NCERT Solution for Coordinate Geometry

(i) (2, 3), (4, 1) (ii) (–5, 7), (–1, 3) (iii) (a, b), (–a, –b)

Sol. (i) Here x_{1} = 2, y_{1} = 3, x_{2} = 4 and y_{2} = 1

∴ The required distance

Sol. Part-I

Let the points be P(0, 0) and Q(36, 15).

Sol. Let the points be A(1, 5), B (2, 3) and (�2, �11)

A, B and C are collinear, if

AB + BC = AC

AC + CB = AB

But AB + BC ≠ AC

AC + CD ≠ AB

BA + AC ≠ BC

∴ A, B and C are not collinear.

Sol. Let the points be A (5, –2), B (6, 4) and (7, –2).

We have AB = BC ≠ AC

∴ΔABC is an isosceles triangle.

Sol. Let the number of horizontal columns represent the x-coordinates whereas the vertical rows represent the y-coordinates.

∴ The points are:

A(3, 4), B(6, 7), C(9, 4) and D(6, 1)

(i) (�1, �2), (1, 0), (�1, 2), (�3, 0)

(ii) (�3, 5), (3, 1), (0, 3), (�1, �4)

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Sol. (i) Let the points be: A(�1, �2), B(1, 0), C(�1, 2) and D(�3, 0).

(ii) Let the points be A (�3, 5), B(3, 1), C(0, 3) and D(�1, �4).

⇒ A, B, C and D are collinear. Thus, ABCD is not a quadrilateral.

(iii) Let the points be A (4, 5), B (7, 6), C (4, 3) and D(1, 2).

Sol. We know that any point on x-axis has its ordinate = 0.

Let the required point be P(x, 0).

Let the given points be A(2, �5) and B(�2, 9)

Sol. The given points are P(2, �3) and Q(10, y).

Sol. Let the points be A(x, y), B (3, 6) and C(�3, 4).

Squaring both sides,

(3 – x)^{2} + (6 – y)^{2} = (–3 – x)^{2} + (4 – y)^{2}

⇒ (9 + x^{2} – 6x) + (36 + y^{2} – 12y) = (9 + x^{2} + 6x) + (16 + y^{2} – 8y)

⇒ 9 + x^{2} – 6x) + 36 + y^{2} – 12y – 9 – x^{2} – 6x – 16 – y^{2} + 8y

⇒ –6x – 6x + 36 – 12y – 16 + 8y = 0

⇒ –12x – 4y + 20 = 0

⇒ –3x – y + 5 = 0

⇒ 3x + y – 5 = 0

which is the required relation between x and y.

Sol. Let the required point be P (x, y).

Here, the end points are:

(–1, 7) and (4, –3)

∵Ratio = 2 : 3 = m_{1} : m_{2}

Sol. Let the given points be A(4, –1) and B(–2, –3).

Let the poi ts P and Q trisect AB.

i.e., AP = PQ = QB

i.e., P divides AB in the ratio of 1 : 2

Q divides AB in the ratio of 2 : 1

Let the coordinates fo P be (x, y).

Sol. Let us consider ‘A‘ as origin, then AB is the x-axis. AD is the y-axis.

Now, the position of green flag-post is

And the position of red flag-post is

or x = 5 and y = (22.5).

Thus, the blue flag is on the 5th line at a distance 22.5 m above AB.

Sol. Let the given points are: A (–3, 10) and B (6, –8).

Let the point P (–1, 6) divides AB in the ratio m_{1} : m_{2}.

Sol. The given points are: A (1, –5) and B (–4, 5).

Let the required ratio = k : 1 and the required point be P (x, y).

Part-I: To find the ratio

Since the point P lies on x-axis,

∴ Its y-coordinates is 0.

Sol. We have the parallelogram vertices

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

Since, the diagonals of a parallelogram bisect each other.

∴ The coordinates of P are:

⇒; 5 + y = 8 T y = 3

∴ The required values of x and y are:

x = 6, y = 3

Sol. Here, centre of the circle is O (2, –3).

Let the end points of the diameter be

A (X, Y) and B (1, 4)

The centre of a circle bisects the diameter.

Hence the coordinates of A are (3, –10).

Sol.

Here, the given points are

A(–2, –2) and B (2, –4)

Let the coordinates of P are (x, y).

Since, the point P divides AB such that

Sol. Here, the given points are:

A(–2, 2) and B (2, 8)

Let P1, P2 and P3 divide AB in four equal parts.

Sol. Let the vertices of the given rhombsu are:

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

∵AC and BD are the diagonals of rhombus ABCD.

(i) (2, 3), (–1, 0), (2, –4) (ii) (–5, –1), (3, –5), (5, 2)

Sol. (i) Let the vertices of the triangle be

A (2, 3), B (–1, 0) and C(2, –4)

Here, x_{1} = 2, y_{1} = 3

x_{2} = –1, y_{2} = 0

x_{3} = 2, y_{3} = –4

(ii) Let the vertices of the triangle be

A(–5, –1), B(3, –5) and C(5, 2)

i.e., x_{1} = –5, y_{1} = –1

x_{2} = 3, y_{2} = –5

x_{3} = 5, y_{3} = 2

(i) (7, –2), (5, 1), (3, k) (ii) (8, 1), (k, –4), (2, –5)

Sol. The given three points will be collinear if the Δ Formed by them has zero area.

(i) Let A (7, –2), B (5, 1) and C(3, k) be the vertics of a triangle.

∴ The given points will be collinear, if are (ΔABC) = 0

or 7(1 – k) + 5 (k + 2) + 3(–2 –1) = 0

⇒ 7 – 7k + 5k + 10 + (–6) – 3 = 0

⇒ 7 – 9 + 5k – 7k = 0

⇒ 8 – 2k = 0

⇒ 2k = 0

The required value of k = 4.

(ii) Let (8, 1), (k, –4) and (2, –5) be the verticles of a triangle.

∴ For the above points being collinear, ar (ΔABC) = 0

i.e., 8 (–4 + 5) + k (–5 –1) + 2[1 – (–4)] = 0

⇒ 8 (+1) + k (–6) + 2(5) = 0

⇒ 8 + (–6k) + 10 = 0

⇒ –6k + 18 = 0

⇒ k = (–18) + (–6) = 3

Thus, k = 3.

Sol. Let the vertices of the triangle be A(0, –1), B(2, 1) and C(0, 3).

Let D, E and F be the mid-points of the sides BC, CA and AB respectively. Then:

Sol. Let A (–4, –2), B (–3, –5), C(3, –2) and D(2, 3) be the vertices of the quadrilateral.

Let us join diagonal BD.

Sol. Here, the vertices of the triangle are A(4, –6), B(3, –2) and C(5, 2).

Let D be the mid-point of BC.

∴ The coordiantes of the mid point D are:

Since AD divides the triangle ABC into two parts i.e., ΔABD and ΔACD,

= 3 sq. units (numerically) ...(2)

From (1) and (2)

ar (ΔABD) = ar (ΔACD))

i.e. A median divides the triangle into two triangles of equal areas.

Sol. Let the required ratio be k : 1 and the point C divides them in the abvoe ratio.

∴ Coordinates of C are:

Since the point C lies on the given line 2x + y – 4 = 0,

∴ We have:

Sol. The given points are:

A (x, y), B (1, 2) and C(7, 0)

The points A, B and C will be collinear if

x (2 – 0) + 1 (0 – y) + 7 (y – 2) = 0

or if 2x – y + 7y – 14 = 0

or if 2x + 6y – 14 = 0

or if x + 3y – 7 = 0

which is the require relation between x and y.

Sol. Let P (x, y) be the center of the circle passing through

A(6, –6), B(3, –7) and C(3, 3).

∴ AP = BP = CP

Taking AP = BP, we have AP^{2} = BP^{2}

⇒ (x – 6)^{2} + (y + 6)^{2} = (x – 3)^{2} + (y + 7)^{2}

⇒ x^{2} – 12x + 36 + y^{2} + 12y + 36 = x^{2} – 6x + 9 + y^{2} + 14y + 49

⇒ –12x + 6x + 12y – 14y + 72 – 58 = 0

⇒ –6x – 2y + 14 = 0

Taking BP = CP, we have BP^{2} = CP^{2}

⇒ (x – 3)^{2} + (y + 7)^{2} = (x – 3)^{2} + (y – 3)^{2}

⇒ x^{2} – 6x + 9 + y^{2} + 14y + 49 = x^{2} – 6x + 9 + y^{2} – 6y + 9

⇒ –6x + 6x + 14y + 6y + 58 – 18 = 0

⇒ 20y + 40 = 0

From (1) and (2),

3x – 2 – 7 = 0

⇒ 3x = 9 ⇒ x = 3

i.e., x = 3 and y = –2

∴ The required centre is (3, –2).

Sol. Let us have a square ABCD such that A(–1, 2) and C(3, 2) are the opposite vertices. Let B(x, y) be an unknown vertex. Since all sides of a square are equal,

∴ AB = BC

⇒ AB^{2} = BC^{2}

⇒ (x + 1)^{2} + (y – 2)^{2} = (x – 3)^{2} + (y – 2)^{2}

⇒ 2x + 1 = –6x + 9

⇒ 8x = 8 ⇒ x = 1

...(1)

Since each angle of a square = 90°,

∴ABC is a right angled triangle.

∴Using Pythagoras theorem, we have:

AB^{2} + BC^{2} = AC^{2}

⇒ (x + 1)^{2} + (y – 2)^{2} + (x – 3)^{2} + (y – 2)^{2}

= [(3 + 1)^{2} + (2 – 2)^{2}]

⇒ 2x^{2} + 2y^{2} + 2x – 4y – 6x – 4y + 1 + 4 + 9 + 4 = 16

⇒ 2x^{2} + 2y^{2} – 4x – 8y + 2 = 0

⇒ x^{2} + y^{2} – 2x – 4y + 1 = 0

...(2)

Substituting the value of x from (1) into (2) we have:

1 + y^{2} – 2 – 4y + 1 = 0

⇒ y^{2} – 4y + 2 – 2 = 0

⇒ y^{2} – 4y = 0

⇒ y(y – 4) = 0

⇒ y = 0 or y = 4

Hence, the requried other two vertices are: (1, 0) and (1, 4).

(i) Taking A as origin, find the coordinates of the vertices of the triangle.

(ii) What will be the coordinates of the vertices of ∴PQR if C is the origin?

Also calculate the areas of the triangles in these cases. What do you observe?

Sol. (i) By taking A as the origin and AD and AB as the coordinate axes, we have P(4, 6), Q(3, 2) and R(6, 5) as the vertices of ∴PQR.

(ii) By taking C as the origin and CB and CD as the coordinate axes, then the vertices of ΔPQR are

P (12, 2), Q (13, 6) and R(10, 3)

Now, ar (ΔPQR)

[wnen P(4, 6), Q(3, 2) and R(6, 5) are the vertices]

Thus, in both cases, the area of ΔPQR is the same.

(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.

(iii) Find the coordinates of points Q and R on medians Be and CF respectively such that BQ : QE

= 2 : 1 and CR : RF = 2 : 1.

(iv)What do you observe?

(v) If A (x_{1}, y_{1}), B(x_{2} , y_{2}) and C(x_{3} , y_{3}) are the vertices of ΔABC, find the coordiantes of the centroid of the triangle.

Sol. We have the vertices of ΔABC as A(4, 2) B(6, 5) and C(1, 4).

(i) Since AD is a median

(iv)What do you observe?

(ii) Since AP : PD = 2 : 1 i.e., P divides AD in the ratio 2 : 1.

∴ Coordinates of P are:

(iv)What do you observe?

(iii) BQ : QE = 2 : 1 ⇒ [The point Q divides BE in the radio 2 : 1]

∴ Coordinates of Q are:

(iv)We observe that P, Q and R represent the same point.

(v) Here, we have A(x_{x}, y_{y}), B(x_{x}, y_{y}), C(x_{x}, y_{y}) as the vertices of ��ABC. Also AD, BE and CF are its medians.

∴D, E and F are the mid points of BC, CA and AB respectively.

We know, the centroid is a point on a median, dividing it in the ratio 2 : 1.

Concidering the median AD,

P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

Sol. We have a rectangle whose vertices are A(–1, –1), B (–1, 4), C(5, 4) and D(5, –1).

We see that:

PQ = QR = RS = SP

i.e., all sides of PQRS are equal.

∴ It can be a square or a rhombus.

But its diagonals are not equal.

i.e., PR ≠ QS

∴ PQRS is a rhombus.