IfP(y) is any point on the line joining the points A(a,0) and BOHthat.

1.	IfP(y) is any point on the line joining the points A(a,0) and BOHthat. x +2 1
Answer» Equation of the line is Ax+By+C=0 If x-interception isA(a,0)and y-interception isB(0,b))then they need to satisfy line equation Aa+C=0;Bb+C=0 from which you can find A=−C/a;B=−C/b After substituting to line equation −C/ax−C/by=−C or x/a+y/b=1

IfP(y) is any point on the line joining the points A(a,0) and BOHthat. x +2 1

Answer»

Equation of the line is

Ax+By+C=0

If x-interception isA(a,0)and y-interception isB(0,b))then they need to satisfy line equation

Aa+C=0;Bb+C=0

from which you can find

A=−C/a;B=−C/b

After substituting to line equation

−C/a*x−C/b*y=−C

x/a+y/b=1