Question

Determine whether the given points are on the graph of the equation.

$x^2+xy+y^2=4;\left(0,-2\right),\left(1,-2\right),\left(2,-2\right)$

Short answer

The points $\left(0,-2\right)\text{and}\left(2,-2\right)$ lie on the graph of the equation whereas the point $\left(1,-2\right)$ does not lie on the graph.

Step-by-step solution

Step 1. Apply the concept of coordinates.

The set of all points $\left(x,y\right)$ in the Cartesian plane or coordinate plane that satisfy the equation is known as the graph of the equation in x and y coordinates.

Step 2. Substitute point and check.

Consider the provided set of points $\left(0,-2\right),\left(1,-2\right)\text{and}\left(2,-2\right)$and equation of graph $x^2+xy+y^2=4$.

Recall that the set of all points $\left(x,y\right)$ in the Cartesian plane or coordinate plane that satisfy the equation is known as the graph of the equation in x and y coordinates.

Substitute the point $\left(0,-2\right)$in the equation $x^2+xy+y^2=4$ and check whether the left-hand side is equal to the right-hand side.

localid="1643620485383" role="math" $$\begin{gathered} 0^2+0\left(-2\right)+{\left(-2\right)}^2=4 \\ 0+0+4=4 \\ 4=4 \end{gathered}$$

The above equality holds true so, the point $\left(0,-2\right)$lie on the graph of the equation.

Substitute the point $\left(1,-2\right)$in the equation $x^2+xy+y^2=4$and check whether the left-hand side is equal to the right-hand side.

localid="1643620505636" $$\begin{gathered} 1^2+1\left(-2\right)+{\left(-2\right)}^2=4 \\ 1-2+4=4 \\ 3=4 \end{gathered}$$

The above equality does not hold so, the point $\left(1,-2\right)$does not lie on the graph of the equation.

Substitute the point $\left(2,-2\right)$in the equation $x^2+xy+y^2=4$and check whether the left-hand side is equal to the right-hand side.

localid="1643620527958" $$\begin{gathered} 2^2+2\left(-2\right)+{\left(-2\right)}^2=4 \\ 4-4+4=4 \\ 4=4 \end{gathered}$$

The above equality holds true so, the point $\left(2,-2\right)$lie on the graph of the equation.

Thus, the points $\left(0,-2\right)\text{and}\left(2,-2\right)$lie on the graph of the equation whereas the point $\left(1,-2\right)$does not lie on the graph.

Step 3. Graph the equation.

Now, on the Cartesian plane graph the line$x^2+xy+y^2=4$ and verify.