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Can someone please explain how I can get coordinates for B using the slopes that are shown in the question? I'm not sure how I came to my solution where I tried to use y=mx+b. For e.g. to solve for the x coordinate I put $x=(3/2)5$ using slope of AB and using x coordinate from (5,0) and got 15/2 then for the y coordinate of B I used $y=(1/2)(15/2)$ to get 15/4. Therefore got coordinates (15/2, 15/4). I believe I have a flaw in my logic. Could you please point out where I'm going wrong.

Thank you in advance!

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  • $\begingroup$ answer is (15/2, 15/4) $\endgroup$ – GMATnoob Dec 10 '14 at 9:06
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Equation of a line with given slope ($m$) passing through a given point $(a,b)$:

$$(y-b)=m(x-a)$$

Find intersection of two such lines one through origin and other through $A$.

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Step 1: using $y=mx+c $ for the line $OB$ we get,

$y=\frac{1}{2}x+0...(I)$

Step 2: using $y=mx+c$ forthe line $AB$ we get,

$y=\frac{3}{2}x+c$, here we have to find $c$, for that substitute $A(5,0)$ and we get,

$0=\frac{3}{2}\times 5+c\Rightarrow c=-\frac{15}{2}\Rightarrow y=\frac{3}{2}x-\frac{15}{2}...(II)$

solving I and II simultaneously,$\frac{1}{2}x=\frac{3}{2}x-\frac{15}{2}\Rightarrow x=\frac{15}{2}$

Substitute this value of $x$ in any one of I and II to get $y$

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Equation of OB : using $y=mx+c$ is, comes to be, y=$\frac{x}{2}$.

; and, equation of AB : using $(y-a)=m(x-b)$ is, y=$\frac{3(x-5)}{2}$.

by solving them, we get, x=$\frac{15}{2}$ and y=$\frac{15}{4}$

So, the point is ($\frac{15}{2}$,$\frac{15}{4}$)

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