Question

. In what ratio does the point (-4 ,6) divide the line segment joining the points (-6, 10) and (3, 6).​

Answer 1

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In what ratio does the point $\sf (-4,6)$ divide the line segment joining the points $\sf (-6,10)$ and $\sf (3,6)$?

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Here, we are given three points as follows :

• $\green{\textsf{\textbf{A:(-6,10)}}}$
• $\green{\textsf{\textbf{B:(3,6)}}}$
• $\green{\textsf{\textbf{C:(-4,6)}}}$

Now, we shall find in what ratio the point C divides the line joining the points A and C.

We should find the ratio between AC and CB :

Let the required ratio be : $\green{\sf{k:1}}$.

So, now :

• $\sf m_1: k$
• $\sf m_2: 1$
• $\sf x_1: -6$
• $\sf x_2: 3$
• $\sf y_1:10$
• $\sf y_2:6$
• $\sf x:-4$
• $\sf y:6$

Using the section formula :

$\green{:\implies{\pmb{\sf{x = \dfrac{ m_{1} x_{2} + m_{2} x_{1}}{ m_{1} + m_{2}}}}}}$

$\sf: \implies - 4 = \dfrac{k(3) + 1( - 6)}{k + 1}$

$\sf: \implies - 4 = \dfrac{3k - 6}{k + 1}$

$\sf: \implies - 4(k + 1) = 3k - 6$

$\sf: \implies - 4k - 3k = - 6 + 4$

$\sf: \implies - 7k = -2$

$\green{:\implies{\pmb{\sf{k = \dfrac{2}{7}}}}}$

Now, substituting the value of k in k : 1 :

$\sf : \implies \dfrac{2}{7} : 1$

We'll multiply the ratio with 7 to get the non-fractional values :

$\sf: \implies7 (\dfrac{2}{7}) : 1(7)$

$\green{:\implies{\pmb{\sf{2 : 7}}}}$

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Therefore, the point (-4,6) divides the line segment joining the points (-6,10) and (3,6) in the ratio of 2 : 7.

Answer 2

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