How do I put an equation into standard form, Ax+Bx=C?
I have the given points (-1,3) and (2,-4).
Also, how do you write an equation in standard form when perpendicular through another?
[the question is "through (-1,2) and perpendicular to 2x-3y=-5]
first use y2-y1/x2-x1 to find the slope
-4-3/2-(-1)=-7/3
then(assuming the equation is parellel)use point slope form y-y=slope(x-x)
y-3=-7/3(x-(-1))
y-3=-7/3x-7/3
y=-7/3x+2/3
then to put it in standard form it must be in the Ax+By=C form with no fractions and the leading coefficient cannot be negative.
y=-7/3x+2/3
7/3x+y=2/3
3(7/3x+y=2/3)
7x+y=2
the way i would work it is first move the equation to standard form because i find it easier to work it that way.
2x-3y=-5
-3y=-2x-5
y=2/3x+5/3
an equation perpendicular to this equation would have a slope of -3/2 because perpendicular lines have opposite recirocal slopes. now use point slope.
y-2=-3/2(x-(-1))
y-2=-3/2x-3/2
y=-3/2x+1/2
now move the equation back to standard form.
y=-3/2x+1/2
3/2x+y=1/2
2(3/2x+y=1/2)
3x+y=1
To put an equation into standard form, Ax + By = C, you need to manipulate the equation so that the coefficients (A, B, C) are integers and A is positive. Here's how you can do it step by step:
1. Start with your given equation, which can be in any form (slope-intercept, point-slope, etc.). Let's say we have the equation in slope-intercept form: y = mx + b, where m is the slope and b is the y-intercept.
2. Rearrange the equation to get it in the form Ax + By = C. In this case, you need to move all the variables to one side of the equation and simplify. Here are the steps for an example equation:
Let's say you have the equation: 2y - 3x = 4.
Move the x term to the left side: -3x + 2y = 4.
Now, multiply the entire equation by -1 to make the coefficient of x positive: 3x - 2y = -4.
Finally, rearrange the terms so that the x term comes before the y term: 3x + (-2y) = -4.
The equation is now in standard form: 3x - 2y = -4.
Now, let's move on to the second part of your question about finding an equation in standard form when a line is perpendicular to another and passes through a specific point:
3. Start with the given equation that the line is perpendicular to. Let's say you have the equation: 2x - 3y = -5.
4. Find the slope of the given line. The slope of the given line is the coefficient of x when the equation is in the form y = mx + b. In this case, the slope is 2/3 because the equation can be rewritten as y = (2/3)x + (5/3).
5. The slope of a line perpendicular to another line is the negative reciprocal of the original line's slope. In this case, the slope of the perpendicular line is -3/2.
6. Use the point-slope form of a line to write the equation of the line perpendicular to the given line, passing through the given point (-1,2). The point-slope form is y - y1 = m(x - x1), where (x1, y1) are the coordinates of the given point and m is the slope.
Plugging in the values, we get: y - 2 = (-3/2)(x - (-1)).
Simplify: y - 2 = (-3/2)(x + 1).
Expand the right side: y - 2 = (-3/2)x - 3/2.
Move the constant term to the right side: y = (-3/2)x - 3/2 + 2.
Simplify further: y = (-3/2)x - 3/2 + 4/2.
Combine the constants: y = (-3/2)x + 1/2.
7. Now, you have the equation in slope-intercept form. To convert it to standard form, follow the steps mentioned earlier. Multiply the equation by 2 to get rid of the fraction: 2y = -3x + 1.
Rearrange the terms to get the x term before the y term: 3x + 2y = 1.
The equation is now in standard form: 3x + 2y = 1.
I hope this helps! Let me know if you have any further questions.