Posted by **Ron** on Sunday, October 2, 2011 at 2:39am.

a) What is the area of the tri-angle determined by the lines y= − 1/ 2x + 5,y =6x and they-axis?

(b) If b > 0 and m < 0, then the line y = mx +b cuts off a triangle from the first quadrant. Express the area of that tri-angle in terms ofm andb.

(c) The lines y = mx +5, y = x and the y-axis form a triangle in the first quadrant. Suppose this triangle has an area of 10 square units. Findm.

- math -
**Steve**, Monday, October 3, 2011 at 5:03am
a)

Find where the two slanting lines intersect:

-x/2 + 5 = 6x

x = 10/13

That's the height of a triangle with base on the y-axis, length 5.

Area = 5 * 10/13 = 50/13

b)

the height = b

the width is where y=0: x = -b/m

Area = b * -b/m = -b^2/m

c)

where do they intersect?

mx+5 = x

x = 5/(1-m)

That's the height of the triangle with base on the y-axis, length 5

Area = 5/(1-m) * 5 = 25/(1-m)

25/(1-m) = 10

25 = 10m - 10

m = 35/10 = 3.5

- math -
**Ria**, Wednesday, October 12, 2011 at 12:53pm
a) -x/2 +5 = 6x

x=10/13

A= 1/2 (b)(h)

A= 1/2 (10/13)(5)= 25/13

b) height=b

width= x =-b/m

A= 1/2 (-b/m)(b)= -b^2/2m

c) x=5/(1-m)

10=1/2 (5) (5/(1-m))

m= -1/4

## Answer This Question

## Related Questions

- Math; Solve by Graphing - Which description best describes the solution to the ...
- algebra - Two lines, C and D, are represented by the following equations: Line C...
- Math - Line a intersects line b, forming angles of 135 degrees and 45 degrees. A...
- Math - This question concerns the straight line that passes through the points...
- Mathamatics - This question concerns the straight line that passes through the ...
- Geometry - I have four questions: 1. The area of trapezoid ABCD is 60. One base ...
- Calculus - 1. Find the area of the region bounded by the curves and lines y=e^x ...
- maths geometry - Find and write down a proof that the product of the gradients ...
- Trigonometry - Let è be the angle between the x-axis and the line connecting the...
- calculus - Hello, I have two questions that I've been working on, but I don't ...

More Related Questions