Question

What is the equation of the line connecting (4,8) and (7,-2)?

Answer 1

For this case we have that by definition, the equation of a line of the slope-intersection form is given by:

`y = mx + b`

Where:

m: It's the slope

b: It is the cut-off point with the y axis

We have two points through which the line passes, so we can find the slope:

`(x_ {1}, y_ {1}) :( 4,8)\\(x_ {2}, y_ {2}) :( 7, -2)`

`m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {-2-8} {7-4} = \frac {-10} {3} = - \frac {10} {3}`

Thus, the equation is of the form:

`y = - \frac {10} {3} x + b`

We substitute one of the points and find "b":

`8 = - \frac {10} {3} (4) + b\\8 = - \frac {40} {3} + b\\8+ \frac {40} {3} = b\\\frac {3 * 8 + 40} {3} = b\\\frac {24 + 40} {3} = b\\b = \frac {64} {3}`

Finally, the equation is of the form:

`y = - \frac {10} {3} x + \frac {64} {3}`

Answer:

`y = - \frac {10} {3} x + \frac {64} {3}`