pre calc

[pencar]

show that the equation for a line with nonzero x and y intercepts can be written as x/a + y/b =1, where a is the x-intercept and b is the y-intercept.

I know the y=mx +b is the equation for a line with m being slope and b being y-intercept. If a is x-intercept then one point would be (a,0) and if b is y-intercept then one point is (0,b)

Then slope should be -b/a? so y=(-b/a) x+ b ???

 

[DrLee]

show that the equation for a line with nonzero x and y intercepts can be written as x/a + y/ b =1, where a is the x-intercept and b is the y-intercept

We can write y = mx + b as one form of a linear equation. Since we are given that y is not equal to zero when x = 0, it follows by inspection that b is not equal to zero. So we can re-write the standard form as

y - mx = b and then divide both sides of the equation by b since b is not equal to zero.

This gives us

y/b - mx/b = 1

So if we assign a = -b/m, then the equation above becomes:

x/a + y/b = 1 which is what the problem statement required. (Note this assumes m is not equal to zero.)

 

[pencar]

thanks

 

[BigGlenntheHeavy]

\(\displaystyle Given: \ x \ intercept \ = \ (a,0) \ and \ y \ intercept \ = \ (0,b), \ a,b \ \ne \ 0. \ Show \ that \ \frac{x}{a}+\frac{y}{b} \ = \ 1\)

\(\displaystyle Then \ m \ = \ \frac{b-0}{0-a} \ = \ \frac{-b}{a}, \ (definition \ of \ slope).\)

\(\displaystyle Ergo, \ y-0 \ = \ \frac{-b}{a}(x-a), \ (point-slope \ form).\)

\(\displaystyle y \ = \ \frac{-bx}{a}+b, \ now \ since \ b \ \ne \ 0, \ \frac{y}{b} \ = \ \frac{-x}{a}+1, \ or \ \frac{x}{a}+\frac{y}{b} \ = \ 1. \ QED\)

\(\displaystyle Note: \ Dr. \ Lee's \ proof \ is \ a \ fallacy, \ (a \ bunch \ of \ crap) \ to \ wit: \ "So \ if \ we \ assign \ a \ = \ -b/m,\)

\(\displaystyle \ then \ the \ equation \ above \ becomes" \ this \ is \ known \ as \ begging \ the \ question \ (assuming \ what \ is \\)

\(\displaystyle to \ be \ proved).\)