Question

Find The Ratio In Which the point P(2,1) divide the line joining the points A(2,7) and B(2,-3).
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Answer 1

$\LARGE{ \underline{\underline{ \sf{Required \: answer:}}}}$

GiveN:

• The points are A(2,7) and B(2,-3).
• The point which divides them is P(2,1)

To Find:

• The ratio in which P divides AB?

Step-by-step Explanation:

Let the ratio be k:1

Hence, m1 = k and m2 = 1

Also, if we see the points:

• x1 = 2 and y1 = 7
• x2 = 2 and y2 = -3
• x = 2 and y = 1

Using Section formula,

$\cdot{ \boxed{ \rm{x = \frac{m1x2 + m2x1}{m1 + m2} \:}} \boxed{ \rm{ \: y = \frac{m1y2 + m2y1}{m1 + m2} }}}$

Putting the given values,

$\rm{1 = \dfrac{k( - 3) + 1(7)}{k + 1} }$

Cross multiplying,

⇒ k + 1 = -3k + 7

⇒ k + 3k = 6

⇒ 4k = 6

⇒ k = 3/2

The ratio is k : 1

⇒ 3/2 : 1

⇒ 3 : 2

Hence, the ratio is 3 : 2 (Ans)

Answer 2

Given

• A line joining the points A(2,7) and B(2,-3).
• Point P(2,1) divide the line.

To find

• The Ratio in which the point P(2,1) divide the line.

Solution

$\sf\pink{⟶}$ Let the ratio be k:1.

Here

• $x_1 = 2\: and\: x_2 = 2$
• $y_1 = 7\: and\: y_2 = -3$
• $m_1 = k\: and\: m_2 = 1$

$\sf\pink{⟶}$ Using the section formula

$\underline{\boxed{\tt{P(x,y) = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2},\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}}}}$

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$\tt:\implies{P(2.1) = \dfrac{2k + 2}{k + 1},\dfrac{(-3k) + 7}{k + 1}}$

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$\sf\pink{⟶}$ On comparing,

$\tt:\implies{\dfrac{2k + 2}{k + 1} = 2\: and\: \dfrac{(-3k) + 7}{k + 1} = 1}$

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$\tt:\implies{\cancel{2k + 2} = \cancel{2k + 2}\: and\: k + 1 = -3k + 7}$

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$\tt:\implies{k + 1 = -3k + 7}$

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$\tt:\implies{k + 3k = 7 - 1}$

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$\tt:\implies{4k = 6}$

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$\tt:\implies{k = \dfrac{6}{4}}$

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$\tt:\implies{k = \dfrac{3}{2}}$

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$\sf\pink{⟶}$ Therefore, the point P(2,1) divides the line AB in $\dfrac{3}{2}:1$, i.e., 3:2

$\sf\pink{⟶}$ Solution = 3:2

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