The coordinates of points A and C are (0, -3) and (3, 3), respectively

Thanks Bunuel - I didn't get it sorry. How did you arrive at 2 as the co-ordinate for B? Sorry again. Also, I was trying to solve this by using the distance formula for AC.

I just put all five points on a plane and saw that the only acceptable answer is C (2, 1). Look at other 4 points (blue) on the diagram and read my explanation: neither of them can divided AC into ratio 2:1.

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y = mx + n

n = -3 (from point A)
3 = 3m -3 (from point C)
m=2
Equation of line AC: y = 2x - 3

Now put in the given answers. None works except from C and D. Both points lie on the line. From the question we know that AB = 2BC. So lets calculate the middle of the line AC. (Xa+Xb)/2 = 0+3 / 2 = 1.5 and (Ya+Yb)/2 = -3+3 / 2 = 0

SO we have M = (1.5;0) which is answer D. Hence answer C is correct.

Want a really long way to solve this? No fancy formulas no estimation on paper.

We have from the known coordinates that the big triangle has sides 3, 6, 3BC.

So, \(3^2+6^2=3BC^2\)

We get that \(bc=\sqrt{5}\)

Note that now we have a smaller triangle

Then, using the proportions given, so AB=2BC, we can determine that the h of the small triangle is 2.
We can also calculate the other side since we know the hypotenuse (\(\sqrt{5}\))

So: \(\sqrt{5}=2^2+x^2\) X=1

Then we can use the same concept of proportions AB=2BC to determine that\(\sqrt{5}\) segment corresponds
to 1\3rd of the big triangle. So, the distance from origin on X-axis to the side (base) of the small triangle is 2. And that means that C (2,y). We got our X.

Now get the linear equation of the hypotenuse of the big triangle. Using the given points A and B, we get that slope=6/3 or 2 and y-intercept is -3

So y=2x-3

Now plug-in the X=2 to get that Y=1

Answer C. Logical deduction + Pythagoras Theorem only.

Explanantion: It’s a Section Formula problem (internal division case).

B (x’, y’) is such that

x’ = {mb+ na}/(m + n); y’ = {md + nc}/ (m + n).

In the given situation, a = 0, b = 3, c = -3, d = 3, m = 2 and n = 1.

we have x’ = (2 X 3 + 1 X 0)/3 = 2, and y’ = (2 X 3 + 1 X -3)/3 = 1,

the coordinates of point B are (2, 1).

IMO-C

AccipiterQ

Found this on Magoosh on reflected points. Might help you

https://magoosh.com/math/coordinate-geo ... ate-plane/

Hey! So first with all coordinate plane problems you want to draw out the given information on a coordinate plane.

You will find in most cases that coordinate plane problems can be solved simply by drawing out a right triangle from the given line segment/hypotenuse, which in this case is line segment AC with A and C having coordinates (0,-3) and (3,3), respectively.

You can do this by drawing out perpendicular lines to the right from A and down from C that will meet at a point, forming a right triangle.

When you do this you can determine the lengths of the triangle legs using the coordinate values or by simply counting the coordinate plane boxes. You see that A is length 3 and B is length 6.

Next you use the Pythagorean theorem to determine the length of hypotenuse C, the line segment created by points A and C given to us by the problem.

\(a^2 + b^2 = c^2\)

\((3)^2 + (6)^2 = c^2\)

\(9 + 36 = c^2\)

\(C = \sqrt{45}\)

\(\sqrt{45}\) can be broken down to \(3\sqrt{5}\) because \(\sqrt{45}\) = \(\sqrt{9}\) x \(\sqrt{5}\) and \(\sqrt{9}\) can be simplified to 3 leaving you with 3 x \(\sqrt{5}\) or \(3\sqrt{5}\).

So now we know that the length of line segment AC is \(3\sqrt{5}\).

We also know that there is a point B on this line segment that divides the line segment into AB and BC. We are told that AB = 2BC. AB is twice as long as BC.

Since we don't know the value of BC or AB, it is useful here to represent BC with x. This gives us BC = x and AB = 2x. If the entire length of AC = AB + BC, substitute in the x values to get AC = 2x + x = 3x.

Now we know that AC = \(3\sqrt{5} \) and AC = 3x. Combine the two formulas to solve for x.

\(3\sqrt{5} = 3x\)

\(x = \sqrt{5}\)

That means that \(AB = 2x = 2\sqrt{5}\).

Why did we go through this whole process? Because now that we have the length of AB, we have the length of the hypotenuse of a smaller right triangle that is inscribed in the larger original right triangle, and we can use the Pythagorean theorem again to give us an idea of the length of the smaller right triangle legs, which will give us the coordinates of point B.

\(c^2 = a^2 + b^2\)

\((2\sqrt{5})^2 = a^2 + b^2\)

(\(2^2\) x \(\sqrt{5}^2\)) = \(a^2 + b^2\) = (4 x 5) = 20

Now we know that \( a^2 + b^2 = 20\) in the smaller triangle, how can this help us determine the coordinates of point B?

We know that the legs of the original larger right triangle are length 3 and 6. That means they are in a 1:2 ratio.

Because this smaller right triangle is inscribed in the larger one, the two triangles must be similar triangles and their legs must be in the same ratio.

What two values are in a 1:2 ratio that, when squared and added to each other are equal to 20?

2 and 4:

\(2^2 + 4^2 = 4 + 16 = 20\)

Therefore, the lengths of the legs of the smaller right triangle are 2 and 4.

Just like you drew the legs of the original larger right triangle 3 values to the right of point A and then 6 values up (or 6 values down from point C), you can draw these smaller legs 2 values to the left of point A and 4 values up. Since point A has coordinates (0, -3), you get (0 + 2, -3 + 4) or (2, 1).

The final coordinates for point B are therefore (2,1), which is answer choice C.

You could also solve by realizing that AB would be a hypotenuse that is \(\frac{2}{3}\) the length of AC based on the information provided by the question (AB = 2BC) and therefore the length of the smaller legs would be in the same ratio or \(\frac{2}{3}(3) = 2\) and \(\frac{2}{3}(6) = 4\) and then use those lengths to determine the coordinates of point B.

You could also visually solve by drawing the line segment and looking at where \(\frac{2}{3 }\)of the line AC going from A to B would be, but I'm sure you know by now that as much as it can sometimes be helpful in a pinch or when you're not sure, going off visuals alone often backfires on the GMAT.

Haven't you put the two points (1, -root5) and (root5, root5) considering them as 5? I would approximate them to a little more than 2; how can we ensure to the (root5, root5) then?

I created the equation otherwise but both b and c are valid that way; clearly not the most reliable way here.

Hy narren,Can u please help me out with n.I did not get what n is.

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A simple solution to this would be to only check the X coordinate.
The X coordinate travels 3 units, from 0 to 3
Since point B divides AC in the ratio is 2:1, we can easily spot that point B would have X coordinate as 2
The only such option is C

GMATNinja
Can you pls tell another approach for this ques , othe then already mentioned in the forum.

It would be great help.

Thanks in advance

To Find Coordinates of B (x, y)

Draw a line, DC || EB || X Axis, thus triangles ACD and ABE are similar due to AAA Property.

So, \(AB/AC = AE/AD\) => \(2/3 = AE/AD \), As AD = 6, then AE = 4, Thus Coordinate for E will be (0, 1).

So, y=1.

Similarly draw a line || y axis passing through point B, and calculate x = 2.

Answer C

Two points: (0,-3), (3,3)

ratio of length of AB:BC= 2:1

For x co-ordinate of point B:-
Length of line on x-axis: 3-0= 3.
Thus, length of AB on x-axis= 2/(2+1)= 2
Adding length of AB to co-ordinate of A; x- Co-ordinate of point B= 0+2= 2.

Similarly, to find y co-ordinate of point B:-
Length of AB on y-axis; 3-(-3)= 6; 6/(2+1)=4
And y co-ordinate of point B; -3+4=1

Thus co-ordinates of point B: (2,1)

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