WEL-COME

SHRI. PADMARAJE VIDYALAYA, SHIROL

SUBJECT TEACHER SOU. DHERE U.A. B.Sc .B.ed.

GEOMETRY

COORDINATE GEOMETRY

Sub unit - Distance formula

CASE 3- Distance between any two points in XY-plane such that the line segment joining these two points is not parallel two any of the axes.

Y X o B(x2y2) A(x1y1) Let point A(x1y1) & B(x2y2) be any two point in XY plane

From B draw perpendicular BQ on X axis Y X o B(x2y2) A(x1y1) D(x2y1) Q From B draw perpendicular BQ on X axis From A draw perpendicular AD on BQ,B-D-Q

By case 2, AD =|x2-x1| and BD = |y2-y1| Y X o B(x2y2) A(x1y1) Q D(x2y1)

Y X o B(x2y2) A(x1y1) Q D(x2y1) In right angled triangle ADB, by pythagoras theorem, AB2 =AD2+BD2 = (x2-x1)2 + (y2-y1)2 since, the length of line seg. AB is always non negative, we get the distance as AB = √(x2-x1)2 + (y2-y1)2

The distance of any point from origin X o BY distance formula OA = √(x2-x1)2 +(y2-y1)2 Here x1y1 is 0 Therefore OA = √(x-0)2 +(y-0)2 OA =√x2+y2 A(x,y) P(x,0)

Some examples

Q. Find the distance between the given points 1) A(3,-4) , B(-5,6) , Solution : Let A(3,-4) =(x1y1)& B(-5,6) Then by using distance formula, AB = √(x2-x1)2+(y2-y1)2 = √ (-5-3)2+(6-(-2))2 = √ (-8)2 + (6+4)2 = √ 64+100 = √164 = √4*41 = 2 √41 The distance betwween points A & B is 2 √41

Q.Show that point (5,11) is equidistant from the points (-5,13) and (3,1) . Let A(5,ll), B(-5,13) & (3,1) Then by distance formula, AB=√(5-(-50))2 + (11-13)2 = √(5+5)2 + (-2)2 = √ (10)2 + 4 =√ (100+4 = √104 …(1)

AC = √(5-3)2+(11-1)2 = √(2)+(10)2 = √ 4+100 = √104 From (1) and(2),AB=AC. Point A is equidistant from point B and C. Hence, the point (5,11) is equidistance from the points (-5,13)and(3,1).

Thank- you