The line x + y - 3 = 0 intersects the y-axis at the point P and the

IMO :-

-x +3*sqrt(y) + 3= 0 is not a linear equation and question needs correction

A) x+y - sqrt(3) =0 intersects at point P on y axis
this means at x=0; y = sqrt(3)
x=1 ; y = 1+ sqrt(3)
x=2; y = 2+ sqrt(3)

B) -x + 3*sqrt(y)+3 =0 intersects at Point R on y axis
this means at x=0 ; y=1
x=1; y = 4/9
x=2 ; y = 1/9


While combining the two equation ,
3*sqrt(y) + y = sqrt(3) - 3
sqrt(y) [ 3+ sqrt(y) ] = sqrt(3) [ 1- sqrt(3) ]

Bunuel, suggest further steps

The line \(x + y - \sqrt{3} = 0\) intersects the y-axis at the point P

i.e. y-intercept @x=0 will be, \(0 + y - \sqrt{3} = 0\) i.e. \(y = √3\)

So coordinate of P are (0, √3)

line \(-x + 3\sqrt{y} + 3 = 0\) intersects the y-axis at the point R (But given equation is not equation of a line)
i.e. y-intercept @x=0 will be, \(0 + 3\sqrt{y} + 3 = 0\) i.e. \(y = 1\) (although √y = -ve is not proper)

So coordinate of Q are (0, 1)

Bunuel Second line \(-x + 3\sqrt{y} + 3 = 0\) (But given equation is not equation of a line)

It should have been \(-x + \sqrt{3}y + 3 = 0\) instead of \(-x + 3\sqrt{y} + 3 = 0\). Edited. Thank yuo.
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\(x + y - \sqrt{3} = 0\)

Slope = -1
i.e .Angle from X-axis = 135º
Tan 135º = 1/√3


\(-x + \sqrt{3}y + 3 = 0\)
SLope = 1/√3
Tan 30º = 1/√3
i.e. Angle from X-axis = 30º

i.e. Angle between the lines = 135º-30º = 105º

Other angle between Lines = 180º-105º = 75º


Answer: Option D

Hi, might be a stupid question but do we have to know Tan for certain angles in GMAT?

Best regards

balthazarium1

The answer to your question is NO.

But this question was easier to solve using trigonometry so I solved it using it. Actually Trigonometry is NOT needed for any GMAT problem.

We could do this problem as well without using trigonometry but the explanation would have been to lengthy then.

Answer - D

You dont need to know the intercepts, knowing the slopes is enough.

The line equations can be re-written as

y = -x + √3 => This means the slope is -1, which means this line will make 45 deg angle with x axis as Tan 45 = 1

y = (1/√3)*x - √3 => This means slope is +ve (1/√3), which means line will make 30 deg angle with y axis as Tan 30 = 1/√3

One slope negative, and one is positive, the Point Q where these two lines will meet will be the sum of the angles 45+30=75.

Edit: Adding an image which may help in visualizing these angles and why we are adding them.

Thats how I solved it.

Is there a way to arrive at the answer without using trigonometry? If so, please do share the solution. Thanks

Dear Sir,

If I am not asking much, could you show this in line-diagram?
I am not getting this way and even unable to draw the diagram.

Thanks in advance.

Here's an approach without Trig

First draw the lines by finding out the x and y intercepts of each line (for y intercept, put x=0 and for x intercept, put y=0)

Notice that the line x+y-√3=0 [or y=(-1)x+√3] has a slope of -1 which means it makes an angle of 45° with the y-axis (so \(\angle RPQ=45°\))

Then in right triangle BOR, notice that that the legs are √3 and 3 so the hypotenuse should be 2√3 which means triangle BOR is a 30°-60°-90° right triangle since the sides are in the ratio 1:√3:2 (so \(\angle PRQ=60°\))

Now in triangle PQR, \(\angle RPQ + \angle PRQ + \angle PQR = 180°\)

\(45°+60°+\angle PQR=180°\)

\(\angle PQR=75°\)

Answer is (D)

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Hi

Can I please get help in solving this question.

I was able to solve and get the coordinates of P (0, √3) and R (0, -√3). Since the two lines intersect at Q, I equated the two equation and got the coordinates for point Q (3√3-3, -2√3+3). I don't think this is a helpful step.

Separately, I got the slopes of both the lines as -1 and 1/√3, respectively. How to use this info to solve the question further?

Bunuel I did not find a solution from you on the thread. Can you please help. Thank you.

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