Triangle in the x-y plane with given vertices

Question

Suppose a triangle in the x-y plane has vertices (-1,0), (1,0) and (0,2), how can I find tthe equations of the three lines that lie along the sides of the triangle in y = mx + b form?

Answer 1

Given: Suppose a triangle in the x-y plane has vertices (-1,0), (1,0) and (0,2), how can I find tthe equations of the three lines that lie along the sides of the triangle in y = mx + b form?

Let these vertices be called A(-1,0), B(1,0), C(0,2). Use the formula for equation of a straight line passing through two given points (x1, y1) and (x2, y2):

y- y1 = [(y2-y1)/(x2-x1)](x-x1)

Equation of line AB: y-0 = [(0-0)/(1+1)](x-+1)

y = 0(x+1)

y=0.

Equation of line BC: y-0 = [(2-0)/(0-1)](x-1)

y = -2(x-1) i.e y = -2x + 2,

Equation of line AC: y - 0 = [(2-0) / (0 + 1)]((x+1)= 2(x+1)

y = 2x + 2

Answer 2

Hi Worrick,

Your Question: Suppose a triangle in the x-y plane has vertices (-1,0), (1,0) and (0,2), how can I find tthe equations of the three lines that lie along the sides of the triangle in y = mx + b form?

Given: vertices (-1,0), (1,0) and (0,2)

Find: Equation of lines that would go along the sides of the triangle in y = mx + b form

To find the equation of the line that would go through the vertices of the triangle, you can 1st fine the slope of the 2 points, and then use the point slope equation to determine the equation of the line.

m = (y2-y1) / (x2-x1) This is the slope when given two points

(y-y1) = m (x-x1) This is the point slope form of the equation

1) Ill use (-1,0) and (1, 0)

m= (0-0) / (1- (-1)) = 0

(y-0) = 0 (x- (-1))

y = 0 Is the equation of a line that passes through (-1,0) and (1,0)

2) (-1,0) (0,2)

m = (2-0) / (0- (-1)) = 2

(y-0) = 2 (x- (-1))

y = 2x +2

3) (1,0) and (0,2)

m = (2-0) / (0 - 1) = -2

(y-0) = -2 (x-1)

y = -2x +2

Hope this helps.