Question

If \(x+p=0, y+4=0\) and \(x+2 y+4=0\) are concurrent, then \(p=\) (1) 4 (2) \(-2\) (3) \(-4\) (4) 2

Short answer

Answer: The value of p when the lines are concurrent is p = -4.

Step-by-step solution

Rewrite the given equations in the standard form

The given equations are: 1. \(x + p = 0\) 2. \(y + 4 = 0\) 3. \(x + 2y + 4 = 0\)

Solve equation (2) for y

From equation (2), we can solve for y directly: \(y = -4\)

Substitute the value of y in equation (3)

Now that we have a value for y, substitute this value into equation (3) to solve for x: \(x + 2(-4) + 4 = 0\)

Solve the modified equation (3) for x

Now, we'll simplify and solve the equation for x: \(x - 8 + 4 = 0\) \(x - 4 = 0\) \(x = 4\)

Substitute the values of x and y in equation (1)

Now that we have values for x and y, substitute these values into equation (1) to solve for p: \(4 + p = 0\)

Solve equation (1) for p

To find the value of p, simply rearrange the equation: \(p = -4\) So, the correct answer is option (3) p = \(-4\).

Key concepts

Linear Equations

Linear equations are the building blocks of algebra. They represent straight lines when graphed on a coordinate plane and are classified as equations of the first degree because each term is either a constant or the product of a constant and a single variable.

• A linear equation in one variable could look like: \(x + 3 = 0\).
• In two variables, it might appear as: \(x + 2y = 7\).
• Standard form for a two-variable linear equation is typically \(Ax + By = C\), where \(A\), \(B\), and \(C\) are constants.

They have simple solutions. You can find the value of the variable by isolating it on one side of the equation. In more complex systems, you'll often use substitution or elimination methods to find solutions, as seen in the exercise discussed here. A thorough understanding of linear equations is beneficial as they are pervasive in numerous mathematical contexts.

Solving Systems of Equations

When you're dealing with multiple linear equations simultaneously, you are solving a system of equations. The main goal here is to find a set of values that satisfy all equations in the system.

• One common method to solve these systems is substitution, where you solve one equation for a variable and substitute this into the other equation.
• Another method is elimination, which involves adding or subtracting equations to eliminate a variable.

The exercise presented focuses on using substitution. First, solve one equation for one variable, then substitute that value into another equation. The solution gives the specific point where the lines represented by the equations intersect, which is the solution to the system. When the equations are concurrent, like in the original problem, they meet at a single point in the coordinate system.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, blends algebra and geometry to discuss figures and their properties. It provides a link between algebraic equations and geometric curves.

• Coordinates of a point are represented as \((x, y)\) in a plane.
• Lines, represented by linear equations like \(x + 2y = 8\), have slopes and intercepts that describe their position in a plane.
• The point where two lines intersect gives their concurrent meeting point, which also serves as a solution to both equations in a system.

The use of coordinate geometry allows us to visualize systems of equations. The meeting point of lines, as shown in the exercise, is their concurrent point. Therefore, understanding coordinate geometry is fundamental for tackling problems involving concurrent lines, as it provides insight into their spatial relationships.