Question #4ece9

We know that the general equation of a line is:

#y = mx + c #

Where #m# is your gradient

Now we must find #m#:

#m = #(change in #y#)#/#(change in # x) # = #( Delta y) / (Delta x) #

Change in #y: -8 - (3) = -11#

Change in #x: -8 - (-10) = 2#

#=> m = -11/2 #

so hence #y = -11/2 x +c #

Now to find #c#, we must input one of the points into the equation anf then solve for #c#:

Inputing# (-8,-8)#

( we could input either of the points that lie on the line)

#-8 = (-11/2 * -8) +c #

#-8 = 44 + c #

Minusing 44 from both sides:

#-52 = c#

hence our line equation is:

#y= -11/2 x -52 #

Hope this helped

You start by putting the points in the equation that is y = ax +b
In your case, the first point is (-10,3) which means that if x is -10 then y is 3. So you put it in the main formula which results in
3 = a(-10) + b.
Now you have two unknowns but luckily you have two points so you want to see whats the correlation between a and b:
3 = -10 a + b
<=> 3 + 10 a = b.
Then you enter your second point which is (-8,8): 8 = -8 a + b
But because you know what b is you can write it in the "a form" =>
8 = -8 a + 3 + 10 a
<=> -8-3 = -8a +10a
<=> -11 = 2a
<=> -11/2 = a
Now that you know a you can put it in the original line
3 = -10(a) +b
<=> 3 = -10(-11/2) +b
<=> 3 - 55 = b
<=> -52 = b
The last step is to put it in the main equation:
y = -11/2 x -52

Note: I modified the question to delete duplication of the point #(-10,8)# (which might have caused some confusion).

Given 2 points, we will assume that the required equation is that of a straight line.
On a straight line the slope (the difference between the #y# coordinate values divided by the difference between the #x# coordinate values) is the same for any pair of points on the line.

Consider the general point #(x,y)# and the given specific points #(-10,3)# and #(-8,-8)#

Slope#= (y-3)/(x-(-10))=(3-(-8))/(-10-(-8))#

Simplifying:
#color(white)("XXX")(y-3)/(x+10)=11/(-2)#

#color(white)("XXX")(-2)(y-3)=11(x+10)#

#color(white)("XXX")6-2y=11x+110#

#color(white)("XXX")-11x-2y=104#

#color(white)("XXX")11x+2y=-104#
(which is the "standard form" for a linear equation)