Make up equations for three lines whose intersection points form a triangle with vertices at (-4,0), (2,9) and (6,5)

Answer 1

To create three lines whose intersection points form a triangle with vertices at (-4,0), (2,9), and (6,5), we can start by defining the equations of two lines that intersect at the point (-4,0). Let's call these lines L1 and L2.

For L1, we will use the point-slope form:

L1: y - y1 = m1(x - x1)

Using the point (-4,0), we have:

L1: y - 0 = m1(x - (-4))
L1: y = m1(x + 4)

For L2, we will also use the point-slope form:

L2: y - y2 = m2(x - x2)

Using the point (-4,0), we have:

L2: y - 0 = m2(x - (-4))
L2: y = m2(x + 4)

Now, we need to find the equation of the third line, L3, that passes through the points (2,9) and (6,5). We can use the point-slope form again:

L3: y - y3 = m3(x - x3)

Using the points (2,9) and (6,5), we can find the slope m3:

m3 = (y3 - y2) / (x3 - x2)
= (5 - 9) / (6 - 2)
= -4 / 4
= -1

Now, substituting the point (6,5) into the equation of L3:

L3: y - 5 = -1(x - 6)
L3: y - 5 = -x + 6
L3: y = -x + 11

The equations for the three lines are:

L1: y = m1(x + 4)
L2: y = m2(x + 4)
L3: y = -x + 11