What is the equation of the line in standard form that passes through the point (3, –2) and is parallel to the line y = –4x + 1?
You know the point and the slope, so the point-slope form:\[y - y_0 = m(x - x_0)\]
Parallel means Slopes are equal ie m1 = m2
Now the points you have are \((x_0, y_0)\) and \(m\) is the slope.
y = mx + c m = slope
As @Yahoo! said, the slope will be equal.\[y - y_0 = -4(x - x_0)\]Now your point.
Hello?
Oooook so what your saying is the equation would be .. y-2=m(x-3) ??
Im sorry guys, im lost with this stuff..
Okay, I'll start from the top: parallel lines have the same slope. We know the slope of the other line (y = -4x + 1) to be m = -4, because it is in slope-intercept form (y = mx + b). So, our line will also have slope m = -4. Because we know a point our line passes through, and we know the slope, point-slope form is the most convenient. Our known point is (x0, y0) = (3, -2) and point-slope form for a line is \[y-y_0 = m(x-x_0)\] so plugging in what we know, we get \[y-(-2) = -4(x-3)\]\[y+2 = -4x +12\rightarrow y = -4x + 10\]Now, the problem asks for us to put the formula in standard form, which is Ax + By = C, with neither A nor B = 0. So, we add -4x to both sides giving us x and y terms on the left and a constant on the right, which is standard form. What is your answer?
Sorry, we add 4x, not -4x!
Ok so i almost got it, Except for... the problem asks for us to put the formula in standard form, which is Ax + By = C, with neither A nor B = 0. So, we add -4x to both sides giving us x and y terms on the left and a constant on the right, which is standard form. I dont understand the Ax+By=C part..
It means you need to have x and y on the left side of the equals sign, and a number on the right. A and B are constants, and they can't equal 0.
C is a constant, too. It can equal 0. Except in this problem, it doesn't :-)
\[y = -4x + 10\]We need both x and y on the left side, so we add 4x to each side: \[y + 4x = -4x + 10 + 4x\]\[y +4x = 10\]Voilà, our equation in standard form!
Here's a diagram showing the original line (in blue) and our new, parallel line passing through (3, -2) in purple.
Ok so im still kinda confused lol im so sorry, math is like greek to me..
OH! the graph is very helpful, thank you
Millions of Greek kids learn Greek, you can do it too :-)
Lol.. i think i should conquer algebra 2 before greek.. lol
This problem is a nice one because it involves many of the tools in the toolbox, giving you a chance to see where each one is best used, except for standard form. Standard form is convenient if you want to quickly graph a line, because it is easy to determine the x and y-intercepts from standard form, and once you know them, you plot those two points, slap down your ruler and draw a line through them. In standard form: Ax + By = C x-intercept is the point where the line crosses the x-axis, so y = 0, giving us Ax = C, so the x-intercept is at (C/A, 0). Similarly, for the y-intercept, that's the point where the line crosses the y-axis, at x = 0, so By = C, and the y-intercept is at (0, C/B). I trust you can see how I arrived at those results? For the first case, Ax + B(0) = C, Ax = C, x = C/A; for the second case, A(0) + By = C, By = C, y = C/B.
Point-slope form is handy when you know a point and the slope. Slope-intercept form is handy when you know the slope and the y-intercept: if you think about it, that's just a special case of point-slope form where x0 = 0: \[y-y_0 = m(x-x_0)\] \[y = m(x - 0) + y_0\]\[y = mx +y_0\]But y0 is just the y-value of the y-intercept in this case, so y_0 = b, giving us our old friend y=mx+b.
Much of this isn't really difficult, it just takes the right explanation for you to understand it, and then plenty of practice to make it all second-nature. Unfortunately, not every teacher can provide the right explanation for every student, and many students aren't willing to do the practice.
Yes. i undersand it! u should be a teacher!!
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