1. WEL-COME

2. SHRI. PADMARAJE VIDYALAYA, SHIROL

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

4. GEOMETRY

5. COORDINATE GEOMETRY

6. Sub unit -
Distance formula

7. 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.

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

9. From B draw perpendicular BQ on X axisY 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

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

11. 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

12. The distance of any point from originX 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)

13. Some examples

14. 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
= √
= √164
= √4*41
= 2 √41
The distance betwween points A & B is 2 √41

15. 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
= √ …(1)

16. 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).

17. Thank- you