In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.4. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4

Question 1.

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, -2) and 8(3, 7).

Solution:

Let the required ratio be k : 1 and the point C divide them in the above ratio.

∴ Coordinates of C are \(\left(\frac{3 k+2}{k+1}, \frac{7 k-2}{k+1}\right)\)

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

∴ We have \(2\left(\frac{3 k+2}{k+1}\right)+\left(\frac{7 k-2}{k+1}\right)-4\) = 0

⇒ 2(3k + 2) + (7k – 2) = 4 × (k + 1)

⇒ 6k + 4 + 7k – 4k – 4 – 2 = 0

⇒ (6 + 7 – 4)k + (-2) = 0 ⇒ 9k – 2 = 0

⇒ k = \(\frac{2}{9}\)

The required ratio = k : 1 = \(\frac{2}{9}\) : 1 = 2 : 9

Question 2.

Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

Solution:

Let the given points be A(x, y), B( 1, 2) and C(7, 0) are collinear.

The points A, B and C will be collinear if area of ∆ABC = 0

⇒ \(\frac{1}{2}\) [x(2 – 0) + 1(0 – y) + 7(y – 2) = 0

or 2x – y + 7y – 14 = 0

or 2x + 6y – 14 = 0 or x + 3y – 7 = 0, which is the required relation between x and y.

Question 3.

Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).

Solution:

Let the points are A (6, -6), B(3, -7) and C(3, 3)

Let P(x, y) be the centre of the circle. Since, the circle is passing through A, B and C.

∴ 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

⇒ 3x + y – 7 = 0 ……………….. (1)

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

⇒ y = \(\frac{-40}{20}\) = -2

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).

Question 4.

The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.

Solution:

Let we 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} = BC2^{2}

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

⇒ x^{2} + 2x + 1 + y^{2} – 4y + 4

⇒ x^{2} – 6x + 9 + y^{2} – 4y + 4

⇒ 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}]

⇒ [x^{2} + 2x + 1 + y^{2} – 4y + 4] + [x^{2} – 6x + 9 + y^{2} – 4y + 4]

= [4^{2} + 0^{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} – y = 0

⇒ y(y – 4) = 0

⇒ y = 0 or y = 4

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

Question 5.

The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

(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?

Solution:

(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 ∆PQR are P(-12, – 2), Q(-13, – 6) and R(- 10, – 3)

Case I: When P(4, 6), Q(3, 2) and R(6, 5) are the vertices.

∴ ar(∆PQR) = \(\frac{1}{2}\) [4(2 – 5) + 3(5 – 6) + 6(6 – 2)]

= \(\frac{1}{2}\) [-12 – 3 + 24] = \(\frac{9}{2}\) sq. units

Case II: When P(-12, -2), Q(-13, -6) and R(-10, -3) are the vertices.

∴ ar(∆PQR)

= \(\frac{1}{2}\) [-12(- 6 + 3) + (-13)(- 3 + 2) + (-10)(-2 + 6)]

= \(\frac{1}{2}\) [-12(-3) + (-13)(-1) + (-10) × (4)]

= \(\frac{1}{2}\) [36 + 13 – 40] = \(\frac{9}{2}\) sq. units

Thus, in both cases, the area of ∆PQR is the same.

Question 6.

The vertices of a ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{4}\) Calculate the area of the ∆ADE and compare it with the area of ∆ABC. [Recall “The converse of basis proportionality theorem”, and “theorem of similar triangles taking their areas and corresponding sides”]

Solution:

Question 7.

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

(i) The median from A meets BCat 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?

[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2:1.]

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

Solution:

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

(i) Since AD is a median

∴ Coordinates of D are

Also, CR : RF = 2 : 1 i.e., the point R divides CF in the ratio 2 : 1

∴ Coordinates of R are

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

(v) Here, we have A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) are 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.

Considering the median AD, coordinates of

Question 8.

ABCD is a rectangle formed by the points A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1). 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.

Solution:

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

∵ P is mid-point of AB

∴ Coordinates of P are

We see that PQ = QR = RS = SP i.e., all sides of PQRS are equal.

∴ It can be a square or a rhombus.

But PR ≠ QS i.e., its diagonals are not equal.

∴ PQRS is a rhombus.

In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.4. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4

Question 1.

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, -2) and 8(3, 7).

Solution:

Let the required ratio be k : 1 and the point C divide them in the above ratio.

∴ Coordinates of C are \(\left(\frac{3 k+2}{k+1}, \frac{7 k-2}{k+1}\right)\)

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

∴ We have \(2\left(\frac{3 k+2}{k+1}\right)+\left(\frac{7 k-2}{k+1}\right)-4\) = 0

⇒ 2(3k + 2) + (7k – 2) = 4 × (k + 1)

⇒ 6k + 4 + 7k – 4k – 4 – 2 = 0

⇒ (6 + 7 – 4)k + (-2) = 0 ⇒ 9k – 2 = 0

⇒ k = \(\frac{2}{9}\)

The required ratio = k : 1 = \(\frac{2}{9}\) : 1 = 2 : 9

Question 2.

Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

Solution:

Let the given points be A(x, y), B( 1, 2) and C(7, 0) are collinear.

The points A, B and C will be collinear if area of ∆ABC = 0

⇒ \(\frac{1}{2}\) [x(2 – 0) + 1(0 – y) + 7(y – 2) = 0

or 2x – y + 7y – 14 = 0

or 2x + 6y – 14 = 0 or x + 3y – 7 = 0, which is the required relation between x and y.

Question 3.

Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).

Solution:

Let the points are A (6, -6), B(3, -7) and C(3, 3)

Let P(x, y) be the centre of the circle. Since, the circle is passing through A, B and C.

∴ 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

⇒ 3x + y – 7 = 0 ……………….. (1)

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

⇒ y = \(\frac{-40}{20}\) = -2

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).

Question 4.

The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.

Solution:

Let we 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} = BC2^{2}

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

⇒ x^{2} + 2x + 1 + y^{2} – 4y + 4

⇒ x^{2} – 6x + 9 + y^{2} – 4y + 4

⇒ 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}]

⇒ [x^{2} + 2x + 1 + y^{2} – 4y + 4] + [x^{2} – 6x + 9 + y^{2} – 4y + 4]

= [4^{2} + 0^{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} – y = 0

⇒ y(y – 4) = 0

⇒ y = 0 or y = 4

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

Question 5.

The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

(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?

Solution:

(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 ∆PQR are P(-12, – 2), Q(-13, – 6) and R(- 10, – 3)

Case I: When P(4, 6), Q(3, 2) and R(6, 5) are the vertices.

∴ ar(∆PQR) = \(\frac{1}{2}\) [4(2 – 5) + 3(5 – 6) + 6(6 – 2)]

= \(\frac{1}{2}\) [-12 – 3 + 24] = \(\frac{9}{2}\) sq. units

Case II: When P(-12, -2), Q(-13, -6) and R(-10, -3) are the vertices.

∴ ar(∆PQR)

= \(\frac{1}{2}\) [-12(- 6 + 3) + (-13)(- 3 + 2) + (-10)(-2 + 6)]

= \(\frac{1}{2}\) [-12(-3) + (-13)(-1) + (-10) × (4)]

= \(\frac{1}{2}\) [36 + 13 – 40] = \(\frac{9}{2}\) sq. units

Thus, in both cases, the area of ∆PQR is the same.

Question 6.

The vertices of a ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{4}\) Calculate the area of the ∆ADE and compare it with the area of ∆ABC. [Recall “The converse of basis proportionality theorem”, and “theorem of similar triangles taking their areas and corresponding sides”]

Solution:

Question 7.

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

(i) The median from A meets BCat 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?

[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2:1.]

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

Solution:

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

(i) Since AD is a median

∴ Coordinates of D are

Also, CR : RF = 2 : 1 i.e., the point R divides CF in the ratio 2 : 1

∴ Coordinates of R are

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

(v) Here, we have A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) are 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.

Considering the median AD, coordinates of

Question 8.

ABCD is a rectangle formed by the points A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1). 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.

Solution:

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

∵ P is mid-point of AB

∴ Coordinates of P are

We see that PQ = QR = RS = SP i.e., all sides of PQRS are equal.

∴ It can be a square or a rhombus.

But PR ≠ QS i.e., its diagonals are not equal.

∴ PQRS is a rhombus.