What is the equation of the line k that is perpendicular to line y = 2

Same approach but c should be = -1/4.

since a = 1/2 and b = -1/2, putting these values on the line eq: y = mx + c
-1/2 = -1/2 (1/2) + C
C = -1/4

I think that the answer is A.

According to the question, we are suppose to find the equation of the line that is perpendicular to the line y=2x. First of all, we already know that the slope of that line must be -1/2 because the product of 2 slopes that are perpendicular to each other is -1, since we already have 2, then the slope of the line that we're looking for is -1/2.

(1) a=-b, so if you substitute this to the given point (a,b), you will have (-b,b). Let's say that b is 5, so if b is positive, then we will have (-5,5) or if b is negative, then we'll have (5,-5). Either way doesn't matter because you can create the same line either way. So you can just pick your points with the slope of -1/2 and you can have your equation.

(2) doesn't tell us anything about the exact value of (a,b) because there are infinite posibilities such as (2,1) or even (100,99).



So my answer is A

tarek, what you mean to say is that

line y = 3x + 5 is same as y = 3x+ 10 (generic equation being y = mx + c)

which I dont think is correct as the intercept c is different. These are two different parallel lines in X-Y co-ordinate plane. You have to figure out the unique value of the "c" to find the equation.

So answer is C which gives you the "c" intercept as -1/4.

perpendicular line is y=-1/2x +c

1) a = -b -->(1)
(a,b) passing throw this line
y=-1/2x +c
b=-1/2*a+c --(2)

Here we have two equations but three unknowns
Not suffcieint

2) a - b = 1 -->(3)
b=-1/2 *a+c --(2)
two equations and three variables

Not suffcient

combined.
three equations and three variables..

suffcient
C.

In this question you need not to solve it completely..

But be careful, some times question may ask for what is value of (a-b) .?
In this scenario, please try to solve it completely.. because two equations may be suffcient.

e.g a-b+c-d=4 -->(1)
3c=4-3d (2)
here we have only two equations.. but we can find a-b? but can't find a or b or c or d values.
I hope this helps

of course, I agree that line y = 3x + 5 is NOT the same as y = 3x+ 10. But according to my numbers which are (5,-5) or even (-5,5), the y intercept here will be 0 because if b is 0, then (0,0). So if you draw that line, you've basically found your line to make your equation out of it, no? What's the OA so that we can finalize our confusion once and for all...heheh

[quote="GMAT TIGER"][quote="gmatcraze"]Can someone help to verify my approach to this problem? I agree that for DS problems, we need not solve it completely ... but am giving the steps below, in case this was a PS problem. Thanks.

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


guys, i got a question : are not the two statements contradictory?
1 st says a = -b 2.st a - b = 1 a=b ????

Not really, the statements don't seem to be contradicting each other. On the contrary, they are complementing each other for us to get their values.

St 1:
a = -b
Adding b on both sides:
a+b=0 ----I

St 2:
a-b=1-----II

When we solve both equations I and II, we get;

a=1/2 and b=-1/2
Now, we know the coordinate through which the line passes;

Since the line is perpendicular to y=2x, the slope of the line is -1/2. Note: Product of slopes of two perpendicular lines equals -1.
\(2*m_2=-1\)
\(m_2=-1/2\)

Equation of a line with slope=m and passing through \((x_1,y_1)\) is:
\(y-y_1=m(x-x_1)\)

Equation of a line with slope=-1/2 and passing through \((1/2,-1/2)\) is:
\(y-\frac{-1}{2}=\frac{-1}{2}(x-(\frac{1}{2}))\)
\(y+\frac{1}{2}=\frac{-1}{2}x+\frac{1}{4}\)
\(y=\frac{-1}{2}x-\frac{1}{4}\)

The attached image contains the two lines we talked about:

Hey tarek99,
That was great thinking though there is something else you need to think about... A line can be uniquely defined if you have the slope and any one point.
Now a line will pass through (0, 0), (5, -5), (-5, 5) and every such point (where a = -b) but what will be its slope? If a line of slope (-1/2) passes through (5, -5), will it also pass through (-5, 5) or (0, 0)?
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I think the equation of line is least botheration to the designer of this question. The question can be rephrased more simply -
Can you determine a and b with two equation s1 and s2. The answer is C

Posted from my mobile device

If I change the equation of the given line:

'What is the equation of the line that is perpendicular to line y=x + 12 and passes through point (a, b)?'

the answer becomes (A) not (C). Figure out why.(tarek99 has a point)
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How so, Karishma? Even for this equation, I think there can be infinite lines that will be perpendicular to "y=x + 12" and pass through points (1/2, -1/2) or (1,-1) or (100, -100). How will that make "A" alone sufficient?

Hey fluke,
First of all, I think your solutions are pretty good. Great job.
As for this question, a line perpendicular to y = x + 12 will have a slope of -1 (since product of the slopes should be -1)

This line passing through (0,0) or (1/2, -1/2) or (1, -1) or (100, -100) will be the same line since its slope is -1.
Look at the diagram below:


Feel free to contradict.
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Oh yes!! "y=-x" will be the equation if a=-b
I am bit myopic at times, Karishma. Thanks for the clarification.

That's the reason don't trust algebra since x's and y's may block the sight of the bigger things

My decision to not calculate the equation of line in original question since its waste of time on DS question. Not that I can't do it.

I agree the equation of the line with slope -1 bagging the condition a = -b is X+y=0.

But coming to the question posed...
We are asked to find a line with slope -1/2 satisying the conditon a=-b . There are many lines perpendicular to the given equation.
Eq : (0,0) equation : x+2y=0
(1,-1)equation : x+2y+1=0
(2,-2) equation: x+2y+4=0
All of these are different lines with slope -1/2 passing through (a,-a).

So my bet is on option C, only way we can close in on a single line equation.

Official Solution:


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

Since line \(k\) is perpendicular to \(y=2x\), its slope must be the negative reciprocal of the slope of \(y=2x\), which is \(2\). Therefore, the slope of line \(k\) is \(-\frac{1}{2}\). Using the formula for slope, which is the "rise over run" or change in \(y\) divided by change in \(x\), we can find the equation of line \(k\) passing through the point \((a,b)\) with slope \(-\frac{1}{2}\). Thus, we have the equation \(-\frac{1}{2}=\frac{y-b}{x-a}\), which simplifies to \(y=-\frac{x}{2}+\frac{a}{2}+b\).

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

If \(a=b=0\), the equation of \(k\) becomes \(y=-\frac{x}{2}+\frac{a}{2}+b=-\frac{x}{2}\), but if \(a = 1\), and \(b= -1\), the equation of \(k\) becomes \(y=-\frac{x}{2}+\frac{a}{2}+b=-\frac{x}{2}+\frac{1}{2}-1\). These are two different equations, so statement (1) alone is not sufficient.

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

If \(a=1\) and \(b=0\), the equation of \(k\) becomes \(y=-\frac{x}{2}+\frac{a}{2}+b=-\frac{x}{2}+\frac{a}{2}\), but if \(a = 0\), and \(b= -1\), the equation of \(k\) becomes \(y=-\frac{x}{2}+\frac{a}{2}+b=-\frac{x}{2}-1\). Again, these are two different equations, so statement (2) alone is not sufficient.

(1)+(2) We have two distinct linear equations \(a = -b\) and \(a - b = 1\), which we can solve for \(a\) and \(b\), and get the exact equation of line \(k\). Solving gives \(a=\frac{1}{2}\) and \(b=-\frac{1}{2}\). Therefore, the equation of line \(k\) is \(y=-\frac{x}{2}+\frac{a}{2}+b=-\frac{x}{2}+\frac{1}{4}-\frac{1}{2}=-\frac{x}{2}-\frac{1}{4}\). Sufficient.


Answer: C
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