Question

Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form?

Answer 1

Line JK: (y₂-y₁/x₂-x₁)
[(-3)-11] / [1-(-3)] = -14/4 == -7/2
Point slope formula: y-y₁=m(x-x₁)
y-(11)=-7/2[x-(-3)]
y-11= -7/2x-21/2 -7/2*3= -21/2
Add 11 to both sides.

y= -7/2x+1/2...

Answer 2

Hello!

First of all we find the slope by dividing the difference of the y values by the difference of the x values as seen below.


`(-3-11)/(1+3) = -(14)/(4) = (7)/(2)`

The slope of our line is 7/2, or 3.5
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Now we will put the slope and a point on our line into slope intercept form and solve for b. We will use (-3,11).

11=3.5(-3)+b
11=10.5+b
b=0.5

Our final equation is shown below.

y=3.5x+0.5

I hope this helps!