The length of a line segment $A B$ is $38 c m$ . A point $X$ on it divides it the ratio $9 : 10$ . Find the length of the line segments $A X$ and $X B$

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Solution

Let the length of $A X$ and $X B$ be $9 k$ and $10 k$ respectively.
Then $9 k + 10 k = 38$
$\Rightarrow 19 k = 38$
$\Rightarrow k = 2$
So, Length of $A X = 9 k$ $=$ $9 \times 2$ $= 18$ $c m$
Length of $X B = 10 k = 10 \times 2 = 20$ $c m$

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The length of a line segment AB is 38 cm. A point X on it divides it the ratio 9:10. Find the length of the line segments AX and XB

Let the length of AX and XB be 9k and 10k respectively.Then 9k+10k=38⇒19k=38⇒k=2So, Length of AX=9k = 9×2 =18 cmLength of XB=10k=10×2=20 cm

The length of a line segment $A B$ is $38 c m$ . A point $X$ on it divides it the ratio $9 : 10$ . Find the length of the line segments $A X$ and $X B$

Let the length of $A X$ and $X B$ be $9 k$ and $10 k$ respectively.
Then $9 k + 10 k = 38$
$\Rightarrow 19 k = 38$
$\Rightarrow k = 2$
So, Length of $A X = 9 k$ $=$ $9 \times 2$ $= 18$ $c m$
Length of $X B = 10 k = 10 \times 2 = 20$ $c m$