Exercises on Analytical Geometry

Test your knowledge with questions about general aspects of Analytical Geometry involving distance between two points, midpoint, straight line equation, among other topics.

Take advantage of the comments in the resolutions to clarify your doubts and gain more knowledge.

question 1

Calculate the distance between two points: A (-2,3) and B (1,-3).

Correct answer: d (A, B) = 3 square root of 5.

To solve this question, use the formula to calculate the distance between two points.

straight d open parentheses straight A comma straight B closes parentheses space equal to space square root of left parenthesis straight x with straight B subscript space minus straight space x with straight A subscript right parenthesis squared space plus space left parenthesis squared with straight B subscript space minus squared squared space with straight A subscript right parenthesis squared end of source

We substitute the values ​​in the formula and calculate the distance.

straight d open parenthesis straight A comma straight B close parenthesis space equals space square root of left parenthesis 1 space minus space left parenthesis minus 2 right parenthesis right parenthesis squared space plus space left parenthesis minus 3 space minus space 3 right parenthesis squared end of root straight d open square brackets A square comma B closes brackets space equals space square root of left parenthesis 1 space plus space 2 right parenthesis squared space plus space left parenthesis minus 3 space minus space 3 right parenthesis squared end of root straight d open brackets straight A comma straight B closes brackets space equal to space square root of 3 squared space plus space left parenthesis minus 6 right parenthesis squared end of root straight d open parentheses straight A comma straight B closes parentheses space equals space square root of 9 space plus space 36 end of root straight d open parentheses straight A comma straight B closes parentheses space equals space square root of 45

The root of 45 is not exact, so it is necessary to carry out rooting until you can no longer remove any number from the root.

straight d open parentheses straight A comma straight B closes parentheses space equals space square root of 9 space. space 5 end of straight root d opens square brackets A straight comma B closes brackets space equals square root space of 3 squared space. space 5 end of root straight d open parentheses straight A comma B closes parentheses space equal to space 3 square root of 5

Therefore, the distance between points A and B is 3 square root of 5.

question 2

On the Cartesian plane there are points D (3.2) and C (6.4). Calculate the distance between D and C.

Right answer: square root of 13.

Being straight d with DP subscript space equal to space open vertical bar straight x with straight C subscript space minus space straight x with straight D subscript close vertical bar and straight d with CP subscript space equals space open vertical bar straight y with straight C subscript space minus space straight y with straight D subscript close vertical bar, we can apply the Pythagorean Theorem to the DCP triangle.

left parenthesis d with DC subscript right parenthesis squared space equals space open parenthesis d with DP subscript close parenthesis squared space plus space open square brackets d with CP subscript close square brackets left bracket d with DC subscript right square bracket space equal to open brackets square x with straight C subscript space minus straight space x with straight D subscript close square brackets space more space open brackets straight y with straight C subscript space minus straight space y with straight D subscript close square parentheses squared space d with DC subscript space space space space equal to square root space of open parentheses square x with straight C subscript space minus space straight x with straight D subscript closes square parentheses space more space opens parentheses straight y with straight C subscript space minus straight space y with straight D subscript closes parentheses squared end of root

Substituting the coordinates in the formula, we find the distance between the points as follows:

straight d with DC subscript equals space square root of open parentheses straight x with straight C subscript space minus space straight x with straight D subscript closes squared parentheses space plus space open parenthesis y with straight C subscript space minus straight space y with straight D subscript closes squared end of root straight space d with subscript DC equals square root of parenthesis left 6 minus 3 right parenthesis squared space plus space left parenthesis 4 minus 2 right parenthesis squared end of root straight space d with subscript DC equal to square root of 3 to square space plus space 2 squared end of root straight space d with subscript DC equal to square root of 9 space plus space 4 end of root straight space d with subscript DC equal to square root of 13

Therefore, the distance between D and C is square root of 13

See too: Distance between Two Points

question 3

Determine the perimeter of triangle ABC, whose coordinates are: A (3,3), B (–5, –6) and C (4,–2).

Correct answer: P = 26.99.

1st step: Calculate the distance between points A and B.

straight d with AB subscript equals space square root of open parentheses straight x with straight A subscript space minus straight space x with straight B subscript closes squared parentheses space plus space opens square brackets y with straight A subscript space minus straight space y with straight B subscript closes squared parentheses end of root straight d with AB subscript equals square root of 3 minus left parenthesis minus 5 right parenthesis right parenthesis squared space plus space left parenthesis 3 minus left parenthesis minus 6 right parenthesis right parenthesis squared end of straight root d with AB subscript equals square root of 8 squared space plus 9 squared space end of straight root d with AB subscript equals square root of 64 space plus space 81 end of root straight d with AB subscript equals square root of 145 straight d with AB subscript approximately equal 12 comma 04

2nd step: Calculate the distance between points A and C.

straight d with AB subscript equals space square root of open parentheses straight x with straight A subscript space minus straight space x with straight C subscript closes parentheses ao square space plus space open parentheses square y with straight A subscript space minus straight space y with straight C subscript closes squared parentheses end of root straight d with A straight C subscript end of subscript equals square root of left parenthesis 3 minus 4 right parenthesis squared space plus space left parenthesis 3 minus left parenthesis minus 2 right parenthesis right parenthesis squared end of root straight d with A straight C subscript end of subscript equals square root of parenthesis left minus 1 right parenthesis squared space plus space 5 squared end of root straight d with A straight C subscript end of subscript equals square root of 1 space plus space 25 end of root straight d with A straight C subscript end of subscript equal to square root of 26 straight d with A straight C subscript end of subscript approx equal 5 comma 1

3rd step: Calculate the distance between points B and C.

straight d with subscript BC equal to space square root of open parentheses straight x with straight B subscript space minus straight space x with straight C subscript closes squared parentheses space plus space open parentheses straight y with straight B subscript space minus straight space y with straight C subscript closes squared parentheses end of root straight d with BC subscript equals square root of left parenthesis minus 5 minus 4 right parenthesis squared space plus space left parenthesis minus 6 minus left parenthesis minus 2 right parenthesis right parenthesis squared end of straight root d with BC subscript equals square root of left parenthesis minus 9 right parenthesis squared space plus space left parenthesis minus 4 right parenthesis squared end of straight root d with BC subscript equal to square root of 81 space plus space 16 end of straight root d with BC subscript equal to square root of 97 straight d with BC subscript approximately equal space 9 comma 85

4th step: Calculate the perimeter of the triangle.

straight p space equal to straight space L with AB subscript space plus straight L with AC subscript space plus straight space L with BC subscript straight p space equals space 12 comma 04 space plus space 5 comma 1 space plus space 9 comma 85 straight p space equals space 26 comma 99

Therefore, the perimeter of triangle ABC is 26.99.

See too: Triangle Perimeter

question 4

Determine the coordinates that locate the midpoint between A (4,3) and B (2,-1).

Correct answer: M (3, 1).

Using the formula to calculate the midpoint, we determine the x coordinate.

straight x with straight M subscript space equal to space numerator straight x with straight A subscript space plus space straight x with straight B subscript over denominator 2 end of fraction straight x with straight M subscript space equal to space numerator 4 space plus space 2 over denominator 2 end of fraction straight x with straight M subscript space equal to space 6 over 2 straight x with straight M subscript space equal to space 3

The y coordinate is calculated using the same formula.

straight y with straight M subscript space equal to space numerator straight y with straight A subscript space plus straight space y with straight B subscript over denominator 2 end of fraction straight x with straight M subscript space equal to space numerator 3 space plus space left parenthesis minus 1 right parenthesis over denominator 2 end of fraction straight x with straight M subscript space equal to space numerator 3 space minus space 1 over denominator 2 end of fraction straight x with straight M subscript space equal to space 2 over 2 straight x with straight M subscript space equal to space 1

According to the calculations, the midpoint is (3.1).

question 5

Calculate the coordinates of the vertex C of a triangle, whose points are: A (3, 1), B (–1, 2) and the barycenter G (6, –8).

Correct answer: C (16, –27).

The barycenter G (xGyG) is the point where the three medians of a triangle meet. Its coordinates are given by the formulas:

straight x with straight G subscript space equal to numerator space straight x with straight A subscript more straight space x with straight B subscript space plus straight space x with straight C subscript space over denominator 3 end of fraction and straight y with straight G subscript space equal to space numerator straight y with straight A subscript more straight space y with straight B subscript space plus straight space y with straight C subscript space over denominator 3 end of fraction

Substituting the x values ​​of the coordinates, we have:

straight x with straight G subscript space equal to numerator space straight x with straight A subscript more straight space x with straight B subscript space plus space straight x with straight C subscript space over denominator 3 end of fraction 6 space equal to space numerator 3 space plus space left parenthesis minus 1 right parenthesis space plus straight space x with straight C subscript over denominator 3 end of fraction 6 space. space 3 space equals space 3 space minus 1 space plus straight space x with a straight C subscript 18 space equals space 2 space plus straight space x with straight C subscript 18 space minus space 2 space equal to space straight x with straight C subscript straight x with straight C subscript space equal to space 16

Now we do the same process for y values.

straight y with straight G subscript space equal to space numerator straight y with straight A subscript space plus straight space y with straight B subscript space plus straight space y with straight C subscript space over denominator 3 end of fraction minus 8 space equal to space numerator 1 space plus space 2 space plus straight space y with straight C subscript space over denominator 3 end of fraction minus 8 space equal to space numerator 3 space plus straight space y with straight C subscript space over denominator 3 end of fraction minus 8 space. space 3 space equals space 3 space plus straight space y with straight C subscript space minus 24 space minus space 3 space space equal to space straight y with straight C subscript straight y with straight C subscript space equal to space minus 27

Therefore, vertex C has the coordinates (16,-27).

question 6

Given the coordinates of the collinear points A (-2, y), B (4, 8), and C (1, 7), determine what the value of y is.

Correct answer: y = 6.

For the three points to be aligned, the determinant of the matrix below must be equal to zero.

straight D narrow space equals space open vertical bar table row with cell with straight x with straight A subscript end of cell cell with straight y with straight A subscript end of cell 1 row with cell with straight x with straight B subscript end of cell cell with straight y with straight B subscript end of cell 1 row with cell with straight x with straight C subscript end of cell cell with straight y with straight C subscript end of cell 1 end of table close vertical bar space equal to space 0

1st step: replace the x and y values ​​in the matrix.

straight D narrow space equals space open vertical bar table row with cell with minus 2 end of cell straight y 1 row with 4 8 1 row with 1 7 1 end of table close vertical bar

2nd step: write the elements of the first two columns beside the matrix.

straight D narrow space equals space open vertical bar table row with cell with minus 2 end of cell straight y 1 row with 4 8 1 row with 1 7 1 end of table closes vertical bar table row with cell bold less bold 2 end of cell bold y row with bold 4 bold 8 row with bold 1 bold 7 end of table

3rd step: multiply the elements of the main diagonals and add them up.

table row with cell bold less bold 2 end of cell bold italic y bold 1 row with 4 bold 8 bold 1 row with 1 7 bold 1 end of table table row with cell with minus 2 end of cell y row with bold 4 8 row with bold 1 bold 7 end of table space space space space space space space space space space space space space arrow in northwest position arrow in northwest position arrow in northwest position space space space space space space space space space space space Diagonals space main

The result will be:

table row with cell bold minus 2 bold. bold 8 bold. bold 1 end of cell plus cell with bold y bold. bold 1 bold. bold 1 end of cell plus cell with bold 1 bold. bold 4 bold. bold 7 end of cell blank row with cell with less bold bold 16 end of cell blank cell with bolder space bold y end of cell blank cell with more bold space 28 end of cell blank end of table table row with blank row with blank end of table

4th step: multiply the elements of the secondary diagonals and invert the sign in front of them.

table row with cell with minus 2 end of cell straight and bold 1 row with 4 bold 8 bold 1 row with bold 1 bold 7 bold 1 end of table table row with cell with bold less bold 2 end of cell bold y row with bold 4 8 row with 1 7 end of table arrow in northeast position arrow in northeast position arrow in northeast position Diagonals space secondary

The result will be:

table row with cell less bold space bold left parenthesis bold 1 bold. bold 8 bold. bold 1 bold right parenthesis end of cell minus cell bold left parenthesis bold minus bold 2 bold. bold 1 bold. bold 7 bold right parenthesis end of cell minus cell bold left parenthesis bold y bold. bold 4 bold. bold 1 bold right parenthesis end of cell blank row with cell with less space bold 8 end of cell blank cell with bolder space bold 14 end of cell blank cell less bold bold space 4 bold y end of cell blank end of table table row with blank row with blank end of table

5th step: join the terms and solve the addition and subtraction operations.

straight D space equals space minus space 16 space plus space straight y space plus space 28 space minus space 8 space plus space 14 space minus space 4 straight y 0 space equal to space minus space 3 straight y space plus space 18 3 straight y space equal to space 18 space straight space y space equal to space 18 over 3 space straight space y space equal to space 6

Therefore, for the points to be collinear, the value of y must be 6.

See too: Matrices and Determinants

question 7

Determine the area of ​​triangle ABC, whose vertices are: A (2, 2), B (1, 3) and C (4, 6).

Correct answer: Area = 3.

The area of ​​a triangle can be calculated from the determinant as follows:

straight A narrow space equal to 1 half space open vertical bar table row with cell with straight x with straight A subscript end of cell cell with straight y with straight A subscript end of cell 1 row with cell with straight x with straight B subscript end of cell cell with straight y with straight B subscript end of cell 1 row with cell with straight x with straight C subscript end of cell cell with straight y with straight C subscript end of cell 1 end of table close vertical bar space double right arrow space A narrow space equal to 1 half space open vertical bar straight D close bar vertical

1st step: replace the coordinate values ​​in the matrix.

straight D narrow space equals space open vertical bar table line with 2 2 1 line with 1 3 1 line with 4 6 1 end of table close vertical bar

2nd step: write the elements of the first two columns beside the matrix.

straight D narrow space equals space open vertical bar table line with 2 2 1 line with 1 3 1 line with 4 6 1 end of table closes vertical bar table line with bold 2 bold 2 line with bold 1 bold 3 line with bold 4 bold 6 end of table

3rd step: multiply the elements of the main diagonals and add them up.

table row with bold 2 bold 2 bold 1 row with 1 bold 3 bold 1 row with 4 6 bold 1 end of table table row with 2 2 row with bold 1 3 row with bold 4 bold 6 end of table space space space space space space space space space space space space arrow in position northwest arrow in northwest position arrow in northwest position space space space space space space space space space space space Diagonals space main

The result will be:

table row with bold 2 bold cell. bold 3 bold. bold 1 end of cell plus cell with bold 2 bold. bold 1 bold. bold 4 end of cell plus cell with bold 1 bold. bold 1 bold. bold 6 end of cell blank row with bold 6 blank cell with bolder space bold 8 end of cell blank cell with more bold space 6 end of cell blank end of table table row with blank row with blank end of table

4th step: multiply the elements of the secondary diagonals and invert the sign in front of them.

space space space table line with 2 2 bold 1 line with 1 bold 3 bold 1 line with bold 4 bold 6 bold 1 end of table table line with bold 2 bold 2 row with bold 1 3 row with 4 6 end of table arrow in northeast position arrow in northeast position arrow in northeast position Diagonals space secondary

The result will be:

table row with cell less bold space bold left parenthesis bold 1 bold. bold 3 bold. bold 4 bold right parenthesis end of cell minus cell bold left parenthesis bold 2 bold. bold 1 bold. bold 6 bold right parenthesis end of cell minus cell bold left parenthesis bold 2 bold. bold 1 bold. bold 1 bold right parenthesis end of cell blank row with cell with less space bold 12 end of cell blank cell with less bold space bold 12 end of cell blank cell with less bold space bold 2 end of cell blank end of table table row with blank row with blank end of table

5th step: join the terms and solve the addition and subtraction operations.

straight D space equals space plus space 6 space more space 8 space more space 6 space less space 12 space less space 12 space minus space 2 straight D space equals space 20 space minus space 26 straight D space equals space minus 6

6th step: calculate the area of ​​the triangle.

straight A narrow space equals 1 half space open vertical bar straight D close vertical bar straight A narrow space equals 1 half space open vertical bar minus 6 closes straight vertical bar A narrow space equals 1 half space. space 6 straight A narrow space equal to 6 over 2 straight A narrow space equal to space 3

See too: Triangle Area

question 8

(PUC-RJ) Point B = (3, b) is equidistant from points A = (6, 0) and C = (0, 6). Therefore, point B is:

a) (3, 1)
b) (3, 6)
c) (3, 3)
d) (3, 2)
e) (3, 0)

Correct alternative: c) (3, 3).

If points A and C are equidistant from point B, it means that the points are located at the same distance. So, dAB = dCB and the formula to calculate is:

straight d with AB subscript equals straight d with CB subscript square root of open parentheses straight x with straight A subscript space minus straight space x with straight B subscript closes squared parentheses space plus space opens parentheses straight y with straight A subscript space minus straight space y with straight B subscript closes squared parentheses end of root equals square root of open parentheses straight x with straight C subscript space minus straight space x with straight B subscript close squared parentheses space plus space open parentheses square y with straight C subscript space minus straight space y with straight B subscript closes parentheses ao root end square

1st step: replace coordinate values.

square root of open parentheses 6 space minus space 3 closes squared parenthesis space more space open parenthesis 0 minus straight space b closes squared parenthesis end of root equals square root of open parentheses 0 space minus space 3 closes square parentheses space plus space opens parentheses 6 space minus square space b closes parentheses to square end of root square root of 3 squared space plus space open parenthesis minus straight space b close parenthesis squared end of root equals square root of open parentheses minus space 3 closes squared parentheses space more space open parentheses 6 space minus straight space b closes squared parentheses end of square root of 9 space plus straight space b squared end of root space equals space square root of 9 space plus space opens parentheses 6 space minus straight space b closes parentheses ao root end square

2nd step: solve the roots and find the value of b.

open parentheses square root of 9 space plus straight space b squared end of root space closes squared parentheses equals space open parentheses square root of 9 space plus space open parentheses 6 space less straight space b closes squared parentheses end of root closes squared parentheses 9 space plus straight space b squared space equals space 9 space plus space opens parentheses 6 space minus straight space b closes parentheses ao square straight b squared space equals space 9 space minus space 9 space plus space left parenthesis 6 space minus straight space b parenthesis right. left parenthesis 6 space minus square space b right parenthesis square space b squared space equals space 36 space minus space 6 straight b space minus space 6 straight b space plus space straight b squared straight b squared space equal to space 36 space minus space 12 straight b space plus space straight b squared 12 straight b space equal to space 36 space plus straight space b squared space minus straight space b squared 12 straight b space equal to space 36 straight b space equal to space 36 over 12 straight b space equal to space 3

Hence, point B is (3, 3).

See too: Exercises on distance between two points

question 9

(Unesp) The triangle PQR, in the Cartesian plane, with vertices P = (0, 0), Q = (6, 0) and R = (3, 5), is
a) equilateral.
b) isosceles but not equilateral.
c) scalene.
d) rectangle.
e) obtuse angle.

Correct alternative: b) isosceles but not equilateral.

1st step: calculate the distance between points P and Q.

straight d with subscript PQ equal to space square root of open parentheses straight x with straight P subscript space minus space straight x with straight Q subscript closes squared parentheses space plus space open parentheses straight y with straight P subscript space minus straight space y with straight Q subscript closes squared parentheses end of root straight d with PQ subscript equal to square root of left parenthesis 0 minus 6 right parenthesis squared space plus space left parenthesis 0 minus 0 right parenthesis squared end of straight root d with subscript PQ equal to root square of left parenthesis minus 6 right parenthesis squared space plus space 0 end of root straight d with PQ subscript equal to square root of 36 straight d with PQ subscript equal space to space 6

2nd step: calculate the distance between points P and R.

straight d with PR subscript equal to space square root of open parentheses straight x with straight P subscript space minus straight space x with straight R subscript closes parentheses ao square space plus space open parentheses straight y with straight P subscript space minus straight space y with straight R subscript closes squared parentheses end of root straight d with PR subscript equals square root of left parenthesis 0 minus 3 right parenthesis squared space plus space left parenthesis 0 minus 5 right parenthesis squared end of straight root d with PR subscript equals square root of left parenthesis minus 3 right parenthesis squared space plus space left parenthesis minus 5 parenthesis right squared end of root straight d with PR subscript equal to square root of 9 space plus space 25 end of root straight d with PR subscript space equal to root space 34 square

3rd step: calculate the distance between points Q and R.

straight d with QR subscript equal to square root space of open parentheses straight x with straight Q subscript space minus straight space x with straight R subscript closes parentheses ao square space plus space open parentheses square y with straight Q subscript space minus straight space y with straight R subscript closes squared parentheses end of root straight d with QR subscript equals square root of left parenthesis 6 minus 3 right parenthesis squared space plus space left parenthesis 0 minus 5 right parenthesis to square end of straight root d with QR subscript equals square root of left parenthesis 3 right parenthesis squared space plus space left parenthesis minus 5 right squared end of straight root d with QR subscript equal to square root of 9 space plus space 25 end of straight root d with QR subscript space equal to space square root of 34

4th step: judge the alternatives.

a) WRONG. The equilateral triangle has equal three-sided measurements.

b) CORRECT. The triangle is isosceles, as two sides have the same measurement.

c) WRONG. The scalene triangle has the measurements of three different sides.

d) WRONG. The right triangle has a right angle, that is, 90º.

e) WRONG. The obtuse-angled triangle has one of the angles greater than 90º.

See too: Triangle Classification

question 10

(Unitau) The equation of the straight line that passes through points (3.3) and (6.6) is:

a) y = x.
b) y = 3x.
c) y = 6x.
d) 2y = x.
e) 6y = x.

Correct alternative: a) y = x.

To make it easier to understand, we will call point (3,3) A and point (6,6) B.

Taking P(xPyP) as a point that belongs to the line AB, then A, B and P are collinear and the equation of the line is determined by:

straight D narrow space equals space open vertical bar table row with cell with straight x with straight A subscript end of cell cell with straight y with straight A subscript end of cell 1 row with cell with straight x with straight B subscript end of cell cell with straight y with straight B subscript end of cell 1 row with cell with straight x with straight P subscript end of cell cell with straight y with straight P subscript end of cell 1 end of table close vertical bar equal to space 0 space

The general equation of the line passing through A and B is ax + by + c = 0.

Substituting the values ​​in the matrix and calculating the determinant, we have:

straight D narrow space equals space open vertical bar table line with 3 3 1 line with 6 6 1 line with straight x straight y 1 end of table close vertical bar table line bold 3 bold 3 line bold 6 bold 6 line bold x bold y end of table straight D space equals space 18 space plus space 3 straight x space plus space 6 straight y space minus space 6 straight x space minus 3 straight y space minus 18 0 space equals space 3 straight x space plus space 6 straight y space minus space 6 straight x space minus 3 straight y 0 space equal to space 3 straight y space minus space 3 straight x 3 straight x space equal to space 3 straight y straight x space equal to space straight y

Therefore, x = y is the equation of the line that passes through the points (3,3) and (6,6).

See too: Line Equation

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