`star` Cartesian plane into two parts

`star` Graphical Solution of Linear Inequalities in Two Variables

`star` Graphical Solution of Linear Inequalities in Two Variables

`\color{purple} ✍️` We know that a line divides the Cartesian plane into two parts. Each part is known as a `color(green)("half plane.")`

`\color{purple} ✍️` A `color(green)("vertical line")` will divide the plane in `color(green)("left and right half planes")` and a `color(green)("non-vertical line")` will divide the plane into `color(green)("lower and upper half planes")` (Figs. 6.3 and 6.4).

`\color{purple} ✍️` A point in the Cartesian plane will either lie on a line or will lie in either of the half planes I or II.

`\color{purple} ✍️` We shall now examine the relationship, if any, of the points in the plane and the inequalities `ax + by < c` or `ax + by > c.`

`\color{purple} ✍️` A `color(green)("vertical line")` will divide the plane in `color(green)("left and right half planes")` and a `color(green)("non-vertical line")` will divide the plane into `color(green)("lower and upper half planes")` (Figs. 6.3 and 6.4).

`\color{purple} ✍️` A point in the Cartesian plane will either lie on a line or will lie in either of the half planes I or II.

`\color{purple} ✍️` We shall now examine the relationship, if any, of the points in the plane and the inequalities `ax + by < c` or `ax + by > c.`

`\color{red} ✍️` Let us consider the line `color(purple)(ax + by = c, \ \ a ≠ 0, b ≠ 0) .................................. (1)`

`color(red)("There are three possibilities namely :")`

`(i) color(blue)(ax + by = c) \ \ \ \ (ii) color(blue)(ax + by > c) \ \ \ \ \ (iii) color(blue)(ax + by < c.)`

`\color{red} ✍️` In case `(i),` clearly, all points `(x, y)` satisfying `(i)` lie on the line it represents and conversely.

`\color{red} ✍️` Consider case (ii), `color(red)("let us first assume that" \ \ b > 0.)`

Consider a point `color(blue)(P (α,β))` on the line `ax + by = c, b > 0,` so that `aα + bβ = c.` Take an arbitrary point `Q (α , γ)` in the half plane `II` (Fig 6.5).

● we interpret, `γ > β`

or `b γ > bβ` or `aα + b γ > aα + bβ` or `aα + b γ > c` i.e., `Q(α, γ )` satisfies the inequality `ax + by > c`

● Thus, all the points lying in the half plane `II` above the line ax + by = c satisfies the inequality `ax + by > c.` Conversely, let `(α, β)` be a point on line `ax + by = c` and an arbitrary point `Q(α, γ)` satisfying

`ax + by > c`

so that `aα + bγ > c`

`⇒ aα + b γ > aα + bβ`

`color(blue)(⇒ γ > β (as b > 0))`

`color(blue)("This means that the point (α, γ ) lies in the half plane II.")`

`color(red)("There are three possibilities namely :")`

`(i) color(blue)(ax + by = c) \ \ \ \ (ii) color(blue)(ax + by > c) \ \ \ \ \ (iii) color(blue)(ax + by < c.)`

`\color{red} ✍️` In case `(i),` clearly, all points `(x, y)` satisfying `(i)` lie on the line it represents and conversely.

`\color{red} ✍️` Consider case (ii), `color(red)("let us first assume that" \ \ b > 0.)`

Consider a point `color(blue)(P (α,β))` on the line `ax + by = c, b > 0,` so that `aα + bβ = c.` Take an arbitrary point `Q (α , γ)` in the half plane `II` (Fig 6.5).

● we interpret, `γ > β`

or `b γ > bβ` or `aα + b γ > aα + bβ` or `aα + b γ > c` i.e., `Q(α, γ )` satisfies the inequality `ax + by > c`

● Thus, all the points lying in the half plane `II` above the line ax + by = c satisfies the inequality `ax + by > c.` Conversely, let `(α, β)` be a point on line `ax + by = c` and an arbitrary point `Q(α, γ)` satisfying

`ax + by > c`

so that `aα + bγ > c`

`⇒ aα + b γ > aα + bβ`

`color(blue)(⇒ γ > β (as b > 0))`

`color(blue)("This means that the point (α, γ ) lies in the half plane II.")`

`\color{green} ★ \color{green} \mathbf(KEY \ POINTS)`

`"1"` The region containing all the solutions of an inequality is called `color(blue)("the solution region.")`

`2.` In order to identify the half plane represented by an inequality, it is just sufficient to take any point `(a, b)` (not online) and check whether it satisfies the inequality or not.

If it satisfies, then the inequality represents the half plane and shade the region which contains the point, otherwise, the inequality represents that half plane which does not contain the point within it. For convenience, the point `(0, 0)` is preferred.

`3.` If an inequality is of the type `color(blue)(ax + by ≥ c)` or `color(blue)(ax + by ≤ c,)` then the points on the line `color(blue)(ax + by = c)` are also included in the solution region. So draw a dark line in the solution region.

`4.` If an inequality is of the form `ax + by > c` or `ax + by < c,` then the points on the line `ax + by = c` are not to be included in the solution region. So draw a broken or dotted line in the solution region.

`"1"` The region containing all the solutions of an inequality is called `color(blue)("the solution region.")`

`2.` In order to identify the half plane represented by an inequality, it is just sufficient to take any point `(a, b)` (not online) and check whether it satisfies the inequality or not.

If it satisfies, then the inequality represents the half plane and shade the region which contains the point, otherwise, the inequality represents that half plane which does not contain the point within it. For convenience, the point `(0, 0)` is preferred.

`3.` If an inequality is of the type `color(blue)(ax + by ≥ c)` or `color(blue)(ax + by ≤ c,)` then the points on the line `color(blue)(ax + by = c)` are also included in the solution region. So draw a dark line in the solution region.

`4.` If an inequality is of the form `ax + by > c` or `ax + by < c,` then the points on the line `ax + by = c` are not to be included in the solution region. So draw a broken or dotted line in the solution region.

Q 3131401322

Solve `3x + 2y > 6` graphically.

Graph of` 3x + 2y = 6` is given as dotted line in the Fig 6.8. This line divides the `xy`-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes (Fig 6.8) and determine if this point satisfies the given inequality, we note that `3 (0) + 2 (0) > 6` or `0 > 6` , which is false. Hence, half plane I is not the solution region of the given inequality. Clearly, any point on the line does not satisfy the given strict inequality. In other words, the shaded half plane II excluding the points on the line is the solution region of the inequality.

Q 3151401324

Solve `3x - 6 ≥ 0` graphically in two dimensional plane.

Graph of `3x – 6 = 0` is given in the

We select a point, say (0, 0) and substituting it in given inequality, we see that: `3 (0) – 6 ≥ 0 or – 6 ≥ 0` which is false.

Thus, the solution region is the shaded region on the right hand side of the line x = 2.

Q 3171401326

Solve y < 2 graphically.

Graph of `y = 2` is given in the Fig 6.10. Let us select a point, (0, 0) in lower half plane I and putting y = 0 in the given inequality, we see that `1 × 0 < 2 or 0 < 2` which is true. Thus, the solution region is the shaded region below the line `y = 2.` Hence, every point below the line (excluding all the points on the line) determines the solution of the given inequality.