How do you find the slope of the line that passes through (-8,-15), (-2,5)?

The two points will give the gradient of a straight line graph.

Consider the standard for of: #y=mx+c#

Where
#y# is the answer

#m# is the gradient (slope)

#c# is the y-axis intercept (point where it crosses the y axis)
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#color(brown)("Determine the gradient")#

Let point 1 #->P_1->(x_1,y_1) = (-8,-15)#

Let point 2 #->P_2->(x_2,y_2)=(-2,5)#

The gradient (slope) is measured moving from left to right on the x-axis. As #" " x_1 < x_2 " "# (#x_1" is less than "x_2#) then we travel from #P_1" to "P_2# for the gradient

So the gradient #->m=("change in up or down")/("change in along") larr "Left to right"#

#m=("change in y-axis")/("change in x-axis") = (y_2-y_1)/(x_2-x_1)#

#m=(5-(-15))/((-2)-(-8)) " "=" "20/6 = 10/3 color(red)( larr" Question answer")#

So the equation now is #y=10/3x+c#

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#color(brown)("Determine y-intercept (crossing point on the y axis)")#

Using any one of the 2 given points substitute to solve for #c#

#P_1->(x_1,y_1)=(color(blue)(-8),color(green)(-15))#
#color(white)(.)#

#=>color(green)(y)=10/3color(blue)(x)+c" "->" "color(green)(-15)=10/3xx(color(blue)(-8))+c#

#-15=-80/3+c#

Add #80/3# to both sides

#11 2/3 = c#
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#color(brown)("Putting it all together")#

#y=mx+c" "->" "y=10/3x+ 11 2/3#

Note that #11 2/3 = 35/3#