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Write a linear equation that passes through the points(2,-2) and (0,1)

User Galfisher
by
9.2k points

2 Answers

3 votes

Answer:


y=(-3)/(2)x-1

Explanation:

Given the two points, we can use the slope formula to get the slope of our equation.


(y_2-y_1)/(x_2-x_1)=m

Plug in what we know


((1)-(-2))/(0-2)=m


m=(-3)/(2)


y=(-3)/(2)x-b

Plug in one of the points given and solve to find our y-intercept [b]


-2=(-3)/(2)(2)-b


b=1

(we already know our y-intercept since we're given what y is when x is 0 [hence (0,1)], but this is how it is done if we did not know it initially)

Final equation


y=(-3)/(2)x-1

User Md Mahbubur Rahman
by
9.0k points
9 votes

Answer:

Slope intercept: y = -3/2x + 1

Point slope: y + 2 = -3/2 * (x - 2) [Forgot to add the work for this, I will add it if you need it, feel free to ask.]

Explanation:

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

But also

m = y_2 - y_1/x_2 - x_1


So lets substitute

m = 1 - (-2)/0 - (2)

Lets find the slope

m = 3/0 - (2)

m = 3/-2

m = -3/2 (Moved the negative)

Now we find the value of b using the equation of a line.

y = mx + b

y = (-3/2) * x + b

y = (-3/2) * (2) + b

-2 = (-3/2) * (2) + b

Now we find the value of b

Lets rewrite

-3/2 * 2 + b = -2

Cancel the CF of 2

-3 + b = -2

Move the terms without b to the right

b = -2 + 3

b = 1

Now we substitute our values of the slope and y-int into y = mx + b to find the equation.

y = -3/2x + 1

User Exbinary
by
8.4k points

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