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Statement 1 and Statement 2 separately give an infinite number of solutions. Solving the two statements together gives you a unique solution for a and b, so you have the point (a;b). You know the line is perpendicular to y=2x and there is only one way to draw a perpendicular to a line from any given point.
_________________

What is the equation of the line that is perpendicular to line \(y = 2x\) and passes through point (a, b)?

1. \(a = -b\) 2. \(a - b = 1\)

I really think that the answer should be A. Because when a = -b, it means that (-b,b). So let's say that b=5, doesn't matter whether b is positive or negative because you will still end up drawing the same line. Well, what will be its y intercept? well if b is zero, then both x and y will be zero, hence, the y intercept is at the origin. So you're gonna end up drawing the same straight line, so you can make an equation out of it.

What is the equation of the line that is perpendicular to line \(y = 2x\) and passes through point (a, b)?

1. \(a = -b\) 2. \(a - b = 1\)

I really think that the answer should be A. Because when a = -b, it means that (-b,b). So let's say that b=5, doesn't matter whether b is positive or negative because you will still end up drawing the same line. Well, what will be its y intercept? well if b is zero, then both x and y will be zero, hence, the y intercept is at the origin. So you're gonna end up drawing the same straight line, so you can make an equation out of it.

what's the OA to this problem?

My answer is A

Tarek,

please note that a and b are not variables... How can you say that b is zero ..? y=-1/2x+c a=-b b=1/2b+c b=2c From 1: you know that b=2c a=-b. there can be infinite number of solutions from this relation. and passing through (0,0) one of that solutions

I hope this is clear now.
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

What is the equation of the line that is perpendicular to line \(y = 2x\) and passes through point (a, b)?

1. \(a = -b\) 2. \(a - b = 1\)

I really think that the answer should be A. Because when a = -b, it means that (-b,b). So let's say that b=5, doesn't matter whether b is positive or negative because you will still end up drawing the same line. Well, what will be its y intercept? well if b is zero, then both x and y will be zero, hence, the y intercept is at the origin. So you're gonna end up drawing the same straight line, so you can make an equation out of it.

what's the OA to this problem?

My answer is A

Tarek,

please note that a and b are not variables... How can you say that b is zero ..? y=-1/2x+c a=-b b=1/2b+c b=2c From 1: you know that b=2c a=-b. there can be infinite number of solutions from this relation. and passing through (0,0) one of that solutions

I hope this is clear now.

I didn't say that b is for sure zero. All I said that no matter what the value of b is, even if it is zero, we would still be drawing the same line, no? for example, (-1,1), (0,0), (5,-5), EVEN if it is (-10, 10), wouldn't all these points fall on the same line? and we already have the perpendicular slope of -1/2, so what's wrong with this reasoning? i'm still trying to figure out what's wrong with this reasoning and sorry for being persistent but i guess it's the best way to learn here...lol

please note that a and b are not variables... How can you say that b is zero ..? y=-1/2x+c a=-b b=1/2b+c b=2c From 1: you know that b=2c a=-b. there can be infinite number of solutions from this relation. and passing through (0,0) one of that solutions

I hope this is clear now.

I didn't say that b is for sure zero. All I said that no matter what the value of b is, even if it is zero, we would still be drawing the same line, no? for example, (-1,1), (0,0), (5,-5), EVEN if it is (-10, 10), wouldn't all these points fall on the same line? and we already have the perpendicular slope of -1/2, so what's wrong with this reasoning? i'm still trying to figure out what's wrong with this reasoning and sorry for being persistent but i guess it's the best way to learn here...lol

No they are not the same. y intercept changes.. when "b" changes. -a=b=2c y=-1/2x+c for (-1,1) y intercept 2 y=-1/2x+2 --> equation (1) (0,0) y=-1/2x --> equation (2) for (5,-5) y=-1/2x+c --> -5=-1/2*5+c= -5/2 y=-1/2x+-5/2 -----> equation (3)

Does equations (1),(2) and (3) are same???? No.
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

please note that a and b are not variables... How can you say that b is zero ..? y=-1/2x+c a=-b b=1/2b+c b=2c From 1: you know that b=2c a=-b. there can be infinite number of solutions from this relation. and passing through (0,0) one of that solutions

I hope this is clear now.

I didn't say that b is for sure zero. All I said that no matter what the value of b is, even if it is zero, we would still be drawing the same line, no? for example, (-1,1), (0,0), (5,-5), EVEN if it is (-10, 10), wouldn't all these points fall on the same line? and we already have the perpendicular slope of -1/2, so what's wrong with this reasoning? i'm still trying to figure out what's wrong with this reasoning and sorry for being persistent but i guess it's the best way to learn here...lol

No they are not the same. y intercept changes.. when "b" changes. -a=b=2c y=-1/2x+c for (-1,1) y intercept 2 y=-1/2x+2 --> equation (1) (0,0) y=-1/2x --> equation (2) for (5,-5) y=-1/2x+c --> -5=-1/2*5+c= -5/2 y=-1/2x+-5/2 -----> equation (3)

Does equations (1),(2) and (3) are same???? No.

oh perfect! now i see where i went wrong. Thank you so much for the great explanation.