Chapter 7 COORDINATE GEOMETRY

Explanation:

 (i) Finding the distance between points (2, 3), and (4, 1) using the Distance Formula yields

d =

(ii) Using the Distance Formula, we can calculate the distance between the positions (-5, 7) and (-1, 3).

d =

(iii) Finding the distance between points (a, b) and (-a, -b) using the Distance Formula yields

d = 

Explanation:

Think about town A at point (0, 0). Town B will therefore be at points (36, 15).

The separation between (0, 0) and (36, 15)

In section 7.2, A is (4, 0) and B is (6, 0)
AB2 = (6 – 4)2 – (0 – 0)2 = 4

Towns A and B will be separated by 39 kilometers. The two towns of A and B, which were the subject of Section 7.2, are 4 km apart.

Explanation:

Let A = (1, 5), B = (2, 3) and C = (–2, –11)

the distance between AB, BC, and CA using the distance formula.

AB =

BC =

CA =

Since AB + AC ≠ BC, BC + AC AB and AC BC.

Points A, B, and C are not parallel as a result.

Explanation:

Let us take A = (5, –2), B = (6, 4) and C = (7, –2)

AB, BC, and CA distances are calculated using the distance formula.

AB =

BC =

CA =

because of AB Equals BC.

A, B, and C are the vertices of an isosceles triangle as a result.

Explanation:

A = (3, 4), B = (6, 7), C = (9, 4), and D = are the values we have (6, 1)

Finding the distances between AB, BC, CD, and DA using the distance formula yields

AB =

BC =

CD =

DA =

As a result, here, all of ABCD's sides are equal. (1)

Now, we shall check the length of its diagonals.

AC =

BD =

ABCD's diagonals are therefore equal. … (2) 

It is evident from (1) and (2) that ABCD is a square.

Champa is accurate as a result.

Explanation:

(i) Let A = (–1, –2), B = (1, 0), C= (–1, 2) and D = (–3, 0)

Finding the distances between AB, BC, CD, and DA using the distance formula yields

AB =

BC =

CD =

DA =

As a result, the quadrilateral's four sides are equal. … (1)

We shall now measure the diagonals' lengths.

AC =

BD =

As a result, the quadrilateral ABCD's diagonals are also equal. … (2)

From (1) and (2), we can say that ABCD is a square.

(ii) Let A = (–3, 5), B= (3, 1), C= (0, 3) and D= (–1, –4)

Finding the distances between AB, BC, CD, and DA using the distance formula yields

AB =

BC =

CD =

DA =

There is no correlation between the lengths of the various sides.

As a result, we are unable to call the quadrilateral ABCD.

(iii) Let A = (4, 5), B= (7, 6), C= (4, 3) and D= (1, 2)

Finding the distances between AB, BC, CD, and DA using the distance formula yields

AB =

BC =

CD =

DA =

Here, the ABCD quadrilateral's two opposite sides are equal. … (1)

Now that we have the diagonal lengths, we can use them.

AC =

BD =

Here, the ABCD diagonals are not equal. (2)

Because ABCD is not a rectangle, we can infer from (1) and (2) that it is a parallelogram.

Explanation:

Consider the x-axis point at (x, 0), which is equidistant from (2, -5) and (–2, 9).

Applying the distance formula, for the parameters stated, we have:

⇒

Squaring is done on both sides, we get

⇒ =

⇒ −4x + 29 = 4x + 85

⇒ 8x = −56

⇒ x = −7

As a result, the x-axis point that is equally spaced apart from (2, -5) and (-2, 9) is (–7, 0)

Explanation:

Considering the Distance formula, we will have

⇒

⇒

We will do Squaring both sides, after we get

100 =

⇒

By factoring this quadratic equation, we can write

⇒

⇒ y (y + 9) – 3 (y + 9) = 0

⇒ (y + 9) (y − 3) = 0

⇒ y = 3, −9

Explanation:

Given that P and R are equally far apart from Q. The distance formula yields

PQ = RQ

⇒

⇒

We will do Squaring both sides, after we get

⇒ 25 + 16 =

⇒

⇒ x = 4, −4

Therefore, Q is (4, 6) or (–4, 6).

To locate QR using the distance formula, we obtain

Utilizing worth of x = 4 QR =

Utilizing worth of x = –4 QR =

Hence, QR =

Finding PR using the distance formula yields

Utilizing worth of x = 4 PR =

Utilizing worth of x =–4 PR =

Hence, x = 4, –4

QR = , PR = 

Explanation:

Given that (3, 6) and (x, y) are equally spaced apart (–3, 4).

The distance formula allows us to write

⇒

We will be done Squaring both sides after we get

⇒

=

⇒ −6x − 12y + 45

= 6x − 8y + 25

⇒ 12x + 4y = 20

⇒ 3x + y = 5.

Explanation:

Let and

To determine the coordinates of the point that splits the join of (-1, 7) and (4, -3) in the ratio 2:3, we use the Section Formula.

As a result, the point's coordinates are (1, 3), which divide the union of (-1, 7) and (4, -3) in the proportion of 2:3.

Explanation:

Finding the coordinates of the line segment's points of trisection at (4, -1) and (–2, –3).

When provided, AC = CD = DB

We need to determine the locations of points C and D.

Let point C's coordinates be and point D's coordinates be .

Point C splits line segment AB in a clear 1:2 ratio, while point D divides it in a 2:1 ratio.

Finding the coordinates of point C, which splits the join of (4, -1) and (-2, -3), using the Section Formula yields the following results:

Finding the coordinates of point D, which splits the join of (4, -1) and (-2, -3) in a 2:1 ratio, using the Section Formula yields the following results:

Hence, the coordinates of point C are (2, ) and the coordinates of point D are (0, ).

Explanation:

Niharika crosses the finish line in position 14 of the distance AD on the second line.

There are one hundred planters. That indicates that she pauses at the 25th flower pot.

Hence, the location's coordinates are as follows: (2 m, 25 m).

Preet posts a red flag while running fifteenth in the distance AD on the eighth line. There are one hundred planters. That indicates that she stops at the 20th flower pot.

Hence, the location's coordinates are as follows: (8, 20).

Finding the distance between the points (2 m, 25 m) and (8 m, 20 m) using the Distance Formula yields

At the exact midpoint of the line segment connecting the two flags, Rashmi puts a blue flag.

To determine the coordinates of this point using the section formula, we obtain

Rashmi posts her flag there, hence the location's coordinates are as follows: (5, ).

This indicates that she posts her flag in the fifth position after traveling =  22.5 metres.

Explanation:

Let the line segment (-1, 6) that connects (-3, 10) and (6, -8) be divided in k:1.

We obtain using the Section formula

⇒ −k – 1 = (−3 + 6k)

⇒ −7k = −2

⇒ k =

As a result, the ratio is :1, which equals 2:7.

Line segment joining points (-3, 10) and (6, -8) is thus divided in 2:7 by point (-1, 6).

Explanation:

Assume that the point of division, with coordinates of (x, 0), splits the line segment connecting A (1, -5) and B (-4, 5) in the ratio k:1.

The formula in Section gives us

… (1)

⇒ 5 = 5k

⇒ k = 1

Adding the value of k to (1) gives us

As a result, the line segment connecting A (1, -5) and B (-4, 5) is divided in 1:1 at point (, 0) on the x-axis.

Explanation:

Let us take  A = (1, 2), B = (4, y), C = (x, 6) and D = (3, 5)

We are aware that parallelogram diagonals cut each other in half. This indicates that the midpoints of diagonals AC and BD would have the same coordinates. … (1)

The midpoint of AC's coordinates is as follows using the Section formula:

The midpoint of BD’s coordinates is as follows using the Section formula:

According to the situation (1), we have

⇒ (1 + x) = 7

⇒ x = 6

Again, according to situation(1), we also get

⇒ 8 = 5 + y

⇒ y = 3

Hence, x = 6 and y = 3

Explanation:

We are looking for point A's coordinates. The diameter is AB, the center's coordinates are (2, -3), and point B's coordinates are (1, 4).

Suppose point A's coordinates are (x, y). The section formula yields

⇒ 4 = x + 1

⇒ x = 3

The section formula yields

⇒ −6 = 4 + y

⇒ y = −10

Hence, point A's coordinates are (3, –10).

Explanation:

A = (–2, –2) and B = (2, –4)

It is Therefore given that AP=AB

PB = AB – AP = AB − AB = AB

So, we consider AP: PB = 3: 4

Let us consider the coordinates of P to be (x, y)

taking the Section formula to find the coordinates of P, we have

Hence, the Coordinates of point P are

Explanation:

 A = (–2, 2) and B = (2, 8)

Let P, Q, and R be the locations on the line segment AB that divide it into four equally sized sections.

Let us consider the coordinates of point, ,

We consider AP = PQ = QR = RS.

This signifies that line segment P is divided by the AB in 1:3.

To determine the coordinates of point P using the Section formula, we get

Therefore, AP = PQ = QR = RS.

This indicates that point Q is AB's midpoint.

To determine point Q's coordinates using the Section formula, we obtain

AP = PQ = QR = RS, so.

Hence, line segment AB is divided by point R in the ratio 3:1.

To determine the coordinates of point P using the Section formula, we get

Hence, P = (–1, ), Q= (0, )and R = (1, )

Explanation:

Let A = (3, 0), B = (4, 5), C = (–1, 4) and D = (–2, –1)

Using the distance formula, we can determine the diagonal AC's length.

Using the distance formula, we can determine the diagonal BD's length.

As Area of rhombus = ½ (product of its diagonals)

= = 24 sq. units

Explanation:

(i) (2, 3), (–1, 0), (2, –4)

Area of Triangle =

= [2 {0 − (−4)} – 1 (−4 − 3) + 2 (3 − 0)]

= [2 (0 + 4) – 1 (−7) + 2 (3)]

= (8 + 7 + 6) = sq. units

(ii) (–5, –1), (3, –5), (5, 2)

Area of Triangle =

= [−5 (−5 − 2) + 3 {2 − (−1)} + 5 {−1 − (−5)}]

= [−5 (−7) + 3 (3) + 5 (4)]

= (35 + 9 + 20)

= (64) = 32 sq. units

Explanation:

(i) (7, –2), (5, 1), (3, k)

As a result of the supplied points' collinearity, the triangle's area is equal to zero.

Area of Triangle =

⇒ [7 (1 − k) + 5 {k − (−2)} + 3 (−2 − 1)]

= (7 − 7k + 5k + 10 − 9) = 0

⇒ (7 − 7k + 5k + 1) = 0

⇒ (8 − 2k) = 0

⇒ 8 − 2k = 0

⇒ 2k = 8

⇒ k = 4

(ii) (8, 1), (k, –4), (2, –5)

The supplied points are collinear, which indicates that the area of the triangle they create is 0.

Area of Triangle =

⇒ [8 {−4 − (−5)} + k (−5 − 1) + 2 {1 − (−4)}]

= (8 − 6k + 10) = 0

⇒ (18 − 6k) = 0

⇒18 − 6k = 0

⇒ 18 = 6k

⇒ k = 3

Explanation:

Let A = (0, –1) =, B = (2, 1) = and

C = (0, 3) =

Area of △ABC =

⇒ Area of △ABC

= [0 (1 − 3) + 2 {3 − (−1)} + 0 (−1 − 1)] =

= 4 sq. units

The midpoints of sides AB, AC, and BC are P, Q, and R, respectively.

Finding the vertices of P, Q, and R using the Section Formula yields the

We apply same formula, Area of △PQR = [1 (1 − 2) + 0 (2 − 0) + 1 (0 − 1)] =

= 1 sq. unit (numerically)

Now, 

Explanation:

Area of Quadrilateral ABCD

= Area of Triangle ABD +

Area of Triangle BCD … (1)

We use a formula to find the area of the triangle:

Area of ABD

=

= [−4 (−5 − 3) – 3 {3 − (−2)} + 2 {−2 − (−5)}]

= (32 – 15 + 6)

= (23) = 11.5 sq units … (2)

To find the area of a triangle once more, use the formula:

Area of △BCD =

= [−3 (−2 − 3) + 3 {3 − (−5)} + 2 {−5 − (−2)}]

= (15 + 24 − 6)

= (33) = 16.5 sq units … (3)

Putting (2) and (3) in (1), we get

Area of Quadrilateral ABCD = 11.5 + 16.5 = 28 sq units

Explanation:

We have △ABC, and we know its vertices.

It must be demonstrated that ar(△ABD) = ar(△ACD).

If point D's coordinates are, then (x, y)

To determine D's coordinates using the section formula, we obtain

Hence, point D's coordinates are (4, 0)

Finding the area of a triangle using a formula:

Area of △ABD =

= [4 (−2 − 0) + 3 {0 − (−6)} + 4 {−6 − (−2)}]

= (−8 + 18 −16)

= (−6) = −3 sq units

No area may be negative.

Hence, we only take its numerical value into account.

△ABD 's area is therefore 3 square units. (1) Again using a formula to determine the area of the triangle:

Area of △ACD =

= [4 (2 − 0) + 5 {0 − (−6)} + 4 {−6 −2 )}]

= (8 + 30 − 32) = ½ (6) = 3 sq units … (2)

From (1) and (2), we get ar(△ABD) = ar(△ACD)

Hence verified.

Explanation:

Let us consider the line will divide the line segment joining A and B (3, 7) in the ratio at at point C. Then, the coordinates of C are

Besides C lies on, hence

Hence, an internal ratio of 2: 9 is needed.

Explanation:

The given  points A B (1, 2) and C (7, 0)  are collinear if

Area of triangle = 0

Explanation:

Let us take P to be the center of the circle that is passing through points A B and C (3, 3). Then AP = BP = CP.

Considering AP = BP

……….(i)

Reconsidering BP = CP

Taking the value of in eq. (i),

Therefore, the centre of the circle is 

Explanation:

Let us consider ABCD to be square and B to be the unknown vertex.

AB = BC

……….(i)

In ABC,

……….(ii)

Taking the value of in eq. (ii),

= 0 or 4

Therefore, the needed vertices of the square are (1, 0) and (1, 4).

Explanation:

(i) With A as the origin, the coordinate axes are AD and AB. The statements P, Q, and R are (4, 6), (3, 2), and (6, 5) respectively.

(ii)Using the coordinate axes CB and CD, and C as the origin. It is obvious that (12, 2), (13, 6), and (10, 3) respectively, provide the points P, Q, and R.

As known the area of the triangle =

Area of PQR (First case) =

=

= = sq. units

Area of PQR (Second case) =

=

= = sq. units

Therefore, the areas are equal in both cases.

Explanation:

Therefore,

DE BC[By Thales theorem]

ADE ABC

……….(i)

AS, Area (ABC) =

= sq. units……….(ii)

From eq. (i) and (ii),

Area (ADE) = Area (ABC) = sq. units

Area (ADE): Area (ABC) = 1: 16

Explanation:

Consider A (4, 2), B (6, 5) and C (1, 4) be the vertices of ABC.

(i) Therefore AD is the median of ABC.

BC has the mid point as D

Its coordinates are =

(ii) Therefore P will divide AD in the ratio 2: 1

Its coordinates are =

(iii) Therefore BE is the median of ABC.

AD has a midpoint as E.

Its coordinates are =

Therefore Q will divide BE in the ratio 2: 1.

Its coordinates are =

Therefore CF is the median of ABC.

AB has a midpoint as F.

Its coordinates are =

Therefore R will divide CF in the ratio 2: 1.

Its coordinates are =

(iv) We notice that the medians AD, BE, and CF is present at the place where the points P, Q, and R coincide. The centroid of the triangle is located at this location.

(v) The midpoints of BC, CA, and AB are D, E, and F, respectively, according to the question.

Coordinates of D are

Coordinates of a point will divide AD in the ratio 2: 1 are

=

The coordinates of E are

The coordinates of a point will divide BE in the ratio 2: 1 are

=

In the same way, the coordinates of a point dividing CF in the ratio 2: 1 are

Therefore, the point is common to AD, BE, and CF and divides them in the ratio 2: 1.

Thus the  median of a triangle are concurrent and the coordinates of the centroid are .

Explanation:

We use the distance formula, PQ =

= =

QR = = =

RS = = =

SP = = =

PQ = QR = RS = SP

Now, PR = = = 6

And SQ = = = 5

PR SQ

since the diagonals are not equal but all the sides are equal.

PQRS is a rhombus.