How do you write the equation given (0,0); (-2,7)?

First, we must determine the slope of the line running through the two points given in the problem. The slope can be found by using the formula: #m = (color(red)(y_2) - color(blue)(y_1))/(color(red)(x_2) - color(blue)(x_1))#

Where #m# is the slope and (#color(blue)(x_1, y_1)#) and (#color(red)(x_2, y_2)#) are the two points on the line.

Substituting the values from the points in the problem gives:

#m = (color(red)(7) - color(blue)(0))/(color(red)(-2) - color(blue)(0)) = -7/2#

Now, we can use the point-slope formula to write an equation for the line. The point-slope formula states: #(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))#

Where #color(blue)(m)# is the slope and #(color(red)(x_1, y_1))# is a point the line passes through.

Substituting the slope we calculated and the values from the first point in the problem gives:

#(y - color(red)(0)) = color(blue)(-7/2)(x - color(red)(0))#

#y = color(blue)(-7/2)x#

We can also substitute the slope we calculated and the values from the second point in the problem giving:

#(y - color(red)(7)) = color(blue)(-7/2)(x - color(red)(-2))#

#(y - color(red)(7)) = color(blue)(-7/2)(x + color(red)(2))#

We can also take the first equation and write this in slope-intercept form. The slope-intercept form of a linear equation is: #y = color(red)(m)x + color(blue)(b)#

Where #color(red)(m)# is the slope and #color(blue)(b)# is the y-intercept value.

#y = color(red)(-7/2)x + color(blue)(0)#

The find the equation of a line or gradient, first find the slope via the formula: #m=(y_2-y_1)/(x_2-x_1)#

If we let #(0,0)->(color(red)(x_1),color(blue)(y_1))# and #(-2,7)->color(red)(x_2),color(blue)(y_2))# then,

#m=color(blue)(7-0)/color(red)(-2-0)=-7/2larr# This is the slope of the line/gradient

Now that we have found the slope we can find the equation via the point-slope formula:

#y-y_1=m(x-x_1)#;

by substituting #-7/2# for #m# and any of the two coordinates given. I will use #(0,0)# be make things easier. We let #(0,0)->(x_1,y_1)#

Thus,

#y-0=-7/2(x-0)#

If we simplify this, we simply get

#y=-7/2xlarr# This is our equation

Graph:

graph{-7/2x [-10, 10, -5, 5]}