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, 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.