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Q10 The distance between the points (4,3) and (2,y) is 8 . What is/are the coordinate(s) of y ?

Q1010\newlineThe distance between the points (4,3) (4,3) and (2,y) (2, y) is 88 . What is/are the coordinate(s) of y y ?

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Q. Q1010\newlineThe distance between the points (4,3) (4,3) and (2,y) (2, y) is 88 . What is/are the coordinate(s) of y y ?
  1. Identify Given Points and Formula: Identify the given points and the distance formula.\newlineThe given points are (4,3)(4,3) and (2,y)(2,y), and the distance between them is 88. The distance formula is:\newlineDistance =((x2−x1)2+(y2−y1)2)= \sqrt{((x_2 - x_1)^2 + (y_2 - y_1)^2)}\newlineWe need to find the value of yy such that the distance between the points is 88.
  2. Plug Known Values: Plug the known values into the distance formula.\newline8=((2−4)2+(y−3)2)8 = \sqrt{((2 - 4)^2 + (y - 3)^2)}\newlineSimplify the equation by squaring both sides to eliminate the square root.\newline82=(2−4)2+(y−3)28^2 = (2 - 4)^2 + (y - 3)^2\newline64=(−2)2+(y−3)264 = (-2)^2 + (y - 3)^2
  3. Simplify Equation: Simplify and solve for (y−3)2(y - 3)^2.64=4+(y−3)264 = 4 + (y - 3)^2Subtract 44 from both sides to isolate (y−3)2(y - 3)^2.64−4=(y−3)264 - 4 = (y - 3)^260=(y−3)260 = (y - 3)^2
  4. Find Possible Values: Find the possible values of yy. Since (y−3)2=60(y - 3)^2 = 60, we take the square root of both sides to solve for y−3y - 3. y−3=±60y - 3 = \pm\sqrt{60} y−3=±4⋅15y - 3 = \pm\sqrt{4\cdot15} y−3=±215y - 3 = \pm2\sqrt{15} Now, solve for yy by adding 33 to both sides. y=3±215y = 3 \pm 2\sqrt{15}

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