three lines will form a triangle if they are not parallel and all three lines do not intersect at same point. they are not parallel if they have different slopes. So we will find slopes.
It is quick to say first equation has slope of 4/3 by looking at the coefficients of x and y.
second is straight line and had slope 0
third would have slope -2
they have different slopes so let's check intersections. these will also be vertices if they form triangle.
1st & 2nd equations will intersect when they are equal. So so that means thwit x and y values will be the same. If if we know their x and y values are the same we can substitute y = 3 of second equation for y in first.
4x - 3 (3) = -5
4x -9 = -5
4x = 4
x = 1
So intersect at (1,3)
second and third we can do the same thing for their intersection.
2x + (3) = 15
2x = 12
x = 6
So (6,3)
Now now to find 1st & 3rd intersection. I will find what y equals in third and sub into first.
2x + y = 15
y = -2x + 15
sub in first now
4x - 3 (-2x + 15) = -5
4x + 6x -45 = -5
10x = 40
x = 4
we have x value. we can sub in and find y. I will use 3rd equation:
2 (4) + y = 15
8 + y = 15
y = 7
So so last one is (4,7)
all intereactions are different so it will be a triangle with those vertices.
So (1,3) (6,3) (4,7)
we could find each side length to see what kind of triangle. i will use Pythagorean theorem to find side length. they hypotenuse will be the actual side and the a and b values will. be the change in x and y values from the vertices in the y = 3 line. we known the length of the y = 3 side because it runs parallel to x axis and therefore shares it's values. so length of that line is the change in x which is :
6 - 1 = 5
so first side:
( 4 - 1)^2 + (7 - 3)^2 = c^2
9 + 16 = c^2
c = 5.
So we have 2 sides with length 5. It is at least isosceles. however lets check last one cuz maybe equilateral.
(6 - 4)^2 + (7 - 3)^2 = c^2
4 + 16 = c^2
c = sqrt (20).
It is isosceles since only 2 sides equal
