The distance from the external point ^2 = The external segment of the secant * the full length of the secant.
tangent distance = 6
External portion of the secant = 3
Full length of the secant = x + 3

6^2 = 3*(x + 3) Remove the brackets.
36 = 3x + 9 Subtract 9 from both sides.
36 - 9 = 3x
27 = 3x Divide by 3
x = 9 <<<<<<<Answer.

7 0

3 years ago

Find the value of x so that (-2, 4) is the midpoint between (x, 2) and (-5,3).

QveST [7]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{x}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-5}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{-5+x}{2}~~,~~\cfrac{3+2}{2} \right)~~=~~\stackrel{midpoint}{(-2,4)}\implies \begin{cases} \cfrac{-5+x}{2}=-2\\[1em] -5+x=-4\\ \boxed{x=1} \end{cases}$

6 0

3 years ago

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