Chapter ISG, Problem 6.6.8P

To determine

A system of inequalities that correctly depicts the points inside a right angled triangle.

Answer to Problem 6.6.8P

One possible set of inequalities are-

  $y>0$ ; $x>0$ ; $\frac{x}{a}+\frac{y}{b}<1$

Explanation of Solution

Given: A right angled triangle with one vertex at the origin. The shaded region contains all the points inside this triangle.

Concept Used: For any right angled triangle the Pythagoras theorem holds true which states that the square of the largest side of a right angled triangle, which is the hypotenuse, is equal to the sum of the squares of the two other sides.

The coordinate axes have the equations $x=0$ and $y=0$ . These lines divides the Cartesian plane into two parts. Depending on which quadrant a point lies, the inequalities are determined. Any point in the shaded region of an inequality should satisfy the inequality otherwise the point doesn’t lie in the shaded region of the inequality or the inequality sign must be reversed to obtain the correct inequation for the region. This is used as a test to correctly determine the inequalities for a given shaded region.

Also, any line having x and y intercepts of a and b units respectively has the equation-

  $\frac{x}{a}+\frac{y}{b}=1$.

Calculations: A right angled triangle with one vertex at the origin can be drawn in several ways. Consider the following right angled triangle, right angled at the origin.

  

Equation of the x -axis is $y=0$ . Since, the shaded region of the triangle lies above this line in the first quadrant, hence, all points will have a positive y -coordinate.

Thus the inequality of the shaded region is-

  $y>0$.

Equation of the y -axis is $x=0$ . Since, the shaded region of the triangle lies on the right hand side of this line in the first quadrant, hence, all points will have a positive x -coordinate.

Thus the inequality of the shaded region is-

  $x>0$.

Equation of line passing through $(a,0)$ and $(0,b)$ is given by-

  $\frac{x}{a}+\frac{y}{b}=1$

Since, (0,0) must lie in the shaded region of this inequality, thus the correct inequality that depicts the shaded region is-

  $\frac{x}{a}+\frac{y}{b}<1$ (Since $\frac{0}{a}+\frac{0}{b}<1$ )

Therefore one possible set of inequalities that describes all the points inside a right angled triangle with one vertex at the origin is-

  $y>0$ ; $x>0$ ; $\frac{x}{a}+\frac{y}{b}<1$

Depending on the orientation of the triangle in the Cartesian plane, the inequalities are subject to change.