# Coordinate Geometry

## 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. Solution: 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.

Solution: 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, x1 = 1, x2 = 2, y1 = 5 and y2 = 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, x1 = 2, x2 = -2, y1 = 3 and y2 = -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, x1 = 1, x2 = -2, y1 = 5 and y2 = -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.