bisector of AB. Therefore, the graph of x – y = is the perpendicular bisector of AB (see Fig. . ).
Example : Find a point on the y -axis which is equidistant from the points A( , ) and B(– , ). Solution : We know that a point on the y -axis is of the form ( , y ). So, let the point P( , y ) be equidistant from A and B. Then ( – ) + ( – y ) = (– – ) + ( – y ) + + y – y = + + y – y y = y = Fig.
. Fig. . So, the required point is ( , ).
Let us check our solution : AP = ( – ) ( – ) BP = (– – ) ( – ) Note : Using the remark above, we see that ( , ) is the intersection of the y -axis and the perpendicular bisector of AB. EXERCISE . . Find the distance between the following pairs of points : (i) ( , ), ( , ) (ii) (– , ), (– , ) (iii) ( a , b ), (– a , – b ) .
Find the distance between the points ( , ) and ( , ). Can you now find the distance between the two towns A and B discussed in Section . . .
Determine if the points ( , ), ( , ) and (– , – ) are collinear. . Check whether ( , – ), ( , ) and ( , – ) are the vertices of an isosceles triangle. .
In a classroom, friends are seated at the points A, B, C and D as shown in Fig. . . Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees.
Using distance formula, find which of them is correct. . Name the type of