# Thread: equation of line ...

1. ## equation of line ...

Find the equation of line that passes through (6,8)
and make with both axes a triangle with minimum area.

My try:
let the equation be y=mx+b
area = 1/2 xb
m=(b-8)/-6

then .....

2. $\displaystyle y = mx + b$

$\displaystyle A = \dfrac{xb}{2}$

$\displaystyle m = \dfrac{b-8}{-6}$

Good! Now, you need to find A' to get the minimum area. But for that, you'll need to get A in terms of either x alone, or b alone.

$\displaystyle y = \dfrac{8-b}{6} x + b$

At point (x, y) = (6, 8), find x in terms of b, or vice versa.

Then, substitute this into your expression for the area.

Differentiate the area, and solve for A' = 0.

You'll get a value of b or x depending on what you chose previously.

Substitute this value into the equation y = mx + b and use the point (6, 8) again to find the other value you need.

3. Originally Posted by Unknown008
$\displaystyle y = mx + b$

$\displaystyle A = \dfrac{xb}{2}$

$\displaystyle m = \dfrac{b-8}{-6}$

Good! Now, you need to find A' to get the minimum area. But for that, you'll need to get A in terms of either x alone, or b alone.

$\displaystyle y = \dfrac{8-b}{6} x + b$

At point (x, y) = (6, 8), find x in terms of b, or vice versa.

Then, substitute this into your expression for the area.

Differentiate the area, and solve for A' = 0.

You'll get a value of b or x depending on what you chose previously.

Substitute this value into the equation y = mx + b and use the point (6, 8) again to find the other value you need.
when substitue x=6 ---> y=8 no relation to use it in A=1/2xb

how can I find a relation ?

4. Oops, right. I tried to go from where you went and failed to see something more.

$\displaystyle m = \dfrac{8-b}{6} = \dfrac{b-0}{0-x}$

There, can you continue now?

5. $\displaystyle \frac {8-b}{6}=\frac {-b}{x}$

$\displaystyle x=\frac {-6b}{8-b}$

$\displaystyle area = A =\frac {-3b^2}{8-b}$

$\displaystyle \frac {dA}{db}=0$

$\displaystyle 3b^2-48=0$

$\displaystyle x=\pm 4$

Then when x=4 or -4 we can find x and then the Area.

IS THIS RIGHT ??

6. Not quite.

I'll change A to make things simpler. $\displaystyle A = \dfrac{3b^2}{b-8}$

$\displaystyle \dfrac{dA}{db} = \dfrac{(b-8)6b - 3b^2}{(b-8)^2}$

This gives:

$\displaystyle \dfrac{dA}{db} = \dfrac{3b^2 - 48b}{(b-8)^2}$

You missed the 'b' in 48b.

7. you are right , thanks

b=16
x=12
m=-4/3

y=-(4/3)x+16

thanks a lot