2 Algebra Questions, Need Help Please

**1. Find the area of a triangle formed by the (x,y)-axes and the line with equation 4x +3y - 12 = 0.**

^ With this one I started out by drawing a graph and using slope-intercept form to create a right triangle from the line that was drawn. I got a base of 3 and a height of 4, which would be 12/2 = 6. This is probably super wrong but just throwing it out there...

**2. Y varies jointly as x and z and inversely as the product of w and p. If y = 3/28 when x = 3, z = 10, w = 7, and p = 8, find y when x = 5, z = 12, w = 9, and p =10.**

Any help is appreciated, THANKS!

Re: 2 Algebra Questions, Need Help Please

1. The y intercept is 4 and the x intercept is 3, so it's actually a height of 4 and a base of 3. But your area of 6 square units is correct.

Re: 2 Algebra Questions, Need Help Please

Quote:

Originally Posted by

**StephenSkinner** **1. Find the area of a triangle formed by the (x,y)-axes and the line with equation 4x +3y - 12 = 0.**

^ With this one I started out by drawing a graph and using slope-intercept form to create a right triangle from the line that was drawn. I got a base of 3 and a height of 4, which would be 12/2 = 6. This is probably super wrong but just throwing it out there...

The only problem I see is "which would be 12/2= 6". **What** would be 6? Do you know the formula for the area of a triangle?

Quote:

**2. Y varies jointly as x and z and inversely as the product of w and p. If y = 3/28 when x = 3, z = 10, w = 7, and p = 8, find y when x = 5, z = 12, w = 9, and p =10.**

Any help is appreciated, THANKS!

Re: 2 Algebra Questions, Need Help Please

I was just trying to say that the height of the triangle is 4 and the base is 3 so the equation would look like "(4x3/2) = 12/2 = 6" So the area is 6. As far as the other problem goes though, I have no clue what to do.