Math, asked by peepeepoopoo, 11 months ago

find the ratio in which the line 3x+4y-9=0 divides the line segments joining the points A(1,3) and B(2,7)​

Answers

Answered by streetburner
5

Answer:

6:25

Step-by-step explanation:

Let the line 3x + 4y - 9 = 0 divides the line segment joining the points (1,3) and (2,7) in the ratio k:1

then the coordinate of the point is :

(2k+1)/(k+1) , (7k+3)/(k+1)

Since this the intersection point, it also lies on the line 3x + 4y - 9 = 0. So:

3(2k+1)/(k+1) + 4(7k+3)/(k+1) - 9 = 0

k = -6/25

Since k is negative, the line 3x + 4y - 9 = 0 , divides the line segment joining the points (1,3) and (2,7) externally in the ratio 

6:25

Answered by Anonymous
5

Solution

Given :-

 \sf \implies \: Equation \: 3x + 4y - 9 = 0

 \sf \implies Point \: (1,3) \: and \: (2,7)

Let

 \sf \implies \: Ratio \: = P \ratio \: 1

Using section formula

 \implies \sf \: p \bigg( \dfrac{x_2m + nx_1}{m + n}  , \:  \dfrac{y_2m + y_1n}{m + n}  \bigg)

Where

 \sf \implies \: x_1 = 1, y_1 = 3 \\  \sf \implies \: x_2 = 2,y_2 = 7

 \sf \implies \: m = p \:  \: and \: n = 1

Now put the value on formula

 \sf  \implies\: p \bigg( \dfrac{2 \times p + 1 \times1}{p + 1}  , \:  \dfrac{7 \times p + 3 \times 1}{p + 1}  \bigg)

 \sf  \implies\: p \bigg( \dfrac{2p + 1 }{p + 1}  , \:  \dfrac{7 p + 3 }{p + 1}  \bigg)

Now put the value of x and y on Given equation

 \sf \implies \: 3x + 4y - 9 = 0

 \sf \implies3 \bigg( \dfrac{2p + 1}{p + 1}  \bigg) + 4 \bigg( \dfrac{7p + 3}{p + 1}  \bigg) - 9 = 0

 \sf\implies \:  \dfrac{6p + 3}{p + 1}  +  \dfrac{28p + 12}{p + 1}  - 9 = 0

Taking Lcm

 \sf \implies \:  \dfrac{6p + 3 + 28p + 12 - 9p - 9}{p + 1}  = 0

 \sf \implies\: 6p + 3 + 28p + 12 - 9p - 9 = 0

 \sf \implies \: 34p + 15 - 9p - 9 = 0

 \sf \implies 25p  + 6 = 0

 \sf \implies \: p =  \dfrac{ - 6}{25}

Answer

The ratio is -6/25 or -6:25

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