NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.3

NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.3 are provided to help students understand and solve problems related to coordinate geometry. This exercise builds on the concepts of distance formula and midpoint theorem and introduces students to applications of coordinate geometry in real-world scenarios.

Exercise 7.3 focuses on:

NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.3

Coordinate Geometry

Medium

Q.

$\begin{array}{l}\text{Find the area of the triangle whose vertices are :}\\ \text{ (i) (}2,3),(-1,0),(2,-4)\\ \text{ (ii) }(-5,-1),(3,-5),(5,2)\end{array}$

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Coordinate Geometry

Medium

Q.

In each of the following find the value of ‘k’, for which the points are collinear.
(i)  (7, –2), (5, 1), (3, k)
(ii) (8, 1), (k, – 4), (2, –5)

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Coordinate Geometry

Medium

Q.

$\begin{array}{l}\text{Find the area of the triangle formed by joining the}\\ \text{mid-points of the sides of the triangle whose }\\ \text{vertices are }(0,-1),(2,1)\text{ and (0, 3). Find the}\\ \text{ratio of this area to the area of the given triangle.}\end{array}$

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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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• Finding the distance between two points and using it to calculate various geometrical properties.

• Using coordinate geometry to solve problems related to locus, which helps in visualizing geometric figures.

• Understanding geometric relationships and solving word problems involving coordinate geometry.

The solutions are explained in a step-by-step manner, ensuring that students can apply these concepts confidently in their Class 10 exams.

Q1. Find the area of the triangle whose vertices are:

(i)

$(2, 3), (−1, 0), (2, −4)$

(2,3),(−1,0),(2,−4)
(ii)

$(-5, −1), (3, −5), (5, 2)$

(−5,−1),(3,−5),(5,2)

Solution:

The area of a triangle with vertices

$(x_1, y_1), (x_2, y_2), (x_3, y_3)$

(x1​,y1​),(x2​,y2​),(x3​,y3​) is given by 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​)∣

(i) For the triangle with vertices

$(2, 3), (−1, 0), (2, −4)$

(2,3),(−1,0),(2,−4):

$$\text{Area} = \frac{1}{2} \left| 2(0 - (-4)) + (-1)((-4) - 3) + 2(3 - 0) \right|$$

Area=21​∣2(0−(−4))+(−1)((−4)−3)+2(3−0)∣

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

=21​∣2(4)+(−1)(−7)+2(3)∣

$$= \frac{1}{2} \left| 8 + 7 + 6 \right| = \frac{1}{2} \times 21 = 10.5 \, \text{square units}$$

=21​∣8+7+6∣=21​×21=10.5square units

(ii) For the triangle with vertices

$(-5, −1), (3, −5), (5, 2)$

(−5,−1),(3,−5),(5,2):

$$\text{Area} = \frac{1}{2} \left| (-5)(-5 - 2) + 3(2 - (-1)) + 5((-1) - (-5)) \right|$$

Area=21​∣(−5)(−5−2)+3(2−(−1))+5((−1)−(−5))∣

$$= \frac{1}{2} \left| (-5)(-7) + 3(3) + 5(4) \right|$$

=21​∣(−5)(−7)+3(3)+5(4)∣

$$= \frac{1}{2} \left| 35 + 9 + 20 \right| = \frac{1}{2} \times 64 = 32 \, \text{square units}$$

=21​∣35+9+20∣=21​×64=32square units

Q2. In each of the following find the value of ‘k’, for which the points are collinear.

(i)

$(7, −2), (5, 1), (3, k)$

(7,−2),(5,1),(3,k)
(ii)

$(8, 1), (k, −4), (2, −5)$

(8,1),(k,−4),(2,−5)

Solution:

Three points are collinear if the area of the triangle formed by them is zero.

(i) For points

$(7, −2), (5, 1), (3, k)$

(7,−2),(5,1),(3,k):
Using the area formula for collinearity:

$$\text{Area} = \frac{1}{2} \left| 7(1 - k) + 5(k - (-2)) + 3((-2) - 1) \right| = 0$$

Area=21​∣7(1−k)+5(k−(−2))+3((−2)−1)∣=0

$$7(1 - k) + 5(k + 2) + 3(-3) = 0$$

7(1−k)+5(k+2)+3(−3)=0

$$7 - 7k + 5k + 10 - 9 = 0$$

7−7k+5k+10−9=0

$$-2k + 8 = 0 \quad \Rightarrow \quad k = 4$$

−2k+8=0⇒k=4

(ii) For points

$(8, 1), (k, −4), (2, −5)$

(8,1),(k,−4),(2,−5):
Using the area formula for collinearity:

$$\text{Area} = \frac{1}{2} \left| 8(-4 - (-5)) + k((-5) - 1) + 2(1 - (-4)) \right| = 0$$

Area=21​∣8(−4−(−5))+k((−5)−1)+2(1−(−4))∣=0

$$8(1) + k(-6) + 2(5) = 0$$

8(1)+k(−6)+2(5)=0

$$8 - 6k + 10 = 0$$

8−6k+10=0

$$-6k + 18 = 0 \quad \Rightarrow \quad k = 3$$

−6k+18=0⇒k=3

Q3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are $(0, −1), (2, 1)$

$(0,-1),(2,1)$ and $(0, 3)$

$(0,3)$. Find the ratio of this area to the area of the given triangle.

Solution:

The area of a triangle with vertices

$(x_1, y_1), (x_2, y_2), (x_3, y_3)$

(x1​,y1​),(x2​,y2​),(x3​,y3​) is given by:

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

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

Given vertices:

$A(0, −1), B(2, 1), C(0, 3)$

A(0,−1),B(2,1),C(0,3)

The midpoints of sides

$AB$

AB,

$BC$

BC, and

$CA$

CA are:

$$P = \left( \frac{0+2}{2}, \frac{-1+1}{2} \right) = (1, 0)$$

P=(20+2​,2−1+1​)=(1,0)

$$Q = \left( \frac{2+0}{2}, \frac{1+3}{2} \right) = (1, 2)$$

Q=(22+0​,21+3​)=(1,2)

$$R = \left( \frac{0+0}{2}, \frac{3+(-1)}{2} \right) = (0, 1)$$

R=(20+0​,23+(−1)​)=(0,1)

Now, calculate the area of triangle

$PQR$

PQR (mid-point triangle):

$$\text{Area of } PQR = \frac{1}{2} \left| 1(2 - 1) + 1(1 - 0) + 0(0 - 2) \right| = \frac{1}{2} \left| 1 + 1 + 0 \right| = \frac{1}{2} \times 2 = 1$$

Area of PQR=21​∣1(2−1)+1(1−0)+0(0−2)∣=21​∣1+1+0∣=21​×2=1

Now, calculate the area of triangle

$ABC$

ABC:

$$\text{Area of } ABC = \frac{1}{2} \left| 0(1 - 3) + 2(3 - (-1)) + 0((-1) - 1) \right|$$

Area of ABC=21​∣0(1−3)+2(3−(−1))+0((−1)−1)∣

$$= \frac{1}{2} \left| 0 + 8 + 0 \right| = \frac{1}{2} \times 8 = 4$$

=21​∣0+8+0∣=21​×8=4

The ratio of the areas is:

$$\frac{\text{Area of } PQR}{\text{Area of } ABC} = \frac{1}{4}$$

Area of ABCArea of PQR​=41​

Q4. Find the area of the quadrilateral whose vertices, taken in order, are $(-4, −2), (−3, −5), (3, −2)$

$(-4,-2),(-3,-5),(3,-2)$, and $(2, 3)$

$(2,3)$.

Solution:

To find the area of a quadrilateral, divide it into two triangles by drawing a diagonal. Here, we can divide the quadrilateral into triangles

$ABC$

ABC and

$ACD$

ACD.

Using the area formula for each triangle:

For triangle ABC:
Vertices:

$A(-4, -2), B(-3, -5), C(3, -2)$

A(−4,−2),B(−3,−5),C(3,−2)

$$\text{Area of } ABC = \frac{1}{2} \left| -4(-5 - (-2)) + (-3)((-2) - (-2)) + 3((-2) - (-5)) \right|$$

Area of ABC=21​∣−4(−5−(−2))+(−3)((−2)−(−2))+3((−2)−(−5))∣

$$= \frac{1}{2} \left| -4(-3) + (-3)(0) + 3(3) \right| = \frac{1}{2} \left| 12 + 0 + 9 \right| = \frac{1}{2} \times 21 = 10.5$$

=21​∣−4(−3)+(−3)(0)+3(3)∣=21​∣12+0+9∣=21​×21=10.5

For triangle ACD:
Vertices:

$A(-4, -2), C(3, -2), D(2, 3)$

A(−4,−2),C(3,−2),D(2,3)

$$\text{Area of } ACD = \frac{1}{2} \left| -4(-2 - 3) + 3(3 - (-2)) + 2((-2) - (-2)) \right|$$

Area of ACD=21​∣−4(−2−3)+3(3−(−2))+2((−2)−(−2))∣

$$= \frac{1}{2} \left| -4(-5) + 3(5) + 2(0) \right| = \frac{1}{2} \left| 20 + 15 + 0 \right| = \frac{1}{2} \times 35 = 17.5$$

=21​∣−4(−5)+3(5)+2(0)∣=21​∣20+15+0∣=21​×35=17.5

Total Area of the Quadrilateral:

$$\text{Area of quadrilateral} = 10.5 + 17.5 = 28 \, \text{square units}$$

Area of quadrilateral=10.5+17.5=28square units

Q5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for $\triangle ABC$

$△ABC$ whose vertices are $A(4, −6), B(3, −2), C(5, 2)$

$A(4,-6),B(3,-2),C(5,2)$.

Solution:

Given vertices:

$A(4, −6), B(3, −2), C(5, 2)$

A(4,−6),B(3,−2),C(5,2)

Let

$M$

M be the midpoint of side

$BC$

BC, so:

$$M = \left( \frac{3 + 5}{2}, \frac{-2 + 2}{2} \right) = (4, 0)$$

M=(23+5​,2−2+2​)=(4,0)

The area of triangle

$ABM$

ABM is:

$$\text{Area of } ABM = \frac{1}{2} \left| 4(−2 − 0) + 3(0 − (−6)) + 4((−6) − (−2)) \right|$$

Area of ABM=21​∣4(−2−0)+3(0−(−6))+4((−6)−(−2))∣

$$= \frac{1}{2} \left| 4(-2) + 3(6) + 4(-4) \right| = \frac{1}{2} \left| -8 + 18 - 16 \right| = \frac{1}{2} \times 6 = 3$$

=21​∣4(−2)+3(6)+4(−4)∣=21​∣−8+18−16∣=21​×6=3

The area of triangle

$ACM$

ACM is:

$$\text{Area of } ACM = \frac{1}{2} \left| 4(2 − 0) + 5(0 − (−6)) + 4((−6) − 2) \right|$$

Area of ACM=21​∣4(2−0)+5(0−(−6))+4((−6)−2)∣

$$= \frac{1}{2} \left| 4(2) + 5(6) + 4(-8) \right| = \frac{1}{2} \left| 8 + 30 - 32 \right| = \frac{1}{2} \times 6 = 3$$

=21​∣4(2)+5(6)+4(−8)∣=21​∣8+30−32∣=21​×6=3

Since both triangles have the same area, the median divides the triangle into two triangles of equal area.

FAQs: Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.3

Q1. What is the main focus of Exercise 7.3?
Answer:
Exercise 7.3 focuses on applying coordinate geometry concepts to solve problems involving the distance between points, locus of points, and geometrical properties using the distance formula and other geometric principles.

Q2. How do I solve problems involving distance in coordinate geometry?
Answer:
To solve distance-related problems, use the distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

d=(x2​−x1​)2+(y2​−y1​)2​

This formula helps calculate the distance between two points

$(x_1, y_1)$

(x1​,y1​) and

$(x_2, y_2)$

(x2​,y2​) on the coordinate plane.

Q3. What is the concept of locus in coordinate geometry?
Answer:
A locus is the set of all points that satisfy a given condition. For example, the locus of points equidistant from a fixed point forms a circle. The concept of locus is used to represent shapes and geometric figures in coordinate geometry.

Q4. How do I apply the distance formula to solve geometric problems?
Answer:
To apply the distance formula in geometric problems:

1. Identify the coordinates of the points involved.

2. Use the distance formula to find the length of sides, diagonals, or distances between any two points.

3. Solve for the required quantities based on the problem's conditions.

Q5. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide clear, detailed explanations for applying coordinate geometry concepts to real-world problems. By practicing these solutions, students can develop strong problem-solving skills and apply formulas confidently, ensuring success in Class 10 exams.

If you want, I can also provide:

• Step-by-step solutions for distance and locus problems

• Coordinate geometry practice problems

• Quick revision tips for Chapter 7

• Class 10 exam PYQs

These NCERT Solutions for Class 10 Maths will help you confidently solve Exercise 7.3 and strengthen your understanding of coordinate geometry, ensuring excellent preparation for Class 10 exams.