what equation would Y =?

Question

The xy-coordinate plane is givenThe m = - 1 enters the window in the second quadrant, goes down and right, crosses the y-axis at y = 2 , crosses the x-axis at x = 2, passes through the point (6, - 4) , and exits the window in the fourth quadrant.

what equation would Y =?

Answer 1

So we are given the slope of a line (m=-1) and a point through which the line passes through (6,-4). Just by knowing this, we can find the y=mx+b equation of the line by using point-slope form.

The point-slope form equation is as follows:

Given a line's slope, which we can call m, and a point through which the line passes through, which we can call (x1,y1), the equation of the line is,

y-y1 = m(x-x1)

We know that m=-1 and that (x1,y1) = (6,-4), since the problem gives us these values for the slope and the point through which the line passes through. So, we can plug these values into the point-slope equation.

y - (-4) = -1(x-6)

Now, since we want only the y term on the left side, we can solve for y in our equation.

first we distribute the -1 term on the right side and make the double negative a positive on the left side.

y + 4 = -x+6

Then, we subtract 4 from both sides

y = -x+2

This is our final answer

Note that we did not need to use the information regarding the line's x-intercept and y-intercept that was provided, since the point-slope equation only requires one point through which the line passes through. Nevertheless, we can verify that our solution is correct by setting y = 0 and x = 0 and seeing if it matches the provided x and y intercepts.

Setting x=0 gives us, y = 0+2 -> y = 2. This matches the problem description since the line crosses the y-axis at y=2.

Setting y = 0 gives us, 0 = -x+2 -> x = 2. This also matches the problem description since the line crosses the x-axis at x=2.