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Basic Mathematics – I

Notes the XY-plane, then they will lie on a line, i.e., three points are collinear (Figure 5.8) if and only
if slope of AB = slope of BC.

Figure 5.8

Example: Three points P (h, k), Q (x , y ) and R (x , y ) lie on a line. Show that (h – x )
1 1 2 2 1
(y – y ) = (k – y ) (x – x ).
2 1 1 2 1
Solution:
Since points P, Q and R are collinear, we have
y k y y
Slope of PQ = Slope of QR, i.e., 1 2 1
x h x x
1 2 1
k y y y
or 1 2 1 ,
h x 1 x 2 x 1
or (h x ) (y y ) = (k y ) (x x ).
1 2 1 1 2 1

Example: In Figure 5.9, time and distance graph of a linear motion is given. Two positions
of time and distance are recorded as, when T = 0, D = 2 and when T = 3, D = 8. Using the concept
of slope, find law of motion, i.e., how distance depends upon time.
Figure 5.9

Solution:

Let (T, D) be any point on the line, where D denotes the distance at time T. Therefore, points
(0, 2), (3, 8) and (T, D) are collinear so that

8 2 D 8
or 6(T 3) = 3(D 8)
3 0 T 3
or D = 2(T + 1),

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