Question

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

Answer 1

Let the given line divides the line segment at the ratio of k:1. If the coordinates of the dividing point is (x', y') then,
x'=(k.2+1.1)/(k+1)
=(2k+1)/(k+1) and
y'=(k.7+1.3)/(k+1)
=(7k+3)/(k+1)
Now, (x',y') lies on 3x+4y-9=0
∴, 3(2k+1)/(k+1)+4(7k+3)/(k+1)-9=0
or, (6k+3+28k+12-9k-9)/(k+1)=0
or, 25k+6=0
or, k=-6/25
∴, the required ratio is 6:25 and the given straight line cuts the line segment externally (since there is a negative sign appears in the ratio) Ans.

Answer 2

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