Question

show that the straight line x(p+2q)+y(p+3q)=p+q passes through a fixed point for different values of p and q

Answer 1

Answer on Question #47277 – Math – Analytic Geometry

Show that the straight line $x(p + 2q) + y(p + 3q) = p + q$ passes through a fixed point for different values of $p$ and $q$.

Solution:

A line is an infinite geometrical figure. If we extend a line segment at both ends, we get a line. A line is always represented by two arrows at its ends, to indicate its infinite nature.

Mathematically, a line can be represented by a linear equation, that is, an equation of degree one. The most general form of a straight line is

$$ax + by + c = 0; \ a^2 + b^2 \ne 0$$

All the points $(x, y)$ that lie on the line satisfy the equation on that line, and conversely, if a point $(x, y)$ satisfies the equation of a line, it lies on that line.

We can simplify our equation by opening the parenthesis:

$$xp + 2xq + yp + 3yq - p - q = 0$$

Combine like terms:

$$(xp + yp - p) + (2xq + 3yq - q) = 0$$ $$p(x + y - 1) + q(2x + 3y - 1) = 0$$

Thus, the given line passes through the intersection of the lines $x + y - 1 = 0$ and $2x + 3y - 1 = 0$.

Now we have to solve the system of the obtained equations.

$$\left\{ \begin{array}{l} x + y - 1 = 0 \\ 2x + 3y - 1 = 0 \end{array} \right.$$

Solve system of equations by elimination by addition. We multiply the first equation by -2 and add two equations.

$$\left\{ \begin{array}{l} -2x - 2y + 2 = 0 \\ 2x + 3y - 1 = 0 \end{array} \right.$$ $$-2x - 2y + 2 + 2x + 3y - 1 = 0$$

Simplify by combining like terms:

$$-2y + 2 + 3y - 1 = 0$$ $$y = -1$$

Now we find the value of $x$. We can substitute the value of $y$ either in the first or second equation. In our case we choose the first equation.

$$x + (-1) - 1 = 0$$ $$x - 2 = 0$$ $$x = 2$$

The coordinate of the find point will be equal to $(2, -1)$.

We can also check obtained solution. Substitute the values of $x$ and $y$ into the original equations.

$$\left\{ \begin{array}{l} 2 + (-1) - 1 = 0 \\ 2(2) + 3(-1) - 1 = 0 \end{array} \right.$$

Simplify the system of equations.

$$\left\{ \begin{array}{l} 2 - 2 = 0 \\ 4 - 3 - 1 = 0 \end{array} \right.$$ $$\left\{ \begin{array}{l} 0 = 0 \\ 0 = 0 \end{array} \right.$$

Thus we got the true statement. The required coordinate of point are equal $(2, -1)$.

We can also consider different values of $p$ and $q$. For example, we put $p = 3$ and $q = 1$. Substitute into original equation of the line.

$$x(3 + 2) + y(3 + 3) = 3 + 1$$

Simplify by opening the parenthesis.

$$5x + 6y = 4$$

Now substitute the find coordinate of the $x$ and $y$.

$$5(2) + 6(-1) = 4$$ $$10 - 6 = 4$$ $$4 = 4$$

We also can consider other values of $p$ and $q$. For example, $p = -2$ and $q = -1$. As in previous part we substitute into original equation. We obtained the following result.

$$x(-2 + 2(-1)) + y(-2 + 3(-1)) = -2 - 1$$

Simplify by opening the parenthesis.

$$-4x - 5y = -3$$

Now substitute the find coordinate of the $x=2$ and $y=-1$.

$$\begin{array}{l} -4(2) - 5(-1) = -3 \\ -8 + 5 = -3 \\ -3 = -3 \\ \end{array}$$

We again got the true statement.

In this way our system has a unique solution therefore for any values of $p$ and $q$ line $x(p + 2q) + y(p + 3q) = p + q$ passes through a fixed point with coordinates $(2,-1)$.

www.AssignmentExpert.com