The point which lies on the line 3x - 4y + 5 = 0 is (1) (2, 1) (2) (0, 0) (3) (1, 2) (4) (-2, 5). Which of the following points lie inside the triangle formed by x + (1) (2, 2) (2)... The point which lies on the line 3x - 4y + 5 = 0 is (1) (2, 1) (2) (0, 0) (3) (1, 2) (4) (-2, 5). Which of the following points lie inside the triangle formed by x + (1) (2, 2) (2) (3, 3) (3) (0, 0)? If (a, a^2) falls inside the angle made by the linear equations y.
Understand the Problem
The question involves identifying points that either lie on a given line or inside a triangle formed by certain linear equations. It is asking for both the point on the line defined by the equation and which of the specific points are inside the triangle formed by another set of linear equations.
Answer
The point on the line is $(1, 2)$ and the point inside the triangle is $(2, 2)$.
Answer for screen readers
The point that lies on the line $3x - 4y + 5 = 0$ is $(1, 2)$.
The point inside the triangle formed by the equations is $(2, 2)$.
Steps to Solve
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Find the equation of the line Start with the line equation given by $3x - 4y + 5 = 0$. Rearranging gives: $$ 4y = 3x + 5 $$ $$ y = \frac{3}{4}x + \frac{5}{4} $$
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Test candidate points for the line Plug each candidate point into the line equation to see if it satisfies the equation (i.e., makes it true):
- For point (2, 1): $$ 3(2) - 4(1) + 5 = 6 - 4 + 5 = 7 \quad \text{(not on the line)} $$
- For point (0, 0): $$ 3(0) - 4(0) + 5 = 5 \quad \text{(not on the line)} $$
- For point (1, 2): $$ 3(1) - 4(2) + 5 = 3 - 8 + 5 = 0 \quad \text{(on the line)} $$
- For point (-5, 2): $$ 3(-5) - 4(2) + 5 = -15 - 8 + 5 = -18 \quad \text{(not on the line)} $$
- For point (-2, 5): $$ 3(-2) - 4(5) + 5 = -6 - 20 + 5 = -21 \quad \text{(not on the line)} $$
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Identify points inside the triangle The triangle is formed by the equations $x + y = 5$, $x = 0$, and $y = 0$. Find their intersection points:
- Intersection of $x + y = 5$ and $x = 0$ gives point (0, 5).
- Intersection of $x + y = 5$ and $y = 0$ gives point (5, 0).
- Intersection of $x = 0$ and $y = 0$ gives point (0, 0).
The vertices of the triangle are (0, 0), (0, 5), and (5, 0).
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Test candidate points for inside the triangle To check if a point $(a, b)$ is inside the triangle, it must satisfy the following inequalities:
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$x + y < 5$ (for the line $x + y = 5$)
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$x > 0$ (for the vertical line $x = 0$)
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$y > 0$ (for the horizontal line $y = 0$)
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For point (2, 2): $$ 2 + 2 < 5 \quad \text{(true), } 2 > 0 \quad \text{(true), } 2 > 0 \quad \text{(true)} $$ This point is inside the triangle.
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For point (3, 3): $$ 3 + 3 < 5 \quad \text{(false)} $$ This point is outside the triangle.
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For point (0, 0): It lies exactly on the vertex of the triangle, hence it is not considered inside.
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The point that lies on the line $3x - 4y + 5 = 0$ is $(1, 2)$.
The point inside the triangle formed by the equations is $(2, 2)$.
More Information
The point $(1, 2)$ satisfies the line equation, while $(2, 2)$ is inside the triangle formed by the sections established by the respective equations.
Tips
- A common mistake is checking if a point is on a line without properly substituting values into the equation. Always ensure to evaluate the equation thoroughly.
- Misjudging points as inside the triangle when they are actually on the edges or vertices. Points need to strictly satisfy the inequalities to be considered inside.
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