Question

i. Point R divides seg PQ externally in the ratio 3 : 1, and P-Q-R. Find the ratio in which point
Q divides seg PR​

Answer 1

Therefore Q divides segment PR in 2:1 ratio internally.

Step-by-step explanation:

Let the co-ordinate of P be(x₁,y₁) and the co-ordinate Q be (x₂,y₂)

Point R divides the line segment PQ externally in the ratio 3:1

Then the co-ordinate of R is $(\frac{3.x_2-1.x_1}{3-1} ,\frac{3.y_2-1.y_1}{3-1} )$

                                              $=(\frac{3x_2-x_1}{2} ,\frac{3y_2-y_1}{2} )$

The distance between (a,b) and (c,d) is $\sqrt{(a-c)^2+(b-d)^2}$ units

Now the distance between P and Q is

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2$ units

The distance between Q and R is

$\sqrt{(\frac{3x_2-x_1}{2} -x_2)^2+(\frac{3y_2-y_1}{2} -y_2)^2}$

$=\sqrt{(\frac{x_2-x_1}{2})^2+(\frac{y_2-y_1}{2} )^2}$  units

Therefore Q divides the line segment PR in the ratio

=The length of PQ : The length of RQ

$=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}:\sqrt{(\frac{x_2-x_1}{2})^2+(\frac{y_2-y_1}{2} )^2}$

$=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}:\sqrt{\frac{1}{4} (x_2-x_1)^2+\frac{1}{4} (y_2-y_1)^2}$

$=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}:\frac{1}{2} \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$=1:\frac{1}{2}$

=2:1

Therefore Q divides segment PR in 2:1 ratio internally.