How do you find the inequalities of a right isosceles triangle?

Question

The following system of inequalities describes a right isosceles triangle. 

X>0

y>0

y<mX+4

a. Find m

b. Find the area of the triangle

 

 

How do I do this problem?!

Answer 1

Elise,

For this problem, you have to use the coordinate plane and graph.

 

First, graph the two inequalities, x >0 and y> 0. The intersection region is the first quadrant.

 

Now when you plot the y-intercept of the third inequality, y < mx + 4, which is b = 4 or  (0, 4), you notice that for the triangle to be isosceles the legs have to measure 4.

 

The second leg, therefore, is the segment from (0,0) to (4,0) on the x-axis.

 

a)  The slope , m, of the line that passes through (0,4) and (4,0)

 

    m = (0  -4)/(4  - 0)

   m  =  -1

 

b)  Area = (1/2)(4)(4)

 

            =  8 square units

Answer 2

Hi Elise,

 

The region you are looking at is bounded by the x-axis the y-axis and the line y = mx+4

x > 0, y > 0, and y < mx +4

 

(a)

at y = 0,

 

0 = mx +4

x = -4/m but we know that X > 0

 

-4/m > 0  is true if and only if m ≤ -1

 

(b) Note that from y = mx + 4, if x = 0, y =4

 

The height of the triangle is h

h = y = 4

Base of trian is

b = x = -4/m

Area, A = hb/2  = (4*-4/m)/2

 

A = -8/m   , m ≤ -1