Question

points P and Q trisect the line segment joining the points A(-2,0) and B(0,8) such that P is near to A. find the coordinates of P and Q​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Coordinates of P is   $(\,\frac{-4}{3}\,,\,\frac{8}{3}\,)$  and coordinates of Q is  $(\,\frac{-2}{3}\,,\,\frac{16}{3}\,)$.

Step-by-step explanation:

Given: coordinates of A ( -2 , 0 ) and B ( 0 , 8 ) of line segment AB

To find: Coordinates of P & Q such that they trisect the line segment AB

Figure is attached.

Since P & Q trisecting the line segment AB, then

the ratio in which point P intersect AB is 1 : 2

& the ratio in which point Q intersect AB is 2 : 1

let coordinates of P be ( p , q ) and Q be ( x , y )

Now we use Section Formula,

$(x,y)\,=\,(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})$

using this formula we get

for Point P,

$P\,(p,q)\,=\,(\frac{1\times0+2\times(-2)}{1+2},\frac{1\times8+2\times0}{1+2})$  

$P\,(p,q)\,=\,(\frac{-4}{3},\frac{8}{3})$

for Point Q,

$Q\,(x,y)\,=\,(\frac{2\times0+1\times(-2)}{1+2},\frac{2\times8+1\times0}{1+2})$  

$Q\,(x,y)\,=\,(\frac{-2}{3},\frac{16}{3})$

Therefore, Coordinates of P is   $(\,\frac{-4}{3}\,,\,\frac{8}{3}\,)$  and coordinates of Q is  $(\,\frac{-2}{3}\,,\,\frac{16}{3}\,)$.