Question

Determine the ratio In which the straight line x-y-2=0 divide the line segment joining (3,-10)&(8,9)

Answer 1

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Answer 2

Answer:

Line x-y-2=0 divide the line segment joining (3,-10) and (8,9) in the ratio 11 :3.

Step-by-step explanation:

To find : Determine the ratio In which the straight line x-y-2=0 divide the line segment joining (3,-10) and (8,9) ?

Solution :

Let A=(3,-10) and B=(8,9)

Let the line x-y-2=0 divides the line segment AB at point C(x,y) in the ratio m:n.

Applying section formula,

$C(x,y)=[\frac{m\times 8+n\times 3}{m+n},\frac{m\times 9+n\times (-10)}{m+n} ]$

$C(x,y)=[\frac{8m+3n}{m+n},\frac{9m-10n}{m+n} ]$

Since point C(x,y) lie on the line x-y-2=0

So, $\frac{8m+3n}{m+n}-\frac{9m-10n}{m+n}-2=0$

$8m+3n-9m+10n-2m-2n=0$

$-3m+11n=0$

$11n=3m$

$\frac{m}{n}=\frac{11}{3}$

Hence, line x-y-2=0 divide the line segment joining (3,-10) and (8,9) in the ratio 11 :3.