Score:

Find the equation of the line with slope 5 that passes through the point $\text{[math]}$

Solution

We use the point–slope form of a line:

$\text{[math]}$

If we substitute the known values, the equation of the line is:

$\text{[math]}$

Therefore, the equation of the line is:

$\text{[math]}$

Find the equation of the line with slope −3 that passes through the point (0,11)

Solution

We use the slope–intercept form of a line:

$\text{[math]}$

Substituting the known values, the equation of the line is:

$\text{[math]}$

Therefore, the equation of the line is:

$\text{[math]}$

Find the equation of the line that passes through the points and $\text{[math]}$

Solution

We find the slope of the line:

$\text{[math]}$

We use the slope–intercept form of a line:

$\text{[math]}$

Substituting the known values, the equation of the line is:

$\text{[math]}$

Therefore, the equation of the line is:

$\text{[math]}$

Find the equation of the line that passes through $\text{[math]}$ and is parallel to the line $\text{[math]}$

Solution

We find the slope of the line $\text{[math]}$

$\text{[math]}$

Since the line that passes through $\text{[math]}$ is parallel to the given line, both lines have the same slope $\text{[math]}$

We substitute the slope and the point into the point–slope equation:

$\text{[math]}$

Therefore, the equation of the line is:

$\text{[math]}$

Find the equation of the line that passes through $\text{[math]}$ and is perpendicular to the line $\text{[math]}$

Solution

We find the slope of the line $\text{[math]}$

$\text{[math]}$

Since the line that passes through the point $\text{[math]}$ is perpendicular to the line $\text{[math]}$ , then its slope is $\text{[math]}$

We substitute the slope and the point through which the line passes into the point-slope equation and obtain

$\text{[math]}$

Therefore, the equation of the line is:

$\text{[math]}$

Determine whether the lines $\text{[math]}$ and $\text{[math]}$ are parallel, perpendicular, or neither.

Solution

We find the slope of the line $\text{[math]}$ .

$\text{[math]}$

We find the slope of the line $\text{[math]}$ .

$\text{[math]}$

Since $\text{[math]}$ , the lines are not parallel.

Since $\text{[math]}$ , the lines are not perpendicular.

Therefore, the lines are neither parallel nor perpendicular.

Determine whether the lines $\text{[math]}$ and $\text{[math]}$ are parallel, perpendicular, or neither.

Solution

We find the slope of the line $\text{[math]}$ .

$\text{[math]}$

We find the slope of the line $\text{[math]}$ .

$\text{[math]}$

Since $\text{[math]}$ , the lines are not parallel.

Since $\text{[math]}$ , the lines are perpendicular.

Determine whether the lines $\text{[math]}$ and $\text{[math]}$ are parallel, perpendicular, or neither.

Solution

We find the slope of the line $\text{[math]}$ .

$\text{[math]}$

We find the slope of the line $\text{[math]}$ .

$\text{[math]}$

Since $\text{[math]}$ , the lines are parallel.

Write all possible forms of the equation of the line that passes through the points $\text{[math]}$ and $\text{[math]}$ .

Solution

We know that the line passes through the points $\text{[math]}$ and $\text{[math]}$ . Therefore, the vector that joins these two points is:

$\text{[math]}$

With this information we can obtain the equations of the line.

Equation of the line through two points:

$\text{[math]}$

Vector equation:

$\text{[math]}$

Parametric equations:

$\text{[math]}$

Continuous equation:

$\text{[math]}$

General equation:

$\text{[math]}$

Explicit equation:

$\text{[math]}$

Point-slope equation:

$\text{[math]}$

Given the parallelogram $\text{[math]}$ with $\text{[math]}$ , find the coordinates of vertex $\text{[math]}$ .

Solution

Before finding the coordinates of the vertex, let us observe the following figure.

representación gráfica de un paralelogramo abcd

We know that the vector from $\text{[math]}$ to $\text{[math]}$ must be equal to the vector from $\text{[math]}$ to $\text{[math]}$ , that is:

$\text{[math]}$

We perform the calculations:

$\text{[math]}$

where $\text{[math]}$ is the x-coordinate of point $\text{[math]}$ , and is its y-coordinate. Thus, we have:

$\text{[math]}$

Therefore, point $\text{[math]}$ is

$\text{[math]}$

Classify the triangle determined by the points: $\text{[math]}$ .

Solution

To classify the triangle, we must first calculate the length of each of its sides. We do this as follows:

$\text{[math]}$

Let us note that the following holds:

$\text{[math]}$

Therefore, the triangle is isosceles. In addition, the following also holds:

$\text{[math]}$

Thus, the triangle is also right-angled. This can be observed in the following figure.

representacion grafica de triangulo en problema de la recta

Find the slope and the y-intercept of the line $\text{[math]}$ .

Solution

We solve for $\text{[math]}$ :

$\text{[math]}$

From this, we can see that the slope is:

$\text{[math]}$

While the y-intercept is:

$\text{[math]}$

Study the relative position of the lines with equations:

a $\text{[math]}$
b $\text{[math]}$
c $\text{[math]}$
d $\text{[math]}$

Solution

Let us note that the coefficients of line 1 and line 2 are proportional:

$\text{[math]}$

Therefore, lines 1 and 2 are coincident (they are the same line).

Likewise, let us note that the coefficients of $\text{[math]}$ and $\text{[math]}$ in lines 1 and 3 are proportional; however, the constant terms are not proportional:

$\text{[math]}$

Therefore, lines 1 and 3 are parallel. Consequently, lines 2 and 3 are parallel (since lines 1 and 2 are the same).

Finally, let us observe that the coefficients of $\text{[math]}$ and $\text{[math]}$ in line 4 are not proportional to the coefficients of any other line:

$\text{[math]}$

Therefore, line 4 intersects lines 1, 2, and 3.

Find the equation of the line $\text{[math]}$ that passes through $\text{[math]}$ and is parallel to the line $\text{[math]}$ .

Solution

Let us observe the following figure of two parallel lines.

representación gráfica de dos rectas paralelas

We know that two lines are parallel if they have the same slope:

$\text{[math]}$

Therefore, line $\text{[math]}$ has the following form (point–slope):

$\text{[math]}$

Setting the equation equal to zero, we obtain:

$\text{[math]}$

Given the quadrilateral $\text{[math]}$ with vertices $\text{[math]}$ and $\text{[math]}$ . Verify that it is a parallelogram and determine its center.

Solution

For the quadrilateral to be a parallelogram, we must have:

$\text{[math]}$ and
$\text{[math]}$

Let us note that:

$\text{[math]}$

Therefore, $\text{[math]}$ .

On the other hand:

$\text{[math]}$

Thus, the quadrilateral is a parallelogram.

Now we must find the midpoint. We know that the diagonals intersect at their midpoint (which is the center of the parallelogram), so it is sufficient to find the midpoint of one diagonal. The midpoint of diagonal $\text{[math]}$ is:

$\text{[math]}$

Thus, the center is point $\text{[math]}$ . Let us observe the figure of the parallelogram.

representacion grafica de un cuadrilatero

Find the equation of the line that passes through the point (2, -3) and is parallel to the line that joins the points (4, 1)and (-2, 2)

Solution

Let $\text{[math]}$ be the line that joins the given points. Then the line $\text{[math]}$ we are looking for is parallel to $\text{[math]}$ . Therefore, they have the same slope:

$\text{[math]}$

Using the point–slope form, the equation of line $\text{[math]}$ is:

$\text{[math]}$

Therefore, the equation of $\text{[math]}$ , written equal to zero, is:

$\text{[math]}$

Points $\text{[math]}$ and $\text{[math]}$ are vertices of an isosceles triangle $\text{[math]}$ , whose vertex $\text{[math]}$ lies on the line $\text{[math]}$ , with $\text{[math]}$ and $\text{[math]}$ being the equal sides. Find the coordinates of vertex $\text{[math]}$ .

Solution

Let us write the coordinates of point $\text{[math]}$ as $\text{[math]}$ . Since $\text{[math]}$ , we have:

$\text{[math]}$

Solving for $\text{[math]}$ :

$\text{[math]}$

Additionally, since sides $\text{[math]}$ and $\text{[math]}$ are equal, we have:

$\text{[math]}$

Squaring both sides:

$\text{[math]}$

Substituting $\text{[math]}$ :

$\text{[math]}$

Expanding and solving the resulting equation (the quadratic terms cancel), we obtain:

$\text{[math]}$

Substituting into the equation for $\text{[math]}$ :

$\text{[math]}$

Therefore, the point is:

$\text{[math]}$

representacion grafica de triangulo isosceles en problema de la recta

The line passes through point $\text{[math]}$ and is parallel to line $\text{[math]}$ . Find $\text{[math]}$ and $\text{[math]}$ .

Solution

Since line $\text{[math]}$ passes through point $\text{[math]}$ , substituting its coordinates gives:

$\text{[math]}$

Since $\text{[math]}$ , the coefficients are proportional:

$\text{[math]}$

Therefore:

$\text{[math]}$

Given triangle $\text{[math]}$ with coordinates $\text{[math]}$ and $\text{[math]}$ , find the equation of the median that passes through vertex $\text{[math]}$ .

Solution

representacion grafica de triangulo con mediana en problema de la recta

The median passes through the midpoint of segment $\text{[math]}$ . We compute this point:

$\text{[math]}$

We now write the equation of the line that passes through points $\text{[math]}$ and $\text{[math]}$ :

$\text{[math]}$

Simplifying, we obtain:

$\text{[math]}$

In a parallelogram, one vertex is known: $\text{[math]}$ , and the intersection point of the diagonals is $\text{[math]}$ . Another vertex is at the origin. Find:

a The other vertices
b The equations of the diagonals
c The lengths of the diagonals

Solution

a) The other vertices

Since is the midpoint of segment $\text{[math]}$ :

$\text{[math]}$

Thus:

$\text{[math]}$

So $\text{[math]}$ .

Since is also the midpoint of segment $\text{[math]}$ :

$\text{[math]}$

Thus:

$\text{[math]}$

In this way, $\text{[math]}$ . Thus, the four vertices are the points $\text{[math]}$ and $\text{[math]}$ .

b The equations of the diagonals.

In this case we only have to use the formula for the line that passes through two points. First, for the diagonal $\text{[math]}$ :

$\text{[math]}$

Simplifying a bit, we obtain $\text{[math]}$ . Then, for the diagonal $\text{[math]}$ we have:

$\text{[math]}$

Which, after simplifying, we obtain $\text{[math]}$ .

c The length of the diagonals.

To calculate the length of the diagonals, it is enough to calculate the distance between the appropriate vertices. For the diagonal $\text{[math]}$ we have:

$\text{[math]}$

While for the diagonal $\text{[math]}$ the length is:

$\text{[math]}$

The graph of the parallelogram is the following:

representacion grafica de paralelogramo y coordenadas del vertice

Summarize with AI:

Did you like this article? Rate it!

5.00 (2 Note(n))

Agostina Babbo

Agostina Babbo is an English and Italian to Spanish translator and writer, specializing in product localization, legal content for tech, and team sports—particularly handball and e-sports. With a degree in Public Translation from the University of Buenos Aires and a Master's in Translation and New Technologies from ISTRAD/Universidad de Madrid, she brings both linguistic expertise and technical insight to her work.

Solved Exercises and Problems on Linear Equations I

What Are First-Degree Equations?

Quadratic Equations: Solved Exercises

Solved Exercises: First-Degree Equations

First-Degree Equation Problems

Exercises on 3×3 Systems of Equations

Exercises and Solved Problems with Quadratic Equations

Problems and Exercises Involving First-Degree Equations

Exercises and Solved Problems Involving Linear Equations I

Cancel reply

Solved Exercises and Problems on Linear Equations I

Theory

What Are First-Degree Equations?

Other Exercises

Quadratic Equations: Solved Exercises

Solved Exercises: First-Degree Equations

First-Degree Equation Problems

Exercises on 3×3 Systems of Equations

Exercises and Solved Problems with Quadratic Equations

Problems and Exercises Involving First-Degree Equations

Exercises and Solved Problems Involving Linear Equations I

Cancel reply