1.

find the point which divides the linesegment joining the points( a+b,a-b)and (a-b,a+b) in the ratio 3:2 internally​

Answer»

Step-by-step EXPLANATION:

Let the POINT A( a+b ,a-b) and B (a-b , a+b ) which divide the linesegment AB at O ( X, Y) in the ratio 3:2 internally .

FORMULA : [mx2+nx1/m+n , my2+ny1/m+n]

put the VALUE

[3(a-b) + 2(a+b)/3+2 , 3(a+b) + 2(a-b) /3+2]

[3a-3b + 2a+2b /5 , 3a+3b + 2a-2b / 5]

[5a-b /5 , 5a+b /5]

[5( a-(b÷5) ) /5 , 5( a+(b÷5) ) /5]

[a-b/5 , a+b/5]

hence, so the point O Which divide the linesegment AB is

X=a-b/5 and Y=a+b/5



Discussion

No Comment Found

Related InterviewSolutions