Question

The diagram shows triangle ABC with vertices A (-2, -1), B (1, 2), and C (4, -1). What are the three inequalities that make up the system that describes the region enclosed by the triangle? Enter each of the inequalities:
BC:

Answer 1

$\large\underline{\sf{Solution-}}$

Given three vertices of the triangle

• Coordinates of A (- 2, - 1)

• Coordinates of B (1, 2)

• Coordinates of C (4, - 1)

We know that,

Equation of line passing through the points (a, b) and (c, d) is given by

$\boxed{ \bf \: y - b \: = \: \dfrac{d - b}{c - a}(x - a)}$

So,

Using thus formula,

Equation of line AB joining the points A (- 2, - 1) and B (1, 2) is

$\rm :\longmapsto\:y + 1 = \dfrac{2 + 1}{1 + 2}(x + 2)$

$\rm :\longmapsto\:y + 1 = \dfrac{3}{3}(x + 2)$

$\rm :\longmapsto\:y + 1 = x + 2$

$\bf\implies \:x - y + 1 = 0 - - (1)$

Now,

Equation of line BC joining the points B (1, 2) and C (4, - 1) is

$\rm :\longmapsto\:y - 2 = \dfrac{ - 1 - 2}{4 - 1}(x - 1)$

$\rm :\longmapsto\:y - 2 = \dfrac{ -3}{3}(x - 1)$

$\rm :\longmapsto\:y - 2 = - x + 1$

$\bf\implies \:x + y - 3 = 0 - - (2)$

Now,

Equation of line AC joining the points A (- 2, - 1) and C (4, - 1) is

$\rm :\longmapsto\:y + 1 = \dfrac{ - 1 + 1}{4 + 2}(x + 2)$

$\bf\implies \:y + 1 = 0 - - - (3)$

To enclosed the triangular region ABC, we must use the rule of origin test.

If origin satisfies the given inequality, then shades towards the origin otherwise away from the origin.

If we look carefully at triangular region, all the lines gave region shaded towards the origin.

That means, origin must satisfy the inequality.

So inequalities are as under,

For equation AB :-

$\bf :\longmapsto\:x - y + 1 \geqslant 0$

For Equation BC :-

$\bf :\longmapsto\:x + y - 3 \leqslant 0$

For Equation AC :-

$\bf :\longmapsto\:y + 1 \geqslant 0$