Question

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

Answer 1

$\large\underline{\sf{Solution-}}$

↝Let us assume given coordinates as A and B.

So,

↝Coordinates of A be (1, 3)

and

↝Coordinates of B be (2, 7).

Let us further assume that the line 3x + y - 9 = 0 divides the line segment joining the points A and B at P in the ratio k : 1.

↝Let coordinates of P be (x, y).

We know,

↝Section Formula which states that

The coordinates of point C (x, y) which divides the line segment joining the points A and B in the ratio m : n internally is given by

$\bf \:( x, y) = \bigg(\dfrac{mx_2 + nx_1}{m + n} , \dfrac{my_2 + ny_1}{m + n} \bigg)$

Here,

$\rm :\longmapsto\:x_1 =1 ,y_1 = 3,x_2=2 ,y_2=7 ,m =k ,n =1$

↝ So, on substituting the values in above formula, we get

$\rm :\longmapsto\:\bf \:( x, y) = \bigg(\dfrac{2k + 1}{k + 1} , \dfrac{7k + 3}{k + 1} \bigg)$

↝ Hence, Coordinates of P is

$\rm :\longmapsto\:\bf \:( x, y) = \bigg(\dfrac{2k + 1}{k + 1} , \dfrac{7k + 3}{k + 1} \bigg)$

Now,

↝ P lies on the line 3x + y - 9 = 0

$\rm :\longmapsto\:3\bigg(\dfrac{2k + 1}{k + 1} \bigg) + \bigg(\dfrac{7k + 3}{k + 1} \bigg) = 9$

$\rm :\longmapsto\:\bigg(\dfrac{6k + 3}{k + 1} \bigg) + \bigg(\dfrac{7k + 3}{k + 1} \bigg) = 9$

$\rm :\longmapsto\:\dfrac{6k + 3 + 7k + 3}{k + 1} = 9$

$\rm :\longmapsto\:\dfrac{13k + 6}{k + 1} = 9$

$\rm :\longmapsto\:13k + 6 = 9(k + 1)$

$\rm :\longmapsto\:13k + 6 = 9k +9$

$\rm :\longmapsto\:13k - 9k = 9 - 6$

$\rm :\longmapsto\:4k = 3$

$\rm :\implies\:k = \dfrac{3}{4}$

Hence, Required ratio is 3 : 4

Additional Information :-

1. Distance Formula :-

${\underline{\boxed{\rm{\quad Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad}}}}$

2. Midpoint Formula :-

${\underline{\boxed{\rm{\quad \dfrac{x_1 + x_2}{2} \; ,\; \dfrac{y_1 + y_2}{2} \quad}}}}$

3. Centroid of a triangle

${\underline{\boxed{\rm{\quad \dfrac{x_1 + x_2 + x_3}{3} \; ,\; \dfrac{y_1 + y_2 + y_3}{3} \quad}}}}$

4. Area of triangle

$\rm\ Area =\dfrac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

5. Condition for 3 points to be collinear

$\boxed{\rm \: x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) = 0}$