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

Answer:

7x + 2y = 1

Step-by-step explanation:

1. Find the slope of the line.

(y2-y1)/(x2-x1) ---->  (11 - (-3))/((-3) -1) ----> (11+3)/ (-3 -1) ----> 14/-4

*lets simplify into (7/-2)*

2. Write into slope intercept form.

y=mx+b (m is the slope) (b is the y intercept)

y = (7/-2)x + b

We now need to find the value of b, this can be done by plugging in values of x and y, and using algebra to solve.

11 = (7/-2)(-3) + b

11 = 10.5 + b

b = 0.5 or 1/2

slope intercept form y = (7/-2)x + 1/2

Now we convert to standard form: ax+by=c

*In this form we our a term cannot be a fraction nor can it be negative*

1. Get x and y on the same side, and since our x term is negative right now, we can make it positive by adding it on both sides.

(7/2)x + y = 1/2

2. We now want to get rid of the fraction in term a, so we multiply the entire function by two, so the denominator cancels out.

2(7/2)x + 2y = 2(1/2)

Simplify and you get

7x + 2y = 1