) and ( x , ). Find the value of x . Solution Slope of the line through the points (– , ) and ( , ) is m −− Slope of the line through the points ( , ) and ( x , ) is m Since two lines are perpendicular, m m = – , which gives or = × = − . .
Collinearity of three points We know that slopes of two parallel lines are equal. If two lines having the same slope pass through a common point, then two lines will coincide. Hence, if A , B and C are three points in the XY-plane, then they will lie on a line, i.e., three points are collinear (Fig . ) if and only if slope of AB = slope of BC.
STRAIGHT LINES Example Three points P ( h, k ), Q ( x , y ) and R ( x , y ) lie on a line. Show that ( h – x ) ( y – y ) = ( k – y ) ( x – x ). Solution Since points P, Q and R are collinear, we have Slope of PQ = Slope of QR, i.e., k or k , or ( h – x ) ( y – y ) = ( k – y ) ( x – x ). Example In Fig .
, time and distance graph of a linear motion is given. Two positions of time and distance are recorded as, when T = , D = and when T = , D = . Using the concept of slope, find law of motion, i.e., how distance depends upon time. Solution Let (T, D) be any point on the line, where D denotes the distance at time T.
Therefore, points ( , ), ( , ) and (T, D) are collinear so that D or (T ) (D