I’m prepping for a test and keep tripping over how to write the equation of a straight line from its graph-say it goes through (−3, 4) and (2, −1)-I tried y=mx+b and point–slope but I’m not sure that’s relevant; how should I do this? Any help appreciated!
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3 Responses
Don’t overthink it: slope m = (−1−4)/(2−(−3)) = −1; plug a point into y = mx + b (say (2, −1)) to get b = 1, so the line is y = −x + 1. Think of it like stairs-go right 1, you go down 1-slope −1; quick refresher: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:forms-of-linear-equations/x2f8bb11595b61c86:writing-lines/v/writing-equations-of-lines-given-two-points
You’re on the right track-both y = mx + b and point–slope are your friends here! First grab the slope from the two points: m = (−1 − 4) / (2 − (−3)) = −5/5 = −1, so the line drops 1 for every step right. Now use point–slope with either point, say (2, −1): y − (−1) = −1(x − 2), so y + 1 = −x + 2, which simplifies to y = −x + 1. Using (−3, 4) gives the same thing: y − 4 = −1(x + 3) → y = −x + 1-nice consistency check. If you like sanity checks, plug in the points: at x = 2, y = −2 + 1 = −1; at x = −3, y = 3 + 1 = 4. Tiny analogy time: imagine walking down a staircase that drops one step for every step forward; if you stand right at x = 0, you’re perched at y = 1-so the line’s equation is y = −x + 1.
Compute the slope m = (−1 − 4)/(2 − (−3)) = −5/5 = −1, then use point–slope: y − 4 = −1(x + 3) ⇒ y = −x + 1.
Think of it as a ramp dropping one unit for each step right, crossing the y-axis at 1; quick review: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:forms-of-linear-equations/xb8d8f9e48a8b31d6:writing-linear-equations/a/writing-linear-equations-using-two-points.