Question

A straight line passes through points (-2,1) and(6,3) find the equation of the line in the form y=mx+c where m and c are constants.

Answer 1

y =
`(1)/(4)` x +
`(3)/(2)`

Explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

calculate m using the slope formula

m =
`(y_(2)-y_(1) )/(x_(2)-x_(1) )`

with (x₁, y₁ ) = (- 2, 1 ) and (x₂, y₂ ) = (6, 3 )

substitute these values into the formula for m

m =
`(3-1)/(6-(-2))` =
`(2)/(6+2)` =
`(2)/(8)` =
`(1)/(4)` , then

y =
`(1)/(4)` x + c ← is the partial equation

to find c , substitute either of the 2 points into the partial equation

using (- 2, 1 )

1 =
`(1)/(4)` (- 2) + c = -
`(1)/(2)` + c ( add
`(1)/(2)` to both sides )

1 +
`(1)/(2)` = c , then c =
`(3)/(2)`

y =
`(1)/(4)` x +
`(3)/(2)` ← equation of line