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The angle made by the line \sqrt{3}x+y+6=0, with y-axis is

1. 60 degrees 2. 30 degrees 3. 45 degrees 4. 120 degrees 5. None of the above

We have the line \(y=-\sqrt{3}*x-6\).

Attachment:

untitled.PNG [ 14.87 KiB | Viewed 3554 times ]

Find the x-intercept of the line (x-intercept of a line the value of x for y=0, the point where the line crosses X-axis): y=0 --> \(x=-2\sqrt{3}\);

Find the y-intercept of the line (y-intercept of a line the value of y for x=0, the point where the line crosses Y-axis): x=0 --> \(y=-6\);

In the right triangle created by the given line and the axis the ratio of lengths of the legs is \(2\sqrt{3}:6=1:\sqrt{3}\) so we have 30°, 60°, and 90° right triangle (in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\). Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°)).

Now, the angle at y-axis is opposite the smallest side \(2\sqrt{3}\) so it equals to 30°.

Answer: B.

Or you can notice that as the slope, which is th ratio of the rise over run is equal to \(-\sqrt{3}\) then the ratio of the length of the leg at y-axis over the length of the leg at x-axis will be \(\sqrt{3}\), which again tells us that we have 30°, 60°, and 90° right triangle.

that was smart. I tried to use m = tan(degree), Is this wrong. If no can you please explain how to compute using this.

Every GMAT geometry question can be solved without trigonometry, so you don't need trigonometry for GMAT.

But yes, \(slope=m=tan(\alpha)\), where \(\alpha\) is the angle line makes with positive x-axis, so as \(m=-\sqrt{3}=tan(\alpha)\) --> \(\alpha=120^{\circ}\) --> angle at y-axis will be 120-90=30. _________________

This question is incomplete, for answer to be 30 degrees, question should ask the smaller angle between line and y-axis. Otherwise answer can be either 150 degrees or 30 degrees. _________________

The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done!

This question is incomplete, for answer to be 30 degrees, question should ask the smaller angle between line and y-axis. Otherwise answer can be either 150 degrees or 30 degrees.

Though I see your point, still only one answer from 150 and 30 is listed among answer choices, so we can safely choose 30 as correct one. _________________

The angle made by the line \sqrt{3}x+y+6=0, with y-axis is

1. 60 degrees 2. 30 degrees 3. 45 degrees 4. 120 degrees 5. None of the above

We have the line \(y=-\sqrt{3}*x-6\).

Attachment:

untitled.PNG

Find the x-intercept of the line (x-intercept of a line the value of x for y=0, the point where the line crosses X-axis): y=0 --> \(x=-2\sqrt{3}\);

Find the y-intercept of the line (y-intercept of a line the value of y for x=0, the point where the line crosses Y-axis): x=0 --> \(y=-6\);

In the right triangle created by the given line and the axis the ratio of lengths of the legs is \(2\sqrt{3}:6=1:\sqrt{3}\) so we have 30°, 60°, and 90° right triangle (in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\). Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°)).

Now, the angle at y-axis is opposite the smallest side \(2\sqrt{3}\) so it equals to 30°.

Answer: B.

Or you can notice that as the slope, which is th ratio of the rise over run is equal to \(-\sqrt{3}\) then the ratio of the length of the leg at y-axis over the length of the leg at x-axis will be \(\sqrt{3}\), which again tells us that we have 30°, 60°, and 90° right triangle.

Can you pls explain how did draw the graph... How did you take the points of x and y..

\(y=mx+b\) is called point-intercept form of equation of a line (in our case it's \(y=-\sqrt{3}*x-6\)), where \(m\) is the slope of the line; \(b\) is the y-intercept of the line; \(x\) is the independent variable of the function \(y\). To draw a graph of a line you need two points (any two distinct points (x,y) which satisfy the equation of a line) and then you just draw a line passing these points. Usually X and Y intercepts are best choices for the points.

X-intercept is the point where a line (a graph) crosses the x-axis. So it's the point on x-axis, any point on x-axis has y-coordinate equal to zero, which means that X-intercept is the point \((x,0)\) - the value of \(x\) when \(y=0\): \(y=mx+b\) --> \(0=mx+b\) --> \(x=-\frac{b}{m}\). So X-intercept of a line \(y=mx+b\) is \(x=-\frac{b}{m}\);

Y-intercept is the point where a line (a graph) crosses the y-axis. So it's the point on y-axis, any point on y-axis has x-coordinate equal to zero, which means that Y-intercept is the point \((0,y)\) - the value of \(y\) when \(x=0\): \(y=mx+b\) --> \(y=m*0+b\) --> \(y=b\). So Y-intercept of a line \(y=mx+b\) is \(y=b\).

In our case for the line given by \(y=-\sqrt{3}*x-6\): X-intercept will be \((-2\sqrt{3}, \ 0)\), as x-intercept of a line is the value of x for y=0, the point where the line crosses X-axis: y=0 --> \(x=-2\sqrt{3}\);

Y-intercept will be \((0, \ -6)\), as y-intercept of a line is the value of y for x=0, the point where the line crosses Y-axis: x=0 --> \(y=-6\).

Can you pls explain how did draw the graph... How did you take the points of x and y..

\(y=mx+b\) is called point-intercept form of equation of a line (in our case it's \(y=-\sqrt{3}*x-6\)), where \(m\) is the slope of the line; \(b\) is the y-intercept of the line; \(x\) is the independent variable of the function \(y\). To draw a graph of a line you need two points (any two distinct points (x,y) which satisfy the equation of a line) and then you just draw a line passing these points. Usually X and Y intercepts are best choices for the points.

X-intercept is the point where a line (a graph) crosses the x-axis. So it's the point on x-axis, any point on x-axis has y-coordinate equal to zero, which means that X-intercept is the point \((x,0)\) - the value of \(x\) when \(y=0\): \(y=mx+b\) --> \(0=mx+b\) --> \(x=-\frac{b}{m}\). So X-intercept of a line \(y=mx+b\) is \(x=-\frac{b}{m}\);

Y-intercept is the point where a line (a graph) crosses the y-axis. So it's the point on y-axis, any point on y-axis has x-coordinate equal to zero, which means that Y-intercept is the point \((0,y)\) - the value of \(y\) when \(x=0\): \(y=mx+b\) --> \(y=m*0+b\) --> \(y=b\). So Y-intercept of a line \(y=mx+b\) is \(y=b\).

In our case for the line given by \(y=-\sqrt{3}*x-6\): X-intercept will be \((-2\sqrt{3}, \ 0)\), as x-intercept of a line is the value of x for y=0, the point where the line crosses X-axis: y=0 --> \(x=-2\sqrt{3}\);

Y-intercept will be \((0, \ -6)\), as y-intercept of a line is the value of y for x=0, the point where the line crosses Y-axis: x=0 --> \(y=-6\).

But why the tan(@) formula works on this. As Y axis has a slope = infinity. So this formula doesn't work and i had chosen last option which says "None of the above". I know this option may not be there in GMAT but why can't we apply a particular formula to reach the same result.

Tan(@) = |(m1 - m2)/(1+m1*m2)|

Where m1 and m2 are the slopes of two lines and "@" is angle between these line.

The angle made by the line \sqrt{3}x+y+6=0, with y-axis is

1. 60 degrees 2. 30 degrees 3. 45 degrees 4. 120 degrees 5. None of the above

Or you can notice that as the slope, which is th ratio of the rise over run is equal to \(-\sqrt{3}\) then the ratio of the length of the leg at y-axis over the length of the leg at x-axis will be \(\sqrt{3}\), which again tells us that we have 30°, 60°, and 90° right triangle.

Hope it's clear.

Hi Bunuel , Sorry to dig in an old post . I am unable to follow your second approach properly (i.e how did you find the ratio length of y to length of x as sqrt3) . Could you please elaborate it more .

The explanation of this question is fantastic and one can quickly solve these kind of questions after understanding it. The only concept that i am not able to understand is how \(2\sqrt{3}:6\) turn into \(1:\sqrt{3}\)

I will really appreciate if anyone can help me in understanding this concept.

Convert the given equation in this form to get y=-rt3x -6.

Since quesn deals in slope you need not calculate intercepts, clearly slope is negative and is tan(rt30 which is 60 degree. It lies in 3rd quadrant Since slope is defined as what angle a line makes with x -axis, the angle with y axis is 90-60 = 30.

If you want to know about intercepts, the above eqn cut x axis when y=0 rt? So when y=0, x = -6/rt3 = 3* 2 / rt3 = -2rt3. Also when x=0, y=-6.

Re: The angle made by the line root{3}x+y+6=0, with y-axis is [#permalink]
12 Oct 2013, 06:01

"The explanation of this question is fantastic and one can quickly solve these kind of questions after understanding it. The only concept that i am not able to understand is how 2\(\sqrt{3}\):6 turn into 1:\(\sqrt{3}\)"

here goes first cancel with 2 on either side of the ratio \(\sqrt{3}\) : 3 on the right 3 can be written as \(\sqrt{3}\)*\(\sqrt{3}\) \(\sqrt{3}\) : \(\sqrt{3}\)* \(\sqrt{3}\) cancel \(\sqrt{3}\) you get 1: \(\sqrt{3}\)

Re: The angle made by the line root{3}x+y+6=0, with y-axis is [#permalink]
12 Nov 2013, 03:07

Another idea: Observe buneul's diagram. The coordinates on the -ve x-axis and -ve y-axis are (-2\(\sqrt{3}\),0) and (0,-6). Now find the distance between them using the formula \(\sqrt{x^2+y^2}\). We get 4\(\sqrt{3}\)=2*2\(\sqrt{3}\). Now we know that the ratio of sides for a 30-60-90 triangle is x:\(\sqrt{3}\)x:2x. From distance calculation above we can see that the following ratio holds perfectly good 30:60:90=2\(\sqrt{3}\): \(\sqrt{3}\)*(2\(\sqrt{3}\)): 2*(2\(\sqrt{3}\)). Comparing we can say that side opposite to 2\(\sqrt{3}\) measures 30 degrees.

gmatclubot

Re: The angle made by the line root{3}x+y+6=0, with y-axis is
[#permalink]
12 Nov 2013, 03:07

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