For every point (a, b) on Line 1, there is a corresponding point

sorry, here is the equation of my interest :

The equation of line 2 is \(\frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1}\) or \(2y = 1 - x\) .

y- intercept is at x=0 as i know. so y= 1/2 when x=0 Am i missing something?

Sorry, you are correct, I had a rough day yesterday I hope it makes sense now.
(edited my post above)

Check this: math-coordinate-geometry-87652.html (chapter "Lines in Coordinate Geometry").

I saw the link. VERY IMPRESSIVE!!!! + kudo from me

GREAT JOB !!!

bunuel/Karishma

This is how I solved and it worked but I don't know why it worked :D Please enligthen me about why it worked...

I plugged in (a,b) in line 1 equeation to get b=2a+1..

then I started plugging in (b,-a) in the answer choices.. the 2nd answer choice resulted in b=2a+1.. and hence I marked B..
I am not sure why this worked.. Please help. .

Responding to a pm:

a and b stand for two numbers which define a co-ordinate on the plane. When you say that (a, b) lies on y = 2x+1, it means the relation between a and b is b = 2a + 1. e.g. if a = 0, b = 1; if a = 1, b = 3... At the end of the day, a line is nothing but a depiction of how one variable changes with another. A line just shows you the relation between 2 variables.

If (b, -a) lies on a line 2y = 1-x, this is just a different way of expressing the same relation between the two numbers a and b.
a and b are the same set of numbers (i.e. if a = 0, b = 1; if a = 1, b = 3...)
So after manipulating the equation a little, you are bound to get b = 2a + 1 only.

As you figured out, the approach is a little un-intuitive. When I looked at the problem, I actually solved it exactly the same way except that I took numbers rather than a and b.

I said, if (a, b) lies on y = 2x + 1, if a = 1, b = 3.
So (3, -1) must lie on the new equation of the line. When I put (3, -1) in the options, I see that only (B) satisfies.

it took me a while to get this

so what i did was create a value for x then sub it into line 1 to find out what y is. now that you have (a,b) create the point (b,-a).

now what you do is sub is points of line 2 into various equations until you find one that makes sense.

 

For every point \((a, b)\) on Line 1, there is a corresponding point \((b, -a)\) on Line 2. Given that the equation of Line 1 is \(y = 2x + 1\), what is the equation of Line 2?

A. \(y = \frac{x}{2} + \frac{1}{2} \)
B. \(y = -\frac{x}{2} + \frac{1}{2}\)
C. \(y = -\frac{x}{2} - \frac{1}{2}\)
D. \(y = \frac{x}{2} - \frac{1}{2}\)
E. \(y = -\frac{x}{2} +1\)


To find the equation of Line 2, we can first identify two points on Line 1 and then determine their corresponding points on Line 2. Points \((0, 1)\) and \((1, 3)\) lie on Line 1, so their corresponding points on Line 2 are \((1, 0)\) and \((3, -1)\), respectively.

Using the form \(y = mx + b\) for the equation of Line 2, we can employ the two points we found to create a system of equations to solve for the slope \(m\) and the y-intercept \(b\). Plugging the coordinates of the points into the equation, we obtain:

\(0 = m(1) + b\)

\(-1 = m(3) + b\)

Upon solving this system of equations, we find that \(m = -\frac{1}{2}\) and \(b = \frac{1}{2}\).

Consequently, the equation of Line 2 is \(y = -\frac{x}{2} + \frac{1}{2}\).


Answer: B­­