Question

find the equation of the line in slope-intercept form that passes through point (5,11) and slope -1/2

Answer 1

Slope-intercept form: y = mx + b [m is the slope, b is the y-intercept or the y value when x = 0 --> (0, y) or the point where the line crosses through the y-axis]

Since you know m = -1/2, plug it into the equation

y = mx + b

`y=-(1)/(2)x+b` To find b, plug in the point (5, 11) into the equation

`11=-(1)/(2) (5)+b`

`11=-(5)/(2) +b` Add 5/2 on both sides

`11+(5)/(2) =b` Make the denominators the same to combine fractions(when add/subtracting)

`(22)/(2)+(5)/(2) =b`

`(27)/(2) =b`

`y=-(1)/(2)x+(27)/(2)`

Answer 2

Slope intercept form
y = mx + b
P1 (5,11)
M = -1/2.

Y - y1 = m(X - X1)
Y - 11 = -1/2(X - 5)
Y - 11 = -1/2X + 5/2
Y - 11 + 11 = -1/2X + 5/2 + 11

Y = -1/2X + 5/2 + 11
Y = -1/2X + 5/2 + 22/2
Y = -1/2X + 27/2

Y = -1/2X + 27/2. I believe this is the equation of the line in slope intercept form. I basically used slope point formula y-y1 = m(X-x1), to find the new equation of the line, inputting the value of the slope along with the point the line would go through.