Question

What is the equation of the line with slope `m= 19/25` that passes through `(16/5 73/10)`?

Answer 1

`y-73/10=19/25(x-16/5)larr` Point-slope form

`y=19/25x+1217/250larr` y=mx+b form

`-19/25x+y=1217/250larr` Standard form

Explanation:

Seeing how we already have the slope and a coordinate, we can find the equation of the line by using the point-slope formula: `y-y_1=m(x-x_1)` where `m` is the slope `(m=19/25)` and `(x_1, y_1)` is a point on the line. Thus, `(16/5,73/10)-> (x_1,y_1)`.

The equation is then...

`y-73/10=19/25(x-16/5)`

...in point slope form.

Since you did not specify in what form the equation should be expressed, the above is an acceptable answer but we could also rewrite the equation is `y=mx+b` form. To do this, we solve for `y`.

`y-73/10=19/25x-304/125`

`ycancel(-73/10+73/10)=19/25x-304/125+73/10`

`y=19/25x-[304/125(2/2)]+[73/10(25/25)]`

`y=19/25x-608/250+1825/250`

`y=19/25x+1217/250larr` The equation in y=mx+b form

Alternatively, the equation could also be expressed in standard form: `Ax+By=C`

`-19/25x+y=cancel(19/25x-19/25x)+1217/250`

`-19/25x+y=1217/250larr` The equation is standard form