2,1) and C(0,6). What is the equation of the two lines.
Please help me and show the solution... Thanks!
Plot the three points and draw in sidesof triangle.Write the equation ol line AC. Take the negative inverse slope of this line and using the point slope formula write the equation of line BK. Solve the two equations to find the coordinates of K.Draw the BK slope diagram fot BK and determine the lenght of BK using the distance formula.
Can you handle this? If not ask for some extra help.
bjh
The first step is to find the equation of the line that contains AC. Use the point-slope formula
$\displaystyle y - y_0 = m(x - x_0)$
where $\displaystyle (x_0, y_0)$ is point A, $\displaystyle (x_1, y_1)$ is point C, and the slope is
$\displaystyle m = \frac{y_1 - y_0}{x_1 - x_0}$.
Plot the three pointsLabel as given.
Draw a horizontal thru Ameeting the y axis @0,-2.Label pointD. ACD is now the slope diagram for line AC Rise is 8, run is 3,slope is 8/3.Distance between A and C is sq.root 73 (d^2=3^2+8^2)
Draw a perpendicular from B to AC meeting AC@K.Slope of BK is the negative inverse of line AC=to -3/8
Equation of line AC from y=mx+b isy=8/3x+6
Equation of BK Use point slope formula where point is 2,1
y-1/x-2=-3/8
8y-8=3x+6
y=-3/8x+14/8
if a=b and a=c b=c
-3/8x+14/8=8/3x+6
x=-1.4 and y=2.3
using these coordinates the slope for BK is 1.3/3.4 from whence the lenght of Bk =3.64. You can now calculate the area of ABC.This is more than you needed for an answer but i believe you needed it
bjh