Find the ratio in which the …
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Sia ? 6 years, 5 months ago
Let A(-2, -3) and B(5, 6) be the given points.
Suppose x-axis divides AB in the ratio k:1 at point P
Then, the coordinates of the point of division are
{tex}P \left[ \frac { 5 k - 2 } { k + 1 } , \frac { 6 k - 3 } { k + 1 } \right]{/tex}
Since P lies on x-axis, and y-coordinates of every point on x-axis is zero.
{tex}\therefore \quad \frac { 6 k - 3 } { k + 1 } = 0{/tex}
{tex}\Rightarrow{/tex} 6k - 3=0
{tex}\Rightarrow{/tex} 6k = 3
{tex}\Rightarrow{/tex} {tex}k = \frac { 3 } { 6 } \Rightarrow k = \frac { 1 } { 2 }{/tex}
Hence, the required ratio is 1:2.
Putting {tex}K = \frac { 1 } { 2 }{/tex}in the coordinates of P.
We find that its coordinates are {tex}\left( \frac { 1 } { 3 } , 0 \right).{/tex}
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