Question

Determining the ratio in which the line 2x+y-4=0 divides the line segment joining the points A(2,-2) and B(3,7).​

Answer 1

A line AB with A(2,-2), B(3,7) is interested by a line 2x + y - 4 = 0.

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☯ Let the ratio at which line is divided be k : 1.

Using section formula,

$:\implies\sf P(x,y) = \bigg( \dfrac{kx_2 + x_1}{k + 1} , \dfrac{ky_2 + y_1}{k + 1} \bigg)$

$:\implies\sf P(x,y) = \bigg( \dfrac{k(3) + 2}{k + 1} , \dfrac{k(7) + (-2)}{k + 1}\bigg)$

$:\implies\sf P(x,y) = \bigg( \dfrac{3k + 2}{k + 1} , \dfrac{7k - 2}{k + 1}\bigg)$

Therefore,

$:\implies\sf x = \dfrac{3k + 2}{k + 1}\qquad\bigg\lgroup\bf eq\;(1) \bigg\rgroup$

$:\implies\sf y = \dfrac{7k - 2}{k + 1}\qquad\bigg\lgroup\bf eq\;(2) \bigg\rgroup$

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☯ Now, Putting value of x and y in the equation of intersecting line we get,

$\star\;{\boxed{\sf{\purple{2x + y - 4 = 0}}}}$

$:\implies\sf 2 \bigg( \dfrac{3k + 2}{k + 1} \bigg) + \dfrac{7k - 2}{k + 1} - 4 = 0$

$:\implies\sf \dfrac{6k + 2}{k + 1} + \dfrac{7k - 2}{k + 1} - 4 = 0$

$:\implies\sf \dfrac{(6k + 2) + (7k - 2) - (4[k + 1])}{k + 1} = 0$

$:\implies\sf \dfrac{(6k + 2) + (7k - 2) - (4k + 4)}{k + 1} = 0$

$:\implies\sf \dfrac{6k + 2 + 7k - 2 - 4k - 4}{k + 1} = 0$

$:\implies\sf \dfrac{9k - 2}{k + 1} = 0$

$:\implies\sf 9k - 2 = 0$

$:\implies\sf 9k = 2$

$:\implies\bf k = \dfrac{2}{9}$

Therefore,

$\dashrightarrow\sf k : 1 = \dfrac{9}{2} : 1$

$\dashrightarrow\sf k : 1 = 9 : 2$

$\therefore\;{\underline{\sf{Hence,\;the\;line\;AB\;is\; divided\;in\;the\;ratio\; \bf{2 : 9}.}}}$

Answer 2

Let , the line 2x + y - 4 = 0 divides the line segment AB in the ratio m : n and the point of intersection be " P(x , y) "

See the attachment for diagram ^^"

We know that , the section formula is given by

$\boxed{ \tt{x = \frac{m x_{2} + nx_{1}}{m + n} \: \: , \: \: y= \frac{m y_{2} + ny_{1}}{m + n} }}$

Thus ,

$\tt \implies x = \frac{m(3) + n(2)}{m + n} \: , \: y = \frac{m(7) + n( - 2)}{m + n}$

$\tt \implies x = \frac{3m + 2n}{m + n} \: , \: y = \frac{7m - 2n}{m + n}$

Since , the point P(x , y) lies in the line 2x + y - 4 = 0

$\therefore$ It satisfies the eq 2x + y - 4 = 0

Thus ,

$\tt \implies2 (\frac{3m + 2n}{m + n}) + ( \frac{7m - 2n}{m + n} ) - 4 = 0$

$\tt \implies \frac{6m + 4n}{m + n} + ( \frac{7m - 2n}{m + n} ) - 4 = 0$

$\tt \implies \frac{6m + 4n + 7m - 2n - 4m - 4n}{m + n} = 0$

$\tt \implies 9m - 2n = 0$

$\tt \implies 9m = 2n$

$\tt \implies \frac{m}{n} = \frac{2}{9}$

Therefore , the ratio is m : n = 2 : 9

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