Question

P, Q & R are collinear points in which the co-ordinates of P & Q are (3,4) &(7, 7) respectively & PR is equal to 10 units.Find the length of P Q & co-ordinares
of R.​

Answer 1

Question -

P, Q and R are collinear points in which the co-ordinates of P and Q are (3,4) and (7,7) respectively and PR is equal to 10 units. Find the length of PQ and co-ordinates of R

Solution -

Given,

P =(3, 4) and Q = (7,7)

By using distance formula,

$PQ = \sqrt{( { x_{2} - x_{1}) }^{2} + {( y_{2} - y_{1})}^{2} }$

$PQ = \sqrt{ {(7 - 3)}^{2} + {(7 - 4)}^{2} }$

$PQ = \sqrt{ {4}^{2} + {3}^{2} }$

$PQ = \sqrt{16 + 9}$

$PQ = \sqrt{25}$

$PQ = 5$

Therefore the length of PQ is 5

But PR = 10 unit

So QR = PR - PQ = 10-5 = 5 unit

So, Q is the mid point of PR

Let (x, y) be the co - ordinates of R

Since,

Q is the mid point of PR,

$\frac{3 + x}{2} = 7$

$3 + x = 7 \times 2$

$3 + x = 14$

$x = 14 - 3$

$x = 11$

And,

$\frac{4 + y}{2} = 7$

$4 + y = 7 \times 2$

$4 + y = 14$

$y = 14 - 4$

$y = 10$

Therefore co-ordinates of R are (11,10)