Question

Find an equation of the line that satisfies the given condition. (Let x be the independent variable and y be the dependent variable.)
The line passing through
(−2, 3)
and parallel to the x-axis

Answer 1

Okay, this is just like the last problem we did, except here you have a different approach to find the slope. If the line is parallel to the x-axis, it is simply \(y = k\) where \(k\) is some constant. That means for any value of \(x\), \(y = k\) and the slope formula will give us \[m = \frac{k - k}{x_2 - x_1} = 0\]

What do you get for the equation of the line?

Answer 2

isn't the -2 da x and the 3 the y?

Answer 3

isnt the answer y=5?

Answer 4

-2 is the x value, 3 is the y value, yes. (sorry, didn't notice you'd answered)

Now does y = 5 look like it will ever go through the point (-2, 3)?

Answer 5

No, it doesn't. Lol. The equation i had used was y=mx+b...y-3=0-(-2) den y-3=2 den i added 3 to 2 and i got 5. wat am i doing wrong?? :/

Answer 6

Go reread my explanation of the slope in this problem. What does m =?

Answer 7

m = 0. we've got a known point of (-2, 3), so using the formula

\[y-y_0 = m(x-x_0)\]\[y-3 = 0(x-(-2))\]\[y-3=0(x+2)\]\[y - 3 = 0\]\[y=3\]

Answer 8

sorry i had seen a "k" not a "y". thanks i seen how u got it now