# Division of Line Segment

This topic is really easy. This is just a matter of small common sense and your skill to substitute.

Formula derivation:

Consider points A(x1,y1) and B(x2,y2) is divided by point P(xp,yp) to a ratio of m:n.

Using the similar triangles we can derive the following formulas:

$x_p=\displaystyle\frac{mx_2+nx_1}{m+n}$

$y_p=\displaystyle\frac{my_2+ny_1}{m+n}$

Note that if m=n. The coordinate of P is the midpoint of points A and B.

Sample Problem 1:

Points A(-1,3) and B(5,7) is bisected by point P. Such that $\overline{AP}:\overline{PB}=1:4$. Find the coordinate of point P.

Solution:

We draft the segment on our paper with the points A, P and B. We can solve the coordinate of P using the formula mentioned above.

Solve for xp :

$x_p=\displaystyle\frac{mx_2+nx_1}{m+n}$

$x_p=\displaystyle\frac{(1)(5)+(4)(-1)}{1+4}$

$x_p=\displaystyle\frac{1}{5}$

Solve for yp :

$y_p=\displaystyle\frac{my_2+ny_1}{m+n}$

$y_p=\displaystyle\frac{(1)(7)+(4)(3)}{1+4}$

$y_p=\displaystyle\frac{19}{5}$

The coordinate of P is      $(\displaystyle\frac{1}{5},\displaystyle\frac{19}{5})$

Sample Problem 2:

Find the coordinates of the point which divides the line segment from (-2,1) and (2,3) in the ratio 3:4.

Solution:

Let the point (-2,1) is A and point (2,3) is B. Since there is no specific location of desired point we have two possible coordinates. The desired points are X and Y since these two points divides the segment AB to 3:4 ratio.

Coordinate of X: Assume in the figure that points Y does not exist.

AX=3 and XB=4

Solve for    $x_x$

$x_x=\displaystyle\frac{mx_2+nx_1}{m+n}$

$x_x=\displaystyle\frac{(3)(2)+4(-2)}{3+4}$

$x_x=\displaystyle\frac{-2}{7}$

Solve for      $y_x$

$y_x=\displaystyle\frac{my_2+ny_1}{m+n}$

$y_x=\displaystyle\frac{(3(3)+4(1)}{3+4}$

$x_x=\displaystyle\frac{13}{7}$

For coordinate of Y: Assume that X does not exist

Solve for $x_y$

$x_y=\displaystyle\frac{mx_2+nx_1}{m+n}$

$x_y=\displaystyle\frac{3(-2)+4(2)}{4+3}$

$x_y=\displaystyle\frac{2}{7}$

Solve for $y_y$

$y_y=\displaystyle\frac{my_2+ny_1}{m+n}$

$y_y=\displaystyle\frac{3(1)+4(3)}{4+3}$

$y_y=\displaystyle\frac{15}{7}$

The two points are: (-2/7 , 13/7) and (2/7 , 15/7)

Sample Problem 3:

The line segment joining P1(1,3) and P2(-2,-4) is extended through each end by a distance equal to thrice its original length. Find the coordinates of the new endpoints.

Solution:

Draft the points P1 and P2 and extend both ends such that the extended segment is thrice the length of the original segment.Let B and C are the new endpoints. Let d the distance between P1 and P2. Then P1 to C is 3d the same distance of 3d from P2 to B.

Solve for coordinates of B:  assume that C does not exist

Solve for     $x_b$

$x_{P2}=\displaystyle\frac{mx_b+nx_{P1}}{m+n}$

$1=\displaystyle\frac{dx_b+(3d)(-2) }{d+3d}$

$1=\displaystyle\frac{dx_b+-6d}{4d}$  ,  we can cancel d here

$1=\displaystyle\frac{x_b+-6}{4}$

$x_b=10$

Solve for     $y_b$

$y_{P2}=\displaystyle\frac{my_b+ny_{P1}}{m+n}$

$3=\displaystyle\frac{dy_b+3d(-4)}{d+3d}$

$3=\displaystyle\frac{dy_b-12d}{4d}$    cancel d

$3=\displaystyle\frac{y_b-12}{4}$

$y_b=24$

Solve for coordinates of C: assume that B does not exist

Solve for     $x_c$

$x_{P1}=\displaystyle\frac{mx_c+nx_{P2}}{m+n}$

$-2=\displaystyle\frac{dx_c+3d(1)}{3d+d}$

$-2=\displaystyle\frac{dx_c+3d}{4d}$  cancel d

$-2=\displaystyle\frac{x_c+3}{4}$

$x_c =-11$

Solve for    $y_c$

$y_{P1}=\displaystyle\frac{my_c+ny_{P2}}{m+n}$

$y_{P1}=\displaystyle\frac{my_c+ny_{P2}}{m+n}$

$-4=\displaystyle\frac{dy_c+3d(3)}{3d+d}$

$-4=\displaystyle\frac{dy_c+9d}{4d}$   cancel d

$-4=\displaystyle\frac{y_c+9}{4}$

$y_c =-25$

The required answer is (10,24) and (-11,-25)