Write the equation of the line which passes through the point (1, –2) and has a slope of -2/3 in standard form

Answer 1

Point-slope is y-y1=m(x-x1). If we plug the given values in, we have y-(-2)=-2/3(x-1). Simplifying gives y+2=-2/3x+2/3 or y=-2/3x-4/3. If we multiply everything by 3, we can get rid of the pesky denominators. Thus 3y=-2x-4. Add the x-term to get into standard form and you'll have 2x+3y=-4. There's your answer.

Answer 2

For this type of question when we're trying to determine the equation of a line, we first need to find the slope of the line and insert it along with a (x,y) coordinate into the Point-Slope Formula.

The point slope formula is,  y-y₁=m(x-x₁). We are given a coordinate point and the slope so we plug in our given values. We can label our given coordinate point ( 1, -2 ) x1 which is 1 and y1 which is -2. Furthermore, we plug x1, y1, and m (our given slope) which is -2/3 into our Point-Slope Formula.

We then get:

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

Simplifying we get:

1. y + 2 = - 2/3x + 2/3

We then subtract the 2 from the left side over to the right to form a complete equation and have it in the form of y = mx + b (Slope-Intercept Form):

1. y= -2/3x - 4/3

You can then multiply the entire equation by 3 to get rid of the 3's in the denominator:

1. 3y = -2x - 4

Lastly we can move the 2x over to have it in standard form, which is ( Ax + By = C):

1. 3y + 2x = - 4 , where this is your final answer.
2. Please let me know if there are any steps i can clarify for you.