Question

$\huge \tt \pink{Question↓ }$
Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

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Answer 1

Step-by-step explanation:

Let k : 1 be the ratio

By section formula,

6k -3 / K + 1 = - 1 - 8k + 10 / K + 1 = 6

- 8k + 10 = 6k + 6

14k = 4

K = 4 /14 = 2 : 7

Answer 2

$\huge \tt \pink{AnsWer↓ }$

We have to check whether the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6). Let P(- 1, 6) divides AB in the ratio m1 : m2.

By using Section formula,

Here x₁ = -3, y₁ = 10, x₂ = 6, y₂ = -8  

  [m(6) + n (-3)]/(m+n) , [m (-8) + n (10)]/(m+n) = (-1,6) 

        (6m - 3n)/(m+n), (-8m+10n)/(m+n) = (-1,6)

 equating the coefficients of x and y

    (6m-3n)/(m+n) = -1       (-8m+10n)/(m+n) = 6

  →   6m -  3n = -1 (m+n)

→     6m - 3n = -1 (m+n)

→     6m-3n=-m-n

→     6m+m=-n+3n

→       7m=2n

 →       m/n= 2/7

→        m:n = 2:7

∴ P(-1, 6) point divides the line segment in the ratio 2 : 7.

∴ P(-1, 6) point divides the line segment in the ratio 2 : 7.