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- ...s a distance <math>r</math> from <math>(h,k)</math>. Using the [[distance formula]], this gives <math>\sqrt{(x - h)^2 + (y - k)^2} = r</math> which is more c ...ath>A</math> and <math>B</math>, respectively. Let <math>M</math> be the [[midpoint]] of <math>AB</math>. The [[perpendicular bisector]] of <math>AM</math> mee9 KB (1,585 words) - 12:46, 2 September 2024
- ...angle]] is a [[cevian]] of the triangle that joins one [[vertex]] to the [[midpoint]] of the opposite side. M = (midpoint(B--C));1 KB (185 words) - 19:24, 6 March 2024
- ...ase [[equilateral triangle]] <math>ABC</math>. Let <math>M</math> be the [[midpoint]] of segment <math>BC</math>. Let <math>h</math> be the distance from <math Applying the distance between a point and a plane formula.6 KB (980 words) - 20:45, 31 March 2020
- and applying the [[quadratic formula]] we get that Let the midpoint of <math>\overline{AB}</math> be <math>M</math> and let <math>FB = x</math>13 KB (2,080 words) - 12:14, 23 July 2024
- ...ength 2 and whose [[endpoint]]s are on adjacent sides of the square. The [[midpoint]]s of the line segments in set <math> S </math> enclose a region whose [[ar ...to <math>\left(\frac{x}{2},\frac{y}{2}\right)</math>. Using the [[distance formula]] we see that <math>d=\sqrt{\left(\frac{x}{2}\right)^2+\left(\frac{y}{2}\ri3 KB (532 words) - 08:22, 11 July 2023
- Euler's formula states that for a convex polyhedron with <math>V\,</math> vertices, <math>E ...is the most recently obtained point, then <math>P_{n + 1}^{}</math> is the midpoint of <math>\overline{P_n L}</math>. Given that <math>P_7 = (14,92)\,</math>,8 KB (1,275 words) - 05:55, 2 September 2021
- If we let <math>M</math> be the midpoint of <math>AC</math>, that mean that <math>AM = \sqrt{10}</math>. Since <mat Also, mark the midpoint <math>M</math> of <math>AC</math>.11 KB (1,747 words) - 20:07, 8 December 2024
- First, we find the height of the solid by dropping a perpendicular from the midpoint of <math>AD</math> to <math>EF</math>. The hypotenuse of the triangle forme .... This tetrahedron has side length <math>2s = 12\sqrt{2}</math>. Using the formula for the volume of a regular tetrahedron, which is <math>V = \frac{\sqrt{2}S6 KB (971 words) - 14:35, 27 May 2024
- ...cts the small circle. This fact can be derived from the application of the Midpoint Theorem to the trapezoid made by dropping perpendiculars from the centers o Since <math>P</math> is the midpoint of segment <math>BC</math>, <math>AP</math> is a median of <math>\triangle14 KB (2,351 words) - 20:06, 8 December 2024
- ...<math>\ell,</math> what is the set of points <math>X</math> such that the midpoint of <math>PX</math> lies on line <math>\ell</math>? The answer to this ques ...>OD</math> is perpendicular to <math>BC</math>. Let <math>M</math> be the midpoint of chord <math>BC,</math> and let <math>N</math> be the intersection of <ma20 KB (3,497 words) - 14:37, 27 May 2024
- Draw the [[midpoint]] of <math>\overline{AC}</math> (the centers of the other two circles), and ...re <math>(0,0)</math> and <math>(-\frac{1}{2},12)</math>. We use the slope formula to calculate the slope, which is <math>-24</math>, leading to an answer of6 KB (1,022 words) - 18:29, 22 January 2024
- ...onometric_identities#Angle_addition_identities | tangent angle subtraction formula]], ...ath>M</math> be the midpoint of <math>BC,</math> let <math>N</math> be the midpoint of <math>AC,</math> and let <math>G</math> be the intersection of these two11 KB (1,722 words) - 08:49, 13 September 2023
- ...x' - x} = \frac{-1}{2} \Longrightarrow x' + 2y' = x + 2y</math>. Also, the midpoint of <math>\overline{PP'}</math>, <math>\left(\frac{x + x'}{2}, \frac{y + y'}4 KB (700 words) - 16:21, 3 May 2021
- ...PC]}{[APB]} = \dfrac{CD}{DB}</math>, this means that <math>D</math> is the midpoint of <math>BC</math>. === Solution 2 (Mass Points, Stewart's Theorem, Heron's Formula) ===14 KB (2,234 words) - 15:31, 22 December 2024
- ..., as shown in the figure. Let <math>d</math> be the distance between the [[midpoint]]s of [[edge]]s <math>AB</math> and <math>CD</math>. Find <math>d^{2}</math C=origin; B=(8,0); D=IP(CR(C,6.5),CR(B,8)); A=(4,-3); P=midpoint(A--B); Q=midpoint(C--D);2 KB (376 words) - 12:49, 1 August 2022
- From here we just need to perform some brutish calculations. Using the formula <math>A = 18\sqrt{133} = \frac{abc}{4R}</math> (where <math>R</math> is the ...{\sqrt {939}}{2}</math>, we can write three equations using the [[distance formula]]:7 KB (1,086 words) - 07:16, 29 July 2023
- Use the [[distance formula]] to determine the lengths of each of the sides of the triangle. We find th ...ath>. Since the [[angle bisector]] of <math>\angle P</math> must touch the midpoint of <math>QS \Rightarrow (-4,-17)</math>, we have found our two points. We r8 KB (1,319 words) - 10:34, 22 November 2023
- ...and the perimeter of the rectangle is equal to <math>2(BC+CD')</math>. The midpoint of <math>\overline{BD'}</math> is <math>(-4-x,0)</math>, and since <math>-4 By the distance formula, this minimum perimeter is <cmath>2\left(\sqrt{(4-\sqrt7)^2+3^2}+\sqrt{(4+\3 KB (601 words) - 08:25, 19 November 2023
- label("$50$",midpoint(D--O),0.5NE); label("$x$",midpoint(A--J),0.5W);8 KB (1,388 words) - 03:38, 27 August 2024
- ...math>, such that <math>AE = CE < BE = DE</math>. Let <math>F</math> be the midpoint of <math>\overline{AB}</math>. Then <math>\overline{OF} \perp \overline{AB} ...c <math>\stackrel{\frown}{BC}</math>. The former can be found by [[Heron's formula]] to be <math>[BCE] = \sqrt{60(60-48)(60-42)(60-30)} = 360\sqrt{3}</math>.3 KB (484 words) - 12:11, 14 January 2023