NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

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Written by Team Trustudies
Updated at 2021-02-11


NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1

1. Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b), (- a, – b)



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Distance formula =(x2?x1)2+(y2?y1)2
(i) Applying Distance Formula to find distance between points (2, 3) and (4,1), we get
d = (4?2)2+(1?3)2=22+(?2)2=4+4=8=22

(ii) Applying Distance Formula to find distance between points (–5, 7) and (–1, 3), we get
d = (?5+1)2+(7?3)2=((?4)2+(4)2=16+16=32=42

(iii) Applying Distance Formula to find distance between points (a, b) and (–a, –b), we get
d = (?a?a)2+(?b?b)2=((?2a)2+(?2b)2=4a2+4b2=2a2+b2

2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

The coordinates of point A are (0, 0) and coordinates of point B are (36, 15).
To find the distance between them, we use Distance formula:
d = (36?0)2+(15?0)2=362?152=1296+225=1521=39km

3. Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Let A = (1, 5), B = (2, 3) and C = (–2, –11)
Using Distance Formula to find distance AB, BC and CA.
AB = (2?1)2+(3?5)2=12?(?2)2=1+5=5
BC = (?2?2)2+(?11?3)2=(?4)2?(?14)2=16+196=212
CA = (?2?1)2+(?11?5)2=(?3)2?(?15)2=9+256=265
Since AB + AC ? BC, BC + AC ? AB and AC ? BC.
Therefore, the points A, B and C are not collinear.

4. Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Let A = (5, –2), B = (6, 4) and C = (7, –2)
Using Distance Formula to find distances AB, BC and CA.
AB = (6?5)2+(4+2)2=12?62=1+36=37
BC = (7?6)2+(?2?4)2=12?(?6)2=1+36=37
CA = (7?5)2+(?2+2)2=22?0=4=2
Since AB = BC.
Therefore, A, B and C are vertices of an isosceles triangle.

5. In a classroom, 4 friends are seated at the points A (3, 4), B (6, 7), C (9, 4) and D (6, 1). Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli. “Don’t you think ABCD is a square?”Chameli disagrees. Using distance formula, find which of them is correct.



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

We have A = (3, 4), B = (6, 7), C = (9, 4) and D = (6, 1)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = (6?3)2+(7?4)2=32+32=9+9=18=32
BC = (9?6)2+(4?7)2=32+32=9+9=18=32
CD = (6?9)2+(1?4)2=32+(?3)2=9+9=18=32
DA = (6?3)2+(1?4)2=32+(?3)2=9+9=18=32
Therefore, All the sides of ABCD are equal here. … (1)
Now, we will check the length of its diagonals.
AC = (9?3)2+(4?4)2=62?02=36=6
BD = (6?6)2+(1?7)2=02?62=36=6
So, Diagonals of ABCD are also equal. … (2)
From (1) and (2), we can definitely say that ABCD is a square.
Therefore, Champa is correct.

6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(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)



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

i) Let A = (–1, –2), B = (1, 0), C= (–1, 2) and D = (–3, 0)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = (1+1)2+(0+2)2=4+4=22
BC = (?1?1)2+(2?0)2=4+4=22
CD = (?3+1)2+(0?2)2=4+4=22
DA = (?3+1)2+(0?2)2=4+4=22
Therefore, all four sides of quadrilateral are equal. … (1)
Now, we will check the length of diagonals.
AC = (?1+1)2+(2+2)2=0+16=4
BD = (?3?1)2+(0+0)2=16+0=4
Therefore, diagonals of quadrilateral ABCD 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)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = (?3?3)2+(1?5)2=36+16=213
BC = (0?3)2+(3?1)2=9+4=13
CD = (?1?0)2+(?4?3)2=1+49=52
DA = (?1+3)2+(?4?5)2=4+81=85
We cannot find any relation between the lengths of different sides.
Therefore, we cannot give any name to the figure ABCD.

(iii) Let A = (4, 5), B= (7, 6), C= (4, 3) and D= (1, 2)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = (7?4)2+(6?5)2=9+1=10
BC = (4?7)2+(3?6)2=9+9=18
CD = (1?4)2+(2?3)2=9+1=10
DA = (1?4)2+(2?5)2=9+9=18
Here opposite sides of quadrilateral ABCD are equal. … (1)
We can now find out the lengths of diagonals.
AC = (4?4)2+(3?5)2=0+4=2
BD = (1?7)2+(2?6)2=36+16=213
Here diagonals of ABCD are not equal. … (2)
From (1) and (2), we can say that ABCD is not a rectangle therefore it is a parallelogram.

7. Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Let the point be (x, 0) on x–axis which is equidistant from (2, –5) and (–2, 9).
Using Distance Formula and according to given conditions we have:
(x?2)2+(0?(?5))2=(x?(?2))2+(0?9)2
=>x2+4?4x+25=x2+4+4x+81
Squaring both sides, we get
= > x2+4?4x+25=x2+4+4x+81
=> ?4x+29=4x+85
=>8x=?56
=>x=?7
Therefore, point on the x–axis which is equidistant from (2, –5) and (–2, 9) is (–7, 0)

8. Find the values of y for which the distance between the points P (2, –3) and Q (10, y) is 10 units.



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Using Distance formula, we have
10=(2?10)2+(?3?y)2
=>10=(?8)2+9+y2+6y
=>10=64+9+y2+6y
Squaring both sides, we get
100 =73+y2+6y
=> y2+6y+27=0

Solving this Quadratic equation by factorization, we can write
=>y2+9y?3y?27=0
=>y(y+9)3(y+9)=0
=>(y+9)(y?3)=0
=>y=3,?9

9. If, Q (0, 1) is equidistant from P (5, –3) and R (x, 6), find the values of x. Also, find the distances QR and PR.



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

It is given that Q is equidistant from P and R. Using Distance Formula, we get
PQ = RQ
PQ=(5?0)2+(?3?1)2=(?5)2+(?4)2=25+16=41
QR= (0?x)2+(1?6)2=(?x)2+(?5)2=x2+25
So, 41=x2+25
On squaring both sides,
41=x2+25
x2=16
x=±4
Coordinates of point R will be (4,6) or (-4,6),
If R is (4,6),
QR = (0?4)2+(1?6)2=42+52=16+25=41
PR = (5?4)2+(?3?6)2=12+92=1+81=82
If R is (-4,6),
QR = (0+4)2+(1?6)2=42+52=16+25=41
PR = (5+4)2+(?3?6)2=92+92=81+81=92

10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4).



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

It is given that (x, y) is equidistant from (3, 6) and (–3, 4).
Using Distance formula, we can write
(x?3)2+(y?6)2=(x?(?3))2+(y?4)2
=>x2+9?6x+y2+36?12y=x2+9+6x+y2+16?8y
Squaring both sides, we get
=> x2+9?6x+y2+36?12y=x2+9+6x+y2+16?8y
=> ?6x?12y+45=6x?8y+25
=>12x+4y=20
=>3x+y=5

NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2

1. Find the coordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2:3.



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Let the points be P(x,y), Using the section formula

P(x,y)=[mx2+nx1m+n,my2+ny1m+n]

x=2(4)+3(?1)2+3=8?35=1
y=2(?3)+3(7)2+3=?6+215=3
Therefore the point is(1,3)

Q.2 Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :


Let P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB
Therefore, point P divides AB internally in the ratio 1:2.
x1 = (1×(?2)+2×4)1+2=(?2+8)3=63=2
y1 = (1×(?3)+2×?1)1+2=(?3?2)3=?53
?P = (x1, y1) = P(2, ?53)
Point Q divides AB internally in the ratio 2:1.
x2 = (2×(?2)+1×4)2+1=(?4+4)3=0
y2 = (2×(?3)+1×?1)2+1=(?6?1)3=?73
The coordinates of the point Q(x2 ,y2 ) = (0, ?73 )

Q.3 To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs 14 th the distance AD on the 2nd line and posts a green flag. Preet runs 15th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

The green flag is 14th of total distance =
14×100=25 m in second line
? the coordinate of green flag = (2 , 25 )
Similarly coordinate of red flag = (8 , 20)
Distance between red and green flag
= (8?2)2+(20?25)2=62+(?5)2=36+25=61m
Now blue flag is posted at the mid point of two flag
Then let, the coordinate of blue flag = (x , y )
?(x,y)=((8+2)2,20+252)=(5,452)=(5,22.5)
? the blue flag is posted in fifth line at a distance of 22.5 m

Q.4 Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6).



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Let the required ratio be k : 1
By section formula ,
x = m1x2+m2x1m1+m2
==> -1 = k×6+1×?3)k+1
==> -k-1 = 6k -3
==> 7k = 2
==> k = 27
The required ratio = k : 1 27 : 1 = 27×7:1×7 = 2 : 7

Q.5 Find the ratio in which the line segment joining A (1, – 5) and B (- 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :


Let p(x , 0) be the point which devides the line segment joining A(1, -5 ) and B(-4,5 ) in the ratio m : 1
Then the using section formula
(x , 0 ) = (m×(?4)+1×1m+1,m×5+1×(?5)m+1)
==> 0 = m×5+1×(?5)m+1
==> 0 = 5m - 5
==> m = 5 /5 = 1
Hence the required ratio is 1 : 1
Since the ratio is 1 : 1 , so P is the mid point
? x = m×(?4)+1×1m+1
x = 1×(?4)+1×11+1=?4+12=?32
?(?32,0) is the required point

Q.6 If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :


Mid point of AC = Mid point BD
==> x+12,6+22=4+32,y+52
==> x+12=72
==> x +1 = 7
==> x = 7-1= 6
==> y+52=82
==> y+5 = 8
==> y = 8 -5 = 3

Q.7 Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, – 3) and B is (1,4).



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Let the coordinates of point A be (x, y).
Mid-point of AB is (2, – 3), which is the centre of the circle.
Coordinate of B = (1, 4)
(2, -3) =(x+12,y+42)
x+12=2==>x+1=4==>x=4?1=3
andy+42=?3==>y+4=?6==>y=?6?4=?10
? x = 3 and y = -10
The coordinates of A(3,-10).

Q.10 Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order.
[Hint: Area of a rhombus = 1/2(product of its diagonals)



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :


Let A(3, 0), B (4, 5), C( – 1, 4) and D ( – 2, – 1) are the vertices of a rhombus ABCD.
Length of diagonal AC
=[3?(?1)]2+(0?4)2=16+16=32=42
Length of diagonal BD
=[(?2)?4]2+(?1?5)2=(?6)2+(?6)2=36+36=72=62
Area of rhombus = 12 × product of diagonals
= 12×42×62=24sq unit

NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.3

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

(i) (2, 3), (-1, 0), (2, -4)

(ii) (-5, -1), (3, -5), (5, 2)



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

We know that formula for area of a triangle whose vertices are (x1,y1),(x2,y2),(x3,y3) is,
= 12 | [x1(y2?y3) + x2(y3?y1) + x3(y1?y2)] |

(i) So, here x1 = 2 , y1 = 3 , x2 = ?1 , y2 = 0 x3 = 2 , y3 = ?4

So, area of triangle = 12 | [2(0+4) ? 1(?4?3) + 2(3?0)] | = 12 | [8+7+6] | = 212

? Area of triangle is 212 sq. units.

(ii) Similarly, here x1 = ?5 , y1 = ?1 , x2 = 3 , y2 = ?5 x3 = 5 , y3 = 2
So, area of triangle = 12 | [(?5)(?5?2) ? 3(2+1) + 5(?1+5)] | = 12 | [35+9+20] | = 32

? Area of triangle is 32 sq. units.

2. 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)



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :


We know that for collinear points area of triangle = 0 ,i.e.,
0 =12[x1(y2?y3) + x2(y3?y1) + x3(y1?y2)]

(i) 7(1?k)+5(k+2)+3(?2?1) = 0
=> 7?7k+5k+10?9 = 0
=> 2k = 8
=> k = 4

(ii) 8(?4+5)+k(?5?1)+2(1+4) = 0
=> 8?6k+10 = 0
=> 6k = 18
=> k = 3

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



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :



The vertices of the triangle ABC are A (0, -1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides AB,BC,CA of this triangle ABC respectively.
Coordinates of D, E, and F are given by :

D = (0+22 , ?1+12) = (1,0)
E = ( 0+22 , 1+32 ) = (1,2)
F = ( 0+02 , 3?12 ) = (0,1)

So, Area of ? DEF = 12 | [1(2?1)+1(1?0)+0(0?2)] | = 1

? Area of ? DEF = 1 sq. units

Area of ? ABC = 12 | [0(1?3) + 2(3+1) + 0(?1?1)] | = 4

? Area of ? ABC = 4 sq. units

So, ? DEF : ? ABC = 1 : 4.

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



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :


Draw a line from B to D



Now we have,
Area of quad. ABCD = | Area of ? DAB | + | Area of ? BCD |

So,
Area of ? DAB = | 12[2(?2+5) ? 4(?5?3) ? 3(3+2)] | =  |12[6+32?15] | = 232 sq. units

Similarly, Area of ? BCD = | 12[?3(?2?3) + 3(3+5) + 2(?5+2)] | = | 12[15+24?6] | = 332 sq. units

? Area of quad. ABCD = 232 + 332 = 562 = 28 sq. units

5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ?ABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

The vertices of triangle ABC are A(4,-6), B (3, – 2) and C (5, 2).

Let D be the mid-point of side BC of ? ABC. Therefore, AD is the median in ? ABC.



So, coordinates of D are ( 3+52 , ?2+22 )
Therefore, coordinates of D = (4,0)

Now,
Area of ? ABD = 12 | [4(?2?0) + 3(0+6) + 4(?6+2)]|
= 12 | [?8+18?16] |
= |?3| = 3 sq. units

Similarly,
Area of ? ACD = 12 | [4(0?2) + 4(2+6) + 5(?6?0)] |
= 12 | [?8+32?30] |
= |?3| = 3 sq. units

? Clearly, Area of ? ABD = Area of ? ACD = 3 sq. units.

Hence, median of a triangle divides it into two triangles of equal areas.

NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.4

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



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Let the line 2x + y - 4 = 0 divides AB in 1 : k ratio, then the coordinates of the point of division is


x = 2+3kk+1 and y = ?2+7kk+1

The point of intersection lie on both lines

? x = 2+3kk+1 and y = ?2+7kk+1 should satisfy 2x + y - 4 = 0
=>