Question

find the ratio in which the line segment joining the points (-3,10) and (6,-8)isdivided by (-1,6)

Answer 1

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

So the point (-1,6) is dividing the line segment in the ratio 2:7.

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Answer 2

Consider points as A(-3,10), B(6,-8), C(-1,6)
Use section formula : $\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}$
Substitute the values of $x_1 , x_2$
so, $\frac{m(-3)+n(6)}{m+n}=-1$
which implies, [tex]-3m+6n=-m-n [/tex]
$-2m=-7n$
implies,
$\frac{m}{n} = \frac{7}{2}$
Therefore, the point divides the line in ratio 7:2