157k views
5 votes
Find the equation of straight line passing through the points (a,0) and (0,b). If the point (1,1) lies on the equation, prove that

1/a + 1/b =1​

2 Answers

5 votes

Answer:

step by step

Explanation:

Line passing (a,0) (0,b)

slope = (y2 - y1) / (x2 - x1) = - b/a

pass (0,b), y interception: b

Line: y = -b/a * x + b

pass(1,1) y=1, x=1

1 = -b/a + b

1/b = (-b/a) /b + b/b

1/b = -1/a + 1

1/a + 1/b = 1

User Alex Che
by
6.0k points
4 votes

Answer:

do the following

Explanation:

the question says that the line is passing thur the point 1,1 ... so when x = 1 so does y ... also when x= a.. then y=0 and when y=b then x =0 .. and this is a straight line... also 1/a + 1/b = 1

if you were to draw on a graph... this .. you would see that when b=2 then x=0 and also when a=2.. then y = 0 ....

and 1/2 + 1/2 =1

so that's the line.... draw in just roughly.... you'll see... then what is the equation of that line.. usually we would use the slope intercept formula for the equations.. there are more than one way to show the line.. is my point.. but.. if we use y = mx + b where m = -1 and b = 2 then...

y= - x+2

That answer got long.. sorry :/

User Dasqueel
by
7.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.