Question

find the ratio in which the straight line 2x+3y-20=0 divides the join of the point 2,3 and 2,10​

Answer 1

SOLUTION

TO DETERMINE

The ratio in which the straight line 2x+3y-20=0 divides the join of the point (2,3) and (2,10)

EVALUATION

Let n : m be the required ratio

Here the given equation of the line is

$\sf{2x + 3y - 20 = 0}$

$\sf{ \implies \: 2x + 3y = 20 \: \: \: \: - - - - (1)}$

Let P be the point where the straight line 2x+3y-20=0 divides the join of the point (2,3) and (2,10) in the ratio n : m

Then the coordinates of the point P is

$= \displaystyle \sf{ \bigg( \frac{2m + 2n}{n + m} , \frac{3m + 10n}{m + n} \bigg)}$

Now the point P lies on the line given by Equation 1

$\displaystyle \sf{ 2\bigg( \frac{2m + 2n}{n + m} \bigg) + 3 \bigg( \frac{3m + 10n}{m + n} \bigg) = 20}$

$\displaystyle \sf{ \implies \: \bigg( \frac{13m + 34n}{n + m} \bigg) = 20}$

$\displaystyle \sf{ \implies \: 13m + 34n = 20n +20 m }$

$\displaystyle \sf{ \implies \: 14n = 7m }$

$\displaystyle \sf{ \implies \: \frac{n}{m} = \frac{7}{14} }$

$\displaystyle \sf{ \implies \: \frac{n}{m} = \frac{1}{2} }$

Hence the required ratio is 1 : 2

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Answer 2

SOLUTION

TO DETERMINE

The ratio in which the straight line 2x+3y-20=0 divides the join of the point (2,3) and (2,10)

EVALUATION

Let n : m be the required ratio

Here the given equation of the line is

2x + 3y - 20 = 0

⟹2x+3y=20−−−−(1)

Let P be the point where the straight line 2x+3y-20=0 divides the join of the point (2,3) and (2,10) in the ratio n : m

Then the coordinates of the point P is

=(2m+2n/n+m , 3m+10n/m+n)

Now the point P lies on the line given by Equation 1

2(2m+2n/n+m) + 3(3m+10n/m+n) = 20

⟹(13m+34n/n+m) = 20

⟹13m+34n=20n+20m

⟹14n=7m

⟹n/m = 7/14

⟹ n/m = 1/2

Hence the required ratio is 1 : 2

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