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Chapter 7: Coordinate Geometry

    Introduction

    Distance Formula

    Section Formula

    Exercise 7.1

    Exercise 7.2

    Summary

    Enhanced Curriculum Support

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Chapter 7: Coordinate Geometry > Section Formula

Section Formula

Suppose a telephone company wants to position a relay tower at point P between points A and B in such a way that the distance of the tower from point B is twice its distance from point A.

If point P lies on line AB, it will divide line AB in the ratio 1:2.

If we take point A as the origin O, and consider 1 km as one unit on both axes, the coordinates of point B will be (36, 15).

We can represent this as shown below.

In order to determine the position of the tower, we must know the coordinates of point P. How do we find these coordinates?

As before, let’s draw some perpendiculars and create familiar geometric shapes, such as triangles or rectangles, so that we can apply known identities and formulas.

In the figure above, let’s first drop a perpendicular from point P to the x-axis.

Now drop a perpendicular from B to x-axis.

Draw a perpendicular from P to BE.

Nice we now have triangles and one rectangle.

We see that the two triangles PAD and BPC are .

We know that in similar triangles, the ratios of the sides are similar. So we have ADPC = APPB = and PDBC = APPB =

Subsituting for P we get x36−x = and y15−y =

Solving these equations we get x = and y = .

Generalizing this, take the below coordinate system

Solving as above by considering similar triangles we get the coordinates of the point P(x, y) which divides the line segment joining the points A(x1,y1) and B(x2,y2), internally, in the ratio m1:m2 are

m1x2+m2x1m1+m2,m1y2+m2y1m1+m2

So, the coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2), internally, in the ratio m1 : m2 are:

(m2x1+m1x2m1+m2, m2y1+m1y2m1+m2) (2)

This is known as the section formula.

This can also be derived by drawing perpendiculars from A, P and B on the y-axis and proceeding as above.

If the ratio in which P divides AB is k : 1, then the coordinates of the point P will be:

(kx2+x1k+1, ky2+y1k+1)

Special Case : The mid-point of a divides the line segment in the ratio 1 : 1. Therefore, the coordinates of the mid-point P of the join of the points A(x1, y1) and B(x2, y2) is:

(1x2+x11+1, 1y2+y11+1) = (x1+x22, y1+y22 )

6. Find the coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally.

Solution : Let P(x, y) be the required point. Using the section formula, we get:

x = ( + )/ =

y = ( + )/ =

Therefore, (7, 3) is the required point.

7. In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and B(3, – 8)?

Solution : Let (– 4, 6) divide AB internally in the ratio m1: m2. Using the section formula, we get:

(-4,6) = (3m1−6m2m1+m2, −8m1+10m2m1+m2)

Recall that if (x, y) = (a, b) then x = and y = .

So, –4 = 3m1−6m2m1+m2 and 6 = −8m1+10m2m1+m2

Now, – 4 = 3m1−6m2m1+m2 gives us

−4m1−4m2=3m1−6m2

Which gives us, m1:m2 = :

Let's verify that the ratio satisfies the y-coordinate also.

−8m1+10m2m1+m2 = −8m1m2+10m1m2+1 (Dividing throughout by m2)

= −8x27+1027+1 =

Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio : .

Alternatively : The ratio m1 : m2 can also be written as (m1m2) : or k : .

Let (– 4, 6) divide AB internally in the ratio k : 1. Using the section formula, we get -

(– 4, 6) = (3k−6k+1, −8k+10k+1)

So, -4 = 3k−6k+1

- 4k - 4 = 3k -

On simplifying, k : 1 = 2 :

Like earlier, we can check for the y-coordinate also.

Note : We can also find this ratio by calculating the distances PA and PB and taking their ratios provided you know that A, P and B are collinear.

8. Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4) (Given in the above figure).

Solution : Let P and Q be the points of trisection of AB i.e., AP = = QB

Therefore, P divides AB internally in the ratio 1 : 2.

Therefore, the coordinates of P, by applying the section formula, are:

(1−7+221+2 , 14+2−21+2) = (Put comma in between the two numbers)

Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are:

(2−7+121+2 , 24+1−21+2) = (Put comma in between the two numbers)

Therefore, the coordinates of the points of trisection of the line segment joining A and B are (–1, 0) and (– 4, 2).

Note : We could also have obtained Q by noting that it is the mid-point of PB. So, we could have obtained its coordinates using the mid-point formula.

9. Find the ratio in which the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4). Also find the point of intersection.

Solution : Let the ratio be k : 1. Then by the section formula, the coordinates of the point which divides AB in the ratio k : 1 are

(−k+5k+1, −4k−6k+1)

This point lies on the y-axis, and we know that on the y-axis the abscissa is .

Therefore, −k+5k+1 = 0

This gives us k =

That is, the ratio is : .

Putting the value of k = 5, we get the point of intersection as .

10. If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.

Solution : We know that diagonals of a parallelogram each other.

So, the coordinates of the mid-point of AC = coordinates of the mid-point of BD

(6+92, 1+42) = (8+p2, 2+32)

(, ) = (8+p2, 52)

152 = 8+p2

p =