Question

Find the ratio in which the line 3x + y – 9 = 0 divides the line joining the points (1, 3) and (2, 7) internally * 1 point 9:6 3:4 4:3 None of these Other:

Answer 1

       $\text{\huge \bf \underline{Answer:}}$

       $\text{It is given that;}$

       $\text{The line }3x + y - 9 = 0 \text{ divides the join of }(1,3) \text{ and }(2,7)$

       $\text{We must find the ratio}$

       $\text{First, let us form the equation of the line passing through }(1,3) \text{ and }(2,7)$

$\implies \text{Applying the two point form}$

       $\dfrac{y - y_1}{y_2-y_1} = \dfrac{x-x_1}{x_2-x_1} \text{ where }(x_1,y_1) \text{ and }(x_2,y_2) \text{ are the points}$

       $\text{In our case; }(x_1,y_1) = (1,3) \text{ and }(x_2,y_2) = (2,7)$

$\implies \dfrac{y - 3}{7 - 3} = \dfrac{x - 1}{2 - 1}$

$\implies \dfrac{y - 3}{4} = \dfrac{x - 1}{1}$

$\implies 1(y - 3) = 4(x-1)$

$\implies y - 3 = 4x-4$

$\implies y - 3 - 4x+4 = 0$

$\implies y - 4x +1 = 0$

$\implies 4x - y - 1 = 0 \text{ ------ (i)}$

       $\text{The other line is }3x + y - 9 = 0 \text{ ------ (ii)}$

       $\text{Now, we have to simultaneously solve (i) and (ii)}$

       $\text{Applying elimination method}$

$\implies \text{Adding (i) and (ii)}$

$\implies 4x - y - 1 + 3x + y - 9 = 0$

$\implies 7x -10 = 0$

$\implies 7x =10$

$\implies x = \dfrac{10}{7}$

      $\text{Substituting in (i)}$

$\implies 4\text{\huge{(}}\dfrac{10}{7}\text{\huge{)}} - y - 1 = 0$

$\implies \dfrac{40}{7} - y - 1 = 0$

$\implies \dfrac{40 - 7y - 7}{7}= 0$

$\implies 40 - 7y - 7 = 0$

$\implies 33 - 7y= 0$

$\implies - 7y= -33$

$\implies 7y= 33$

$\implies y= \dfrac{33}{7}$

$\implies \text{The line }3x + y - 9 = 0 \text{ cuts the join of (1,3) and (2,7) at }\text{\huge{(}}\dfrac{10}{7},\dfrac{33}{7}\text{\huge{)}}$

       $\text{Let the ratio be }1:k$

       $\text{Applying section formula}$

$\implies \text{\huge{(}}\dfrac{mx_2 + nx_1}{m+n}, \dfrac{my_2 + ny_1}{m+n}\text{\huge{)}} \text{ where }m:n \text{ is the ratio}$

       $\text{In our case; }m:n = 1:k$

$\implies \text{\huge{(}}\dfrac{1(2) + k(1)}{1+k}, \dfrac{1(7) + k(3)}{1+k}\text{\huge{)}} = \text{\huge{(}}\dfrac{10}{7}, \dfrac{33}{7}\text{\huge{)}}$

$\implies \text{\huge{(}}\dfrac{2 + k}{1+k}, \dfrac{7 + 3k}{1+k}\text{\huge{)}} = \text{\huge{(}}\dfrac{10}{7}, \dfrac{33}{7}\text{\huge{)}}$

$\implies \dfrac{10}{7} = \dfrac{2+k}{1+k}\: \: ; \: \: \dfrac{33}{7} = \dfrac{7 + 3k}{1+k}$

$\implies 10(1+k) = 7(2+k)\: \: ; \: \: 33(1+k) = 7(7 + 3k)$

$\implies 10+10k = 14+7k\: \: ; \: \: 33 + 33k = 49 + 21k$

$\implies 10+10k - 14-7k = 0\: \: ; \: \: 33 + 33k - 49 - 21k = 0$

$\implies 3k - 4 = 0\: \: ; \: \: 12k - 16 = 0$

$\implies 3k = 4\: \: ; \: \: 12k = 16$

$\implies k = \dfrac43\: \: ; \: \: k = \dfrac{16}{12} = \dfrac43$

$\implies k = \dfrac43$

$\implies 1 : k = 3 : 4$

  $\therefore \: \: \boxed{\boxed{\text{The ratio is }3:4}}$