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$${{\bullet\!\!\! -\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!\bullet\!\!\! -\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!\bullet}\atop O \;\quad\quad M\quad\quad\; P}$$

Given that $OM = x + 8$, $MP = 2x - 6$, $OP = 44$, is $M$ the midpoint of $OP$?

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We have that $$OM + MP = OP$$ so $$(x+8)+(2x-6)=3x+2=44$$ $$3x=42$$ $$x=14$$ Therefore $OM=x+8=22$ and $MP=2\cdot 14-6=28-6=22$. Because the lengths of $OM$ and $MP$ are equal, $M$ is the midpoint of the line $OP$.

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up vote 1 down vote accepted

HINT $$OM + MP = OP$$ Move your mouse over the gray area for the complete solution.

Note that $$OM + MP = OP$$ and hence we get that $3x+2 = 44 \implies x =14$. Hence, $OM = 14 + 8 = 22 = \dfrac{44}2 = \dfrac{OP}2$. Hence, $M$ is indeed the midpoint.

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Thanks for the answer, my only question is how did you find out that OM + MP = OP? – user34548 Jun 28 '12 at 2:19
@Michael Since $M$ is a point on the line, we have the distance from $O$ to $P$ is nothing but the sum of the distance from $O$ to $M$ and then from $M$ to $P$. – user17762 Jun 28 '12 at 2:21
Ah okay. So for my other problem where: PQ = x^2 +3 and QR = 4+2x and PR = 15 then it is safe to say that PQ + QR = 15? and in this case how would you proceed? – user34548 Jun 28 '12 at 2:27
If $PQR$ are along a line and $Q$ is inbetween $P$ and $R$, then $PQ + QR = PR$. If this is the case, then we get that $$x^2 + 3 + 4 + 2x = 15 \implies x^2 + 2x -8 = 0$$ Now can you find $x$, keeping in mind that $QR \geq 0$. – user17762 Jun 28 '12 at 2:29
Yes. So that means that x^2+x = 4. So x = 2? – user34548 Jun 28 '12 at 2:31

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