# Coordinate Geometry

## Distance Formula

### NCERT Exercise

#### 7.1 Part 2

Question 2: Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in section 7.2 and shown in following figure.

**Answer:** In particular, the distance of a point (x, y) from the origin is given by

`sqrt(x^2+y^2)`

So, let 36 be x and 15 be y

Hence, `sqrt(x^2+y^2)`

`=sqrt(36^2+15^2)`

`=sqrt(1296+225)`

`=sqrt(1521)=39` units

Now, as discussed in Section 7.2 in the book, town B is 36 east and 15 north from the town A. Thus, distance between town A and B will be equal to 39 unit. If distance is measured in km, then distance will be equal to 39 km.

Question 3: Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.

**Answer:** Let A = (1, 5), B = (2, 3) and C = (-2, -11)

Thus, distance between AB by distance formula

`sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x_{1} = 1, x_{2} = 2, y_{1} = 5 and y_{2} = 3

So, `AB = sqrt((2-1)^2+(3-5)^2)`

`=sqrt(1^2+(-2)^2)=sqrt(1+4)`

Or, `AB = sqrt5` units

Now, distance between BC by distance formula

`sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x_{1} = 2, x_{2} = -2, y_{1} = 3 and y_{2} = -11

So, `BC = sqrt((-2-2)^2+(-11-3)^2)`

`=sqrt((-4)^2+(-14)^2)`

`=sqrt(16+196)=sqrt(212)`

`=sqrt(4xx53)=2sqrt(53)` units

And distance between AC by distance formula

`sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x_{1} = 1, x_{2} = -2, y_{1} = 5 and y_{2} = -11

So, `AC = sqrt((-2-1)^2+(-11-5)^2)`

`=sqrt((-3)^2+(-16)^2)`

`=sqrt(9+256)=sqrt(265)` units

Now, for collinear lines

`AB + BC = AC`

After substituting the values of AB, BC and AC

`sqrt5+sqrt(53)≠sqrt(265)`

Thus, given points are non-collinear.