the distance between them? Let us draw PR and QS perpendicular to the x -axis from P and Q respectively. Also, draw a perpendicular from P on QS to meet QS at T. Then the coordinates of R and S are ( , ) and ( , ), respectively.
So, RS = units. Also, QS = units and TS = PR = units. Fig. .
Fig. . Therefore, QT = units and PT = RS = units. Now, using the Pythagoras theorem, we have PQ = PT + QT = + = PQ = units How will we find the distance between two points in two different quadrants?
Consider the points P( , ) and Q(– , – ) (see Fig. . ). Draw QS perpendicular to the x -axis.
Also draw a perpendicular PT from the point P on QS (extended) to meet y -axis at the point R. Fig. . Then PT = units and QT = units.
(Why?) Using the Pythagoras Theorem to the right triangle PTQ, we get PQ = = units. Fig. . Let us now find the distance between any two points P( x , y ) and Q( x , y ).
Draw PR and QS perpendicular to the x -axis. A perpendicular from the point P on QS is drawn to meet it at the point T (see Fig. . ).
Then, OR = x , OS = x . So, RS = x – x = PT. Also, SQ = y , ST = PR = y . So, QT = y – y .
Now, applying the Pythagoras theorem in PTQ, we get PQ = PT + QT = ( x – x ) + ( y – y ) Therefore, PQ = Note that since distance is always non-negative, we take only the positive square root. So, the distance between the points P(