Point M(1,y) lies on the line …

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Sia ? 6 years, 4 months ago
Let the required ratio be k:1.
Then, by the section formula, the coordinates of P are
{tex}P \left( \frac { 9 k + 15 } { k + 1 } , \frac { 20 k + 5 } { k + 1 } \right){/tex}
But, this point is given as P(11, y).
{tex}\therefore{/tex}{tex}\frac { 9 k + 15 } { k + 1 } = 11{/tex} {tex}\Rightarrow{/tex} 9k + 15 = 11k + 11 {tex}\Rightarrow{/tex} 2k = 4 {tex}\Rightarrow{/tex} k = 2
So, the required ratio is 2:1
Putting k = 2 in P, we get
{tex}y = \frac { 20 \times 2 + 5 } { ( 2 + 1 ) } = \frac { 45 } { 3 } = 15{/tex}
Hence, y = 15.
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