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Point HH is located at (2,7)(2,-7). Select all of the following that are 77 units from point HH. \newlineChoose all answers that apply:\newlineA x-axis\newlineB (5,7)(-5,-7)\newlineC (7,7)(-7,7)

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Q. Point HH is located at (2,7)(2,-7). Select all of the following that are 77 units from point HH. \newlineChoose all answers that apply:\newlineA x-axis\newlineB (5,7)(-5,-7)\newlineC (7,7)(-7,7)
  1. Distance Formula Application: To find points that are 77 units away from point H(2,7)H(2,-7), we need to consider the distance formula, which is d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, where dd is the distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2). We are looking for points where d=7d = 7.
  2. Check Option A: Let's first consider option A, the xx-axis. The xx-axis has the equation y=0y = 0. To find the distance from point HH to the xx-axis, we only need to consider the yy-coordinate of HH, which is 7-7. The distance from HH to the xx-axis is therefore xx00 units, since the xx-coordinate does not affect the vertical distance to the xx-axis.
  3. Check Option B: Now let's check option B, the point (5,7)(-5,-7). We use the distance formula: d=(2(5))2+(7(7))2d = \sqrt{(2 - (-5))^2 + (-7 - (-7))^2}. Simplifying, we get d=(2+5)2+02d = \sqrt{(2 + 5)^2 + 0^2}, which is d=72d = \sqrt{7^2}. This simplifies to d=7d = 7. So, point (5,7)(-5,-7) is 77 units away from point HH.
  4. Check Option C: Finally, let's check option C, the point (7,7)(-7,7). Again, we use the distance formula: d=(2(7))2+(77)2d = \sqrt{(2 - (-7))^2 + (-7 - 7)^2}. Simplifying, we get d=(2+7)2+(77)2d = \sqrt{(2 + 7)^2 + (-7 - 7)^2}, which is d=92+(14)2d = \sqrt{9^2 + (-14)^2}. This simplifies to d=81+196d = \sqrt{81 + 196}, which is d=277d = \sqrt{277}. The value of 277\sqrt{277} is not 77, so point (7,7)(-7,7) is not 77 units away from point d=(2(7))2+(77)2d = \sqrt{(2 - (-7))^2 + (-7 - 7)^2}00.

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