Find an equation of the line containing the given pair of points

(1/4,-1/2) and (3/4,6)What is the equation of the line y=? Any and all help will be appreciated. Thanks

Use the points given to find the slope.

slope = (y1-y2)/(x1-x2)
Then the straight line formula is
y = mx + b. Plug in one set of points, along with m you just determined, and solve for b.

Im still not understanding

To find the equation of the line containing the given pair of points, you can use the point-slope form of a line given by:

y - y₁ = m(x - x₁)

where (x₁, y₁) are the coordinates of one point on the line, and 'm' is the slope of the line.

Step 1: Find the slope (m) of the line using the formula:

m = (y₂ - y₁) / (x₂ - x₁)

Given the points (1/4, -1/2) and (3/4, 6), we have:
x₁ = 1/4, y₁ = -1/2
x₂ = 3/4, y₂ = 6

Substituting the values into the formula:

m = (6 - (-1/2)) / (3/4 - 1/4)
m = (6 + 1/2) / (3/4 - 1/4)
m = (13/2) / (2/4)
m = (13/2) / (1/2)
m = (13/2) * (2/1)
m = 13

Step 2: Choose one of the given points (let's use (1/4, -1/2)), and substitute the coordinates and the slope into the point-slope form:

y - y₁ = m(x - x₁)
y - (-1/2) = 13(x - 1/4)

Simplify:

y + 1/2 = 13(x - 1/4)

Step 3: Rearrange the equation to solve for y:

y = 13(x - 1/4) - 1/2
y = 13x - 13/4 - 1/2
y = 13x - 13/4 - 2/4
y = 13x - 15/4

Therefore, the equation of the line is y = 13x - 15/4.