How do you divide a line segment equation?

Suppose you have a line segment PQ¯ on the coordinate plane, and you need to find the point on the segment 13  of the way from P to Q.

Let’s first take the easy case where P is at the origin and line segment is a horizontal one.

The length of the line is 6 units and the point on the segment 13  of the way from P to Q would be 2 units away from P, 4 units away from Q and would be at (2,0).

Consider the case where the segment is not a horizontal or vertical line.

The components of the directed segment PQ¯ are 〈6,3〉  and we need to find the point, say X on the segment 13  of the way from P to Q.

Then, the components of the segment PX¯ are 〈(13)(6),(13)(3)〉=〈2,1〉.

Since the initial point of the segment is at origin, the coordinates of the point X are given by (0+2,0+1)=(2,1).

Now let’s do a trickier problem, where neither P nor Q is at the origin.

Use the end points of the segment PQ¯  to write the components of the directed segment.

〈( x2−x1),(y2−y1)〉=〈(7−1),(2−6)〉 =〈6,−4〉

Now in a similar way, the components of the segment PX¯  where X is a point on the segment 13  of the way from P to Qare 〈(13)(6),(13)(−4)〉=〈 2,−1.25〉.

To find the coordinates of the point X add the components of the segment PX¯  to the coordinates of the initial point P.

So, the coordinates of the point X are (1+2,6 −1.25)=(3,4.75).

Note that the resulting segments, PX¯ and XQ ¯, have lengths in a ratio of 1:2.

In general: what if you need to find a point on a line segment that divides it into two segments with lengths in a ratio a:b?

Consider the directed line segment XY¯  with coordinates of the endpoints as X (x1,y1)  and Y(x2,y2).

Suppose the point Z divided the segment in the ratio a:b, then the point is aa+b of the way from X to Y.

So, generalizing the method we have, the components of the segment XZ¯ are 〈(aa+b(x2−x1) ),(aa+b(y2−y1))〉.

Then, the X-coordinate of the point Z is

x1+ aa+b(x2−x1)=x1(a+b)+a(x2−x1)a+ b =bx1+ax2a+b.

Similarly, the Y-coordinate is

y1 +aa+b(y2−y1)=y1(a+b)+a(y2−y1)a +b =by1+ay2a+b.

Therefore, the coordinates of the point Z are (b x1+ax2a+b,by1+ay2a+b).

Example 1:

Find the coordinates of the point that divides the directed line segment MN¯ with the coordinates of endpoints at M(−4,0)  and M(0,4) in the ratio 3:1?

Let L be the point that divides MN ¯  in the ratio 3:1.

Here, (x1,y1)=(−4,0),(x2,y2)= (0,4) and a:b=3:1.

Substitute in the formula. The coordinates of L are

(1(−4)+3(0)3+1, 1(0)+3(4)3+1).

Simplify.

(−4+04,0+124)=(−1 ,3)

Therefore, the point L(−1,3)  divides MN¯  in the ratio 3:1.

Example 2:

What are the coordinates of the point that divides the directed line segment AB¯  in the ratio 2:3?

Let C be the point that divides AB¯  in the ratio 2:3.

Here, (x1,y1) =(−4,4),(x2,y2)=(6,−5) and a:b=2:3.

Substitute in the formula. The coordinates of C are

(3(−4)+2(6)5,3(4)+2(−5)5).

Simplify.

(−12+125,12−105)=(0,25) =(0,0.4)

Therefore, the point C(0,0.4)  divides AB¯  in the ratio 2:3.

You can note that the Midpoint Formula is a special case of this formula when a=b=1.

Suppose a point divides a line segment into two parts which may be equal or not, with the help of the section formula we can find that point if coordinates of a line segment are given and we can also find the ratio in which the point divides the given line segment if coordinates of that point are given.

When a point C divides a line segment AB in the ratio m:n, then we use the section formula to find the coordinates of that point. The section formula has 2 types. These types depend on point C which can be present between the points or outside the line segment.

The two types are:

1. Internal Section Formula
2. External Section Formula

Internal Section Formula

When the point divides the line segment in the ratio m : n internally at point C then that point lies in between the coordinates of the line segment then we can use this formula. It is also called Internal Division.

If the coordinates of A and B are (x1, y1) and (x2, y2) respectively then Internal Section Formula is given as:

Derivation of the Formula

Let A (x1, y1) and B (x2, y2) be the endpoints of the given line segment AB and C(x, y) be the point which divides AB in the ratio m : n.

Then, AC / CB = m / n

We want to find the coordinates (x, y) of C.

Now draw perpendiculars of A, C, B parallel to Y coordinate joining at P, Q, and R on X-axis.

By seeing the above diagram,

AM = PQ = OQ – OP = (x – x1)

CN = QR = OR – OQ = (x2 –  x)

CM = CQ – MQ = (y – y1)

BN = BR – NR = (y2 – y)

Clearly, we can see that ∆AMC and ∆CNB are similar and, therefore, their sides are proportional by AA congruence rule.

AC / CB = AM / CN = CM / BN

Now substituting the values in the above relation

=> m / n = [x – x1 / x2 -x] = [y – y1 / y2 – y]

=> m / n = [x – x1 / x2 -x]  and m / n = [y – y1 / y2 – y]

Solving the 1st condition,

=> m(x2 – x) = n(x – x1)

=> (m + n)x = (mx2 + nx1)

=> x = (mx2 + nx1) / (m + n)

Solving the 1st condition,

=> m(y2 – y) = n(y – y1)

=> (m + n)y = (my2 + ny1)

=> y = (my2 + ny1) / (m + n)

Therefore, the co-ordinates of C (x, y) are

 { (m × x2 + n × x1) / (m + n ) ,  (m × y2 + n × y1) / (m + n ) }

External Section Formula

When the point which divides the line segment is divided externally in the ratio m : n lies outside the line segment i.e when we extend the line it coincides with the point, then we can use this formula. It is also called External Division.

If the coordinates of A and B are (x1,y1) and (x2,y2) respectively then external Section Formula is given as

Derivation of the Formula

To derive the internal section we took a line segment and a point C(x, y) inside the line, but in the case of the external section formula, we have to take that point C(x, y) outside the line segment.

Let A(x1, y1) and B(x2, y2) be the endpoints of the given line segment AB and C(x, y) be the point which divides AB in the ratio m : n externally.

We want to find the coordinates (x, y) of C. For that, draw perpendiculars of A, B, C parallel to Y coordinate joining at P, Q, and R on X axis.

By seeing the above diagram,

AM = PR = OR – OP = (x – x1)

BN = QR = OR – OQ = (x – x2)

similarly,

CM = RC – MR = (y – y1)

CN = CR – NR = (y – y2)

Clearly, we can see that triangle AMC and triangle BNC are similar and, therefore, their sides are proportional by AA congruence rule

AC / BC = AM / BN = CM / CN

Now substituting the values in the above relation

=> m / n = [x – x1 / x – x2] = [y – y1 / y – y2]

=> m / n = [x – x1 / x – x2] and m / n = [y – y1 / y – y2]

Solving the 1st condition,

=> m(x – x2) = n(x – x1)

=> (m – n)x = (mx2 – nx1)

=> x = (mx2 – nx1) / (m – n)

Solving the 2nd condition,

=> m(y – y2) = n(y – y1)

=> (m – n)y = (my2 – ny1)

=> y = (my2 – ny1) / (m – n)

Therefore, the co-ordinates of C (x, y) are 

{ (m × x2 – n × x1) / (m – n) ,  (m × y2 – n × y1) / (m – n ) }

Problems On Section Formula

Problem 1: Find the coordinates of point C (x, y) where it divides the line segment joining (4, – 1) and (4, 3) in the ratio 3 : 1 internally ? 

Solution: 

Given coordinates are A (4, -3) and B (8, 5)

Let C (x, y) be a point which divides the line segment in the ratio of 3 : 1  i.e m : n = 3 : 1 

Now using the formula C(x, y) = { (m × x2 + n × x1) / (m + n ) ,  (m × y2 + n × y1) / (m + n ) } as C is dividing internally.

=> C(x, y) = {(3*4 + 1*4 ) / (3+1), (3 * 3  + 1 *(-1)) / (3+1)}

=> C(x, y) = {16 / 4, 8 / 4}

=> C(x, y) = {4, 2} 

Hence, the coordinates are (4, 2).

Problem 2: If a point P(k, 7) divides the line segment joining A(8, 9) and B(1, 2) in a ratio m : n then find values of m and n.

Solution: 

It is not mentioned that the point is dividing the line segment internally or externally. So, at that time we will consider the internal section as the default.

Given coordinates are A (8, 9) and B (1, 2)

Let the given point P (k, 7) divides the line segment in the ratio of m : 1

Now using the section formula, finding only the x coordinate, 

=> k = (m × x2 + n × x1) / (m + n )

=> k = (m × 1 + 1 × 8) / (m +1)

=> k = (m + 8) / (m + 1)

=> km + k = m + 8       …….(1)

Again using section formula for y coordinate.

=> 7 =  (m × y2 + n × y1) / (m + n )

=> 7 = (m × 2 + 1 × 9) / (m + 1)

=> 7 = (2m + 9) / (m +1)

=> 7m + 7 = 2m +9

=> 5m = 2

=> m = 5 / 2

So the required ratio is 5 : 2

Therefore, value of m is 5 and value of n is 2

Problem 3: A (4, 5) and B (7, -1) are two given points, and the point C divides the line-segment AB externally in the ratio 4 : 3. Find the coordinates of C.

Solution:

Given coordinates are A (4, 5) and B (7, -1)

Let C (x, y) be a point which divides the line segment externally in the ratio of 4 : 3  i.e m : n = 4 : 3 

Now using the formula C(x, y) = { (m × x2 – n × x1) / (m – n) ,  (m × y2 – n × y1) / (m – n ) } as C is dividing internally.

value of x = (mx2 – nx1) / (m – n)

              => (4 * 7 – 3 * 4) / (4 – 3)

              => 16

value of y = (my2 – ny1) / (m – n )

              => (4 * (-1) – 3 * 5) / (4 – 3)

              => -19

Hence, the coordinates are (16, -19).

Problem 4: Line 2x+y−4=0 divides the line segment joining the points A(2,−2) and B(3,7). Find the ratio of line segment in which the line is dividing?

Solution:

Given coordinates are A (2, -2) and B (3, 7).

Line with equation 2x + y – 4 = 0 divides the line segment at point C (x, y)

Let us assume the given line cuts the line segment in the ratio 1 : n.

By section formula,

           => x = (m * x2 + n * x1) / (m + n)

           => x = (3 + 2n) / (1 + n)       ………..1

Similarly,

          => y = (m * y2 + n * y1) / (m + n)

          => y = (7 – 2n) / (1 + n)         ……….2

Now substituting the equations 1 and 2 in the given equation of the line.

          => 2x + y – 4 = 0

          => 2 [(3 + 2n) / (1 + n) ] +  [(7 – 2n) / (1 + n)] – 4 = 0  

          => 6 + 4n + 7 − 2n − 4(1 + n) = 0 

          =>13 + 2n − 4 − 4n = 0

          =>9 − 2n = 0

          => n = 2 / 9

Therefore the ratio at which the line divides is 9 : 2. We can also find the values of x and y by substituting the value of n in the equation 1 and 2.

Problem 5: A(2, 7) and B(–4, –8) are coordinates of the line segment AB. There are two points that trisected the segment. Find the coordinates of them.

Solution: 

The two points trisected the line segment, which means the segment is divided into 3 equal parts.

AS = ST = TB                …………1

=> AS / SB

=> AS / ST + TB

=> AS / (AS + AS)          from equation 1

=> AS / 2 AS

=> 1 / 2

So, S divides the line segment AB in the ratio of 1 : 2

Now applying section formula to find the coordinates of point S

=> x1 = (1 × (-4) + 2 × 2) / (1 + 2)

=> x1 = (-4 + 4) / 3

=> x1 = 0

Similarly, for y coordinate,

=> y1 =  (1 × (-8) + 2 × 7) / (1 + 2)

=> y1 = (14 – 8) / 3

=> y1 = 2

Also.

=> AT / TB

=> (AS +ST) / TB

=> 2 TB / TB           from equation 1

=> 2 / 1

So, T divides the line segment AB in the ratio of 2 : 1

Now applying section formula to find the coordinates of point T

=> x2 = (2 × (-4) + 1 × 2) / (2 + 1)

=> x2 = (-8 + 2) / 3

=> x2 = -2

Similarly, for y coordinate

=> y2 =  (2 × (-8) + 1 × 7) / (2 + 1)

=> y2 = (-16 + 7) / 3

=> y2 = -3

Thus, the coordinates are :

S (x1, y1) = (0, 2)

T (x2, y2) = (-2, -3)