Question

Find the ratio in which the line segment joining the points (8,7) and (-2,2) is divided by (0,3).

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Answer 1

Step-by-step explanation:

let the ratio be k:1

therefore,

p(x, y)= $( \frac{ - 2 \times k + 8 \times 1}{k + 1} ),( \frac{2 \times k + 7 \times 1}{k + 1} )$

$x \: = \: \frac{ - 2k + 8}{k + 1} \\ = > \: 0 = \frac{ - 2k + 8}{k + 1} \\ = > - 2k + 8 = 0 \\ = > 2k = 8 \\ = > k = 4$

$y = \frac{2k + 7}{k + 1} \\ = > 3 = \frac{2k + 7}{k + 1} \\ = > 3k + 3 = 2k + 7 \\ = > 3k - 2k = 7 - 3 \\ = > k = 4$

therefore, ratio = k:1

=4:1

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Answer 2

Given :

The ratio in which the line segment joining the points (8,7) and (-2,2) is divided by (0,3).

To Find :

The ratio.

Solution :

Analysis :

Here the section formula is used. We first have to assume the ratio then substitute the required formulas in the formula and equate it to find the ratio.

Required Formula :

Section Formula,

$\normalsize\boxed{\bf(x,y)=\left(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2}\right)}$

where,

• (x, y) = Coordinates of point formed
• (x₁, y₁) = Coordinates of first point
• (x₂, y₂) = Coordinates of second point

Explanation :

• Let the point (8, 7) be P.
• Let the point (2, -2) be Q.
• Let the point (0, 3) be PQ.

• PQ = (0, 3)

Let us assume that the ratio is k : 1.

• k = m₁
• 1 = m₂

☯ According to the question,

$\\ :\implies\normalsize\sf(x,y)=\left(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2}\right)$

where,

• x = 0
• y = 3
• x₁ = 8
• x₂ = -2
• y₁ = 7
• y₂ = 2
• m₁ = k
• m₂ = 1

Using the required formula and substituting the required values,

$\\ :\implies\normalsize\sf(0,3)=\left(\dfrac{k.-2+1.8}{k+1},\dfrac{k.2+1.7}{k+1}\right)$

$\\ :\implies\normalsize\sf(0,3)=\left(\dfrac{-2k+8}{k+1},\dfrac{2k+7}{k+1}\right)$

Here,

$\\ :\implies\normalsize\sf0=\left(\dfrac{-2k+8}{k+1}\right)\qquad\qquad\Bigg\lgroup\bf eq.(i)\Bigg\rgroup$

and

$\\ :\implies\normalsize\sf3=\left(\dfrac{2k+7}{k+1}\right)\qquad\qquad\Bigg\lgroup\bf eq.(ii)\Bigg\rgroup$

From eq.(i) :

$\\ :\implies\normalsize\sf0=\left(\dfrac{-2k+8}{k+1}\right)$

By cross multiplying,

$\\ :\implies\normalsize\sf0\times k+1=-2k+8$

$\\ :\implies\normalsize\sf0=-2k+8$

Transposing 8 to LHS,

$\\ :\implies\normalsize\sf0-8=-2k$

$\\ :\implies\normalsize\sf-8=-2k$

$\\ :\implies\normalsize\sf\dfrac{-8}{-2}=k$

$\\ :\implies\normalsize\sf\cancel{\dfrac{-8}{-2}}=k$

$\\ :\implies\normalsize\sf4=k$

$\\ \normalsize\therefore\boxed{\bf k=4.}\qquad\bf(eq.(iii))$

From eq.(ii) :

$\\ :\implies\normalsize\sf3=\left(\dfrac{2k+7}{k+1}\right)$

By cross multiplying,

$\\ :\implies\normalsize\sf3(k+1)=2k+7$

Expanding the brackets,

$\\ :\implies\normalsize\sf3k+3=2k+7$

Transposing 2k to LHS and 3 to RHS,

$\\ :\implies\normalsize\sf3k-2k=7-3$

$\\ :\implies\normalsize\sf k=4$

$\\ \normalsize\therefore\boxed{\bf k=4.}\qquad\bf(eq.(iv))$

From eq.(iii) and eq.(iv),

The value of k is same.

Therefore, k = 4.

The ratio is k : 1 = 4 : 1.

The ratio dividing the line is 4 : 1.