A straight line through the point \((2,2)\) intersects the line \(\sqrt{3 x}+y=0\) and \(\sqrt{3 x}-y=0\) at the points \(\mathrm{A}\) and \(B\). The equation to the line \(\mathrm{AB}\) so that the triangle \(\mathrm{OAB}\) is equilateral is (a) \(x=2\) (b) \(\mathrm{y}=2\) (c) \(\mathrm{x}+\mathrm{y}=4\) (d) none

Short Answer

Expert verified
The correct answer is (d) None, as none of the given options form an equilateral triangle OAB when intersecting the lines \(\sqrt{3 x}+y=0\) and \(\sqrt{3 x}-y=0\).

Step by step solution

01

Find the intersection points of the lines with the given point(2,2)

Since both lines intersect at point (2,2), we can substitute the x and y coordinates of this point into the equations for the lines: For the line \(\sqrt{3 x}+y=0\), substitute (2,2): \( \sqrt{3(2)} + 2 = 0 \) For the line \(\sqrt{3 x}-y=0\), substitute (2,2): \( \sqrt{3(2)} - 2 = 0 \) Observe that both equations are satisfied, so the points A and B are the intersections of the lines with point (2,2).
02

Test each option for forming an equilateral triangle OAB

We will now test each of the given options to see which one forms an equilateral triangle OAB. (a) \(x=2\) This equation represents a vertical line passing through the point (2,2). Since it is impossible to form an equilateral triangle on a vertical line, this is not the correct option. (b) \(y=2\) This equation represents a horizontal line passing through the point (2,2). In this case, since points A and B lie on the lines \(\sqrt{3 x}+y=0\) and \(\sqrt{3 x}-y=0\), the points A and B will have the same y-coordinate (2). Also, the distance between the point (0,0) and A≠the distance between the point (0,0) and B. So, it is not possible to form an equilateral triangle OAB on this line. (c) \(x+y=4\) This line passes through the point (2,2), and we need to find the coordinates of points A and B on this line. Substitute the equation of line \(x+y=4\) into the equation of lines \(\sqrt{3 x}+y=0\) and \(\sqrt{3 x}-y=0\): For the line \(\sqrt{3 x}+y=0\): \( \sqrt{3(4-y)} + y = 0 \) For the line \(\sqrt{3 x}-y=0\): \( \sqrt{3(4-y)} - y = 0 \) Solving these equations, we find that neither of the equations has a real solution. This indicates that the intersection of line \(x+y=4\) with the lines \(\sqrt{3 x}+y=0\) and \(\sqrt{3 x}-y=0\) does not exist. Hence, this option is not correct either. (d) None Since none of the given options form an equilateral triangle, the correct answer is (d) None.

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