NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.4

NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.4 are designed to help students strengthen their understanding of coordinate geometry and apply their knowledge to real-world problems. This exercise focuses on the properties of triangles and how to use coordinate geometry to find the area of a triangle formed by three points.

Exercise 7.4 includes problems that require students to:

NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.4

Coordinate Geometry

Medium

Q.

$\begin{array}{l}\mathrm{You}\mathrm{have}\mathrm{studied}\mathrm{in}\mathrm{Class}\mathrm{IX},\left(\mathrm{Chapter}9,\mathrm{Example}3\right),\\ \mathrm{that}\mathrm{a}\mathrm{median}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{divides}\mathrm{it}\mathrm{into}\mathrm{two}\mathrm{triangles}\\ \mathrm{of}\mathrm{equal}\mathrm{areas}.\mathrm{Verify}\mathrm{this}\mathrm{result}\mathrm{for}\mathrm{\Delta ABC}\mathrm{whose}\\ \mathrm{vertices}\mathrm{are}\mathrm{A}\left(4,-6\right),\mathrm{B}\left(3,-2\right)\mathrm{and}\mathrm{C}\left(5,2\right).\end{array}$

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Coordinate Geometry

Medium

Q.

$\begin{array}{l}\text{Determine the ratio in which the line\hspace{0.33em}}2\mathrm{x}+\mathrm{y}-4=0\text{\hspace{0.33em}divides the line segment joining the points A(}2,-2\text{)}\\ \text{and B(3, 7).\hspace{0.17em}}\end{array}$

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Coordinate Geometry

Medium

Q.

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

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Coordinate Geometry

Medium

Q.

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

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Coordinate Geometry

Medium

Q.

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

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Coordinate Geometry

Medium

Q.

The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity.
Saplings of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown
in the following 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 ΔPQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?

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Coordinate Geometry

Medium

Q.

$\begin{array}{l}\text{The vertices of a }\mathrm{\Delta }\text{\hspace{0.17em}ABC are A(4, 6), B(1, 5) and C(7, 2). }\\ \text{A line is drawn to intersect sides AB and AC at D and E }\\ \text{respectively, such that }\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}=\frac{\text{1}}{4}.\text{ Calculate the area }\\ \text{of the }\mathrm{\Delta }\text{\hspace{0.17em}ADE and compare it with the area of }\mathrm{\Delta }\text{\hspace{0.17em}ABC.}\\ \text{(Recall converse of basic proportionality theorem and }\\ \text{Theorem 6.6 related to the ratio of the areas of two}\\ \text{similar triangles).}\end{array}$

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Coordinate Geometry

Medium

Q.

$\begin{array}{l}\text{Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of }\mathrm{\Delta }\text{ABC.}\\ \text{ (i) The median from A meets BC at D. Find the}\\ \text{ coordinates of the point D.}\\ \text{ (ii) Find the coordinates of the point P on AD such that}\\ \text{ AP}:\text{PD}=\text{2}:\text{1}\\ \text{ (iii) Find the coordinates of points Q and R on medians}\\ \text{ BE and CF respectively such that BQ}:\text{QE }=\text{ 2}:\text{1 }\\ \text{ and CR}:\text{RF}=\text{2}:\text{1.}\\ \text{ (iv) What do yo observe?}\\ \text{ [Note: The point which is common to all the three}\\ \text{ medians is called the centroid and this point divides}\\ \text{ each median in the ratio 2}:\text{1.]}\\ \text{ (v) If A(}{\mathrm{x}}_1,{\mathrm{y}}_1\text{), B(}{\mathrm{x}}_2,{\mathrm{y}}_2\text{) and C(}{\mathrm{x}}_3,{\mathrm{y}}_3\text{) are the vertices of }\\ \mathrm{\Delta }\text{ ABC, find the coordinates of the centroid of the}\\ \text{ triangle.}\end{array}$

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Coordinate Geometry

Medium

Q.

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.

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• Find the area of a triangle using the coordinates of its vertices.

• Understand the application of the determinant method to calculate the area of triangles.

• Apply the concepts of distance formula, midpoint theorem, and area formula to solve geometrical problems.

The solutions are explained step-by-step and provide a comprehensive understanding of how to apply coordinate geometry concepts to solve problems efficiently, making it easier for students to prepare for Class 10 exams.

Q1. Determine the ratio in which the line $2x + y - 4 = 0$

$2x+y-4=0$ divides the line segment joining the points A(2, 2) and B(3, 7).

Solution:

Let the line

$2x + y - 4 = 0$

2x+y−4=0 divide the line segment joining A(2, 2) and B(3, 7) in the ratio

$k:1$

k:1. Let the coordinates of the dividing point be

$C(x, y)$

C(x,y).

Using the Section Formula:

$$C(x, y) = \left( \frac{kx_2 + x_1}{k+1}, \frac{ky_2 + y_1}{k+1} \right)$$

C(x,y)=(k+1kx2​+x1​​,k+1ky2​+y1​​)

Substitute

$A(2, 2)$

A(2,2) and

$B(3, 7)$

B(3,7):

$$C(x, y) = \left( \frac{k(3) + 2}{k+1}, \frac{k(7) + 2}{k+1} \right)$$

C(x,y)=(k+1k(3)+2​,k+1k(7)+2​)

Now, since point

$C(x, y)$

C(x,y) lies on the line

$2x + y - 4 = 0$

2x+y−4=0, we substitute these coordinates into the line equation:

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

2(k+1k(3)+2​)+(k+1k(7)+2​)−4=0

Simplifying the equation:

$$\frac{6k + 4}{k+1} + \frac{7k + 2}{k+1} - 4 = 0$$

k+16k+4​+k+17k+2​−4=0

$$\frac{6k + 4 + 7k + 2}{k+1} = 4$$

k+16k+4+7k+2​=4

$$\frac{13k + 6}{k+1} = 4$$

k+113k+6​=4

Cross-multiply:

$$13k + 6 = 4(k + 1)$$

13k+6=4(k+1)

$$13k + 6 = 4k + 4$$

13k+6=4k+4

$$9k = -2$$

9k=−2

$$k = -\frac{2}{9}$$

k=−92​

Thus, the ratio in which the line divides the line segment is

$-2:9$

−2:9, which means the division is external.

Q2. Find a relation between x and y if the points

$(x, y)$

$(x,y)$,

$(1, 2)$

$(1,2)$, and

$(7, 0)$

$(7,0)$ are collinear.

Solution:

Three points are collinear if the area of the triangle formed by them is zero. Using the area formula for a triangle:

$$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$

Area=21​∣x1​(y2​−y3​)+x2​(y3​−y1​)+x3​(y1​−y2​)∣

For points

$(x, y)$

(x,y),

$(1, 2)$

(1,2), and

$(7, 0)$

(7,0), the area is zero:

$$\frac{1}{2} \left| x(2 - 0) + 1(0 - y) + 7(y - 2) \right| = 0$$

21​∣x(2−0)+1(0−y)+7(y−2)∣=0

Simplifying:

$$x(2) + 1(-y) + 7(y - 2) = 0$$

x(2)+1(−y)+7(y−2)=0

$$2x - y + 7y - 14 = 0$$

2x−y+7y−14=0

$$2x + 6y = 14$$

2x+6y=14

$$x + 3y = 7$$

x+3y=7

Thus, the relation between

$x$

x and

$y$

y is

$x + 3y = 7$

x+3y=7.

Q3. Find the centre of a circle passing through the points

$(6, 6)$

$(6,6)$,

$(3, 7)$

$(3,7)$, and

$(3, 3)$

$(3,3)$.

Solution:

The center of a circle passing through three points is the point of intersection of the perpendicular bisectors of any two sides of the triangle formed by these points.

Let the three points be

$A(6, 6)$

A(6,6),

$B(3, 7)$

B(3,7), and

$C(3, 3)$

C(3,3).

To find the perpendicular bisectors:

1. Find the midpoint of

   $AB$AB:

$$\text{Midpoint of } AB = \left( \frac{6 + 3}{2}, \frac{6 + 7}{2} \right) = \left( \frac{9}{2}, \frac{13}{2} \right)$$

Midpoint of AB=(26+3​,26+7​)=(29​,213​)

2. Find the slope of

   $AB$AB:

$$\text{Slope of } AB = \frac{7 - 6}{3 - 6} = \frac{1}{-3} = -\frac{1}{3}$$

Slope of AB=3−67−6​=−31​=−31​

3. Find the slope of the perpendicular bisector (negative reciprocal of the slope of

   $AB$AB):

$$\text{Slope of perpendicular bisector} = 3$$

Slope of perpendicular bisector=3

4. Equation of the perpendicular bisector (using point-slope form of the line equation):

$$y - \frac{13}{2} = 3\left(x - \frac{9}{2}\right)$$

y−213​=3(x−29​)

Simplifying:

$$y = 3x - \frac{15}{2} + \frac{13}{2} = 3x - 1$$

y=3x−215​+213​=3x−1

Now, for the second side, say

$BC$

BC, we can similarly find its perpendicular bisector and solve the system of equations to find the center of the circle.

After calculations, the center is

$(3, -2)$

(3,−2).

Q4. The two opposite vertices of a square are

$(-1, 2)$

$(-1,2)$ and

$(3, 2)$

$(3,2)$. Find the coordinates of the other two vertices.

Solution:

Given two opposite vertices of a square, the center of the square is the midpoint of these vertices.

Midpoint of

$(-1, 2)$

(−1,2) and

$(3, 2)$

(3,2):

$$\text{Midpoint} = \left( \frac{-1 + 3}{2}, \frac{2 + 2}{2} \right) = (1, 2)$$

Midpoint=(2−1+3​,22+2​)=(1,2)

Let the other two vertices be

$B(x_1, y_1)$

B(x1​,y1​) and

$C(x_2, y_2)$

C(x2​,y2​), which are symmetric with respect to the center. The coordinates of these vertices can be found using symmetry or rotation. Through geometric reasoning, the coordinates of the other two vertices are:

$$B(1, 4) \quad \text{and} \quad C(1, 0)$$

B(1,4)andC(1,0)

Thus, the other two vertices of the square are

$(1, 4)$

(1,4) and

$(1, 0)$

(1,0).

Q5. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings 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 following figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

Solution:

(i) Taking A as origin, the coordinates of the vertices of the triangle are:

• $P(4, 6)$

  P(4,6)

• $Q(3, 2)$

  Q(3,2)

• $R(6, 5)$

  R(6,5)

(ii) If C is the origin, the coordinates of the vertices of the triangle change accordingly, and the areas can be calculated using the formula for the area of a triangle:

$$\text{Area of triangle} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$

Area of triangle=21​∣x1​(y2​−y3​)+x2​(y3​−y1​)+x3​(y1​−y2​)∣

From the calculations, it is observed that the areas in both cases are the same.

FAQs: Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.4

Q1. What is the focus of Exercise 7.4?
Answer:
Exercise 7.4 focuses on finding the area of a triangle formed by three points on the coordinate plane. It helps students apply the coordinate geometry formulas to calculate areas using the determinant method and distance formula.

Q2. How do I find the area of a triangle using its vertices?
Answer:
The area of a triangle with vertices

$(x_1, y_1)$

(x1​,y1​),

$(x_2, y_2)$

(x2​,y2​), and

$(x_3, y_3)$

(x3​,y3​) can be found using the formula:

$$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$

Area=21​∣x1​(y2​−y3​)+x2​(y3​−y1​)+x3​(y1​−y2​)∣

This formula is derived from the determinant method.

Q3. What is the determinant method in coordinate geometry?
Answer:
The determinant method is a technique to calculate the area of a triangle when the coordinates of its vertices are known. The formula involves the use of the determinant of a matrix that contains the coordinates of the triangle's vertices.

Q4. Why is the area of a triangle important in coordinate geometry?
Answer:
The area of a triangle is crucial in various geometric problems, and it helps in understanding the relationship between the vertices and sides of a triangle in the coordinate plane. It also forms the basis for solving many practical problems in mathematics, physics, and engineering.

Q5. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide clear, detailed steps for solving area of triangle problems and applying coordinate geometry concepts efficiently. By practicing these solutions, students can develop strong problem-solving skills, making them well-prepared for Class 10 exams.