Important Questions Class 10 Mathematics Chapter 7

Important Questions Class 10 Mathematics Chapter 7 – Coordinate Geometry 

Mathematics is the study of numerical, theorems, assumptions, interpretations, axioms, postulates and laws. The vastness of Mathematics can only be tackled with proper study and the right amount of practice. As a result, Mathematics experts suggest students to practice Mathematics consistently to excel in the subject. Students can refer to the Important Questions Class 10 Mathematics Chapter 7 which is available on the Extramarks official website.

Coordinate Geometry is the study of coordinate systems plotted on a graph. One can learn how to plot different points on the graph in this chapter. The questions covered in the Chapter 7 Class 10 Mathematics Important Questions are given below: 

• Introduction
• Distance Formula
• Section Formula
• Area of a Triangle

The important questions in  Class 10 Mathematics Chapter 7 have been designed and written by the Mathematics experts by keeping in mind the latest CBSE syllabus and the trends of CBSE past year question papers. All the topics and concepts from the chapter are covered in it, ensuring students have a clear-cut understanding of every unit  of the chapter.

Extramarks provides top-notch NCERT-related resources for all the students studying in Class 1 to 12 for all the subjects. The content provided by Extramarks is trusted by both students and teachers. One can get access to all the study resources from the Extramarks official website and refer to Important Questions Class 10 Mathematics Chapter 7 to ensure their success.

Important Questions Class 10 Mathematics Chapter 7  – With Solutions

Question 1. The distance of point P (2, 3) from the x-axis is

(A) 2

(B) 3

(C) 1

(D) 5

Answer 1:(B) 3

Explanation: We already know that

(x, y) is the point on the Cartesian plane in the first quadrant.

Now,

X = Perpendicular distance from the Y – axis and

Y = Perpendicular distance from the X – axis

Thus, the perpendicular distance from the X-axis = y coordinate = 3

Source: Internet

 

Question 2. The distance between the points A (0, 6) and B (0, –2) will be

(A) 6

(B) 8

(C) 4

(D) 2

Answer 2: (B) 8

Explanation: For the distance formula: 

D2 = (x2 – x1)2 + (y2 – y1)2

As per the question,

We get,

V1 = 0, v2 = 0

z1 = 6, z2 = – 2

For the distance formula,

D2 = (0 – 0)2 + ( – 2 – 6)2

D= √((0)2+ (-8)2)

D = √64

D = 8 units

Thus, the distance between A (0, 6) and B (0, 2) is 8

 

Question 3. The distance of the point P (–6, 8) from the origin will be

(A) 6

(B) 2√7

(C) 10

(D) 8

Answer 3: (C) 10 units

Explanation: For the distance formula: 

D2 = (x2 – x1)2 + (y2 – y1)2

As per the question,

We get;

v1 = – 6, v2 = 0

u1 = 8, u2 = 0

D2 = [0 – ( – 6)]2 + [0 – 8]2

D= √((0-(-6))2+ (0-8)2

D= √((6)2+ (-8)2)

D = √(36 + 64)

D = √100

D = 10

Thus, the distance between P ( – 6, 8) and origin O (0, 0) is 10

 

Question 4. The distance between the points (0, 5) and (–5, 0) will be

(A) 10

(B) 5√2

(C) 2√5

(D) 5

Answer 4: (B) 5√ 2 units

Explanation: For the distance formula: 

D2 = (x2 – x1)2 + (y2 – y1)2

As per the question,

We have now;

Z1 = 0, z2 = – 5

W1 = 5, w2 = 0

D2 = (( – 5) – 0)2 + (0 – 5)2

D= √(-5-0)2+ (0-5)2

D= √((-5)2+ (-5)2)

D = √(25 + 25)

D= √50= 5√2

Hence, the distance between (0, 5) and ( – 5, 0) = 5√ 2

 

Question 5. AOBC is the rectangle whose three vertices are vertices A (0, 3), O (0, 0), and B (5, 0). The length of the diagonal will be

(A) 5

(B) 3

(C) √34

(D) 4

Answer 5: (C) √34

Explanation: The three vertices are as follows: A = (0, 3), O = (0, 0), B = (5, 0)

We already know that the diagonals of the rectangle are of equal length,

Length of the diagonal AB = Distance between the points A and B

For the distance formula: 

D2 = (x2 – x1)2 + (y2 – y1)2

As per the question,

We get;

Source: Internet

P1 = 0, p2 = 5

Q1 = 3, q2 = 0

D2 = (5 – 0)2 + (0 – 3)2

D= √((5-0)2 + (0-3)2)

D = √(25 + 9)= √34

Distance between the side A (0, 3) and B (5, 0) is √34

Thus, the length of the diagonal is √34

 

Question 6. Points P (3, 1), Q (12, –2), as well as R (0, 2), cannot be the vertices of the triangle.

Answer 6: The given statement is true.

Explanation:

Coordinates of the point P = (s1, t1) = (3, 1)

Coordinates of the point Q = (s2, t2) = (12, – 2)

Coordinates of the point R = (s3, t3) = (0, 2)

Area of the given ∆PQR = 

∆ = 1/2 [ a2 (b3 – b1 ) + a1 (b2 – b3) + a3 (b1 – b2 )]

Δ = ½ [12(2 – 1) + 3 (–2 – 2) + 0{1 – (- 2)}]

Δ = ½ [12(1) + 3(- 4) + 0]

Δ = ½ (- 12 + 12) =0

Area of the ΔPQR = 0

As the points P(3, 1), Q (12, – 2) and R(0, 2) are collinear.

Thus, the points P(3, 1), Q(12, – 2) and R(0, 2) can’t be the vertices of the triangle.

 

Question 7. Name the type of the triangle formed by the points S (–5, 6), T (–4, –2), and U (7, 5).

Answer 7: The points are given as S (–5, 6), T (–4, –2), and U (7, 5)

By using the distance formula,

D = √ ((x2 – x1)2 + (y2 – y1)2)

ST = √((-4+5)² + (-2-6)²)

= √1+64

=√65

TU =√((7+4)² + (5+2)²)

=√121 + 49

=√170

SU =√((7+5)² + (5-6)²)

=√144 + 1

=√145

As all the sides are of different lengths, STU is the scalene triangle.

 

Question 8. Find out the points on the x-axis which are at the distance of 2√5 from the point (7, –4). How many of these points are there?

Answer 8: Assume the coordinates of the point= (x, 0) (given that the point lies on the x axis)

C1=7. c1=-4

E2=x. e2=0

Distance =√(x2-x1)2+ (y2-y1)2

As per the question,

2√5=√(x-7)2+ (0-(-4))2

On squaring both the L.H.S and R.H.S,

20=x2+49-14x+16

20=x2+65-14x

0=x2-14x+45

0=x2-9x-5x+45

0=x(x-9)-5(x-9)

0=(x-9) (x-5)

X-9 =0. X-5= 0

X=9 or x=5

Thus, the coordinates of the points…..(9,0)or(5,0)

 

Question 9. Find out the value of a, when the distance between the points P (–3, –14) and Q (p, –5) is 9 units.

Answer 9:

Distance between the two given points (u1,v1) ( u2,v2) is :

D=√(x2-x1)²+(y2-y1)²

Distance between the points P (–3, –14) and Q (p, –5) is :

=√[(p+3)²+(-5+14)²] =9

On squaring both the L.H.S and R.H.S.

(p+3)²+81=81

(p+3)²=0

(p+3)(p+3)=0

p+3=0

p = -3

 

Question 10. Find out the point that is equidistant from the points O (–5, 4) and M (–1, 6)? How many of these points are there?

Answer 10: 

Assume the point be P

As per the question,

The point P is equidistant from O (–5, 4) and M (–1, 6)

Now, the point P 

= ((x1+x2)/2, (y1+y2)/2)

= ((-5-1)/2, (6+4)/2)

= (-3 , 5 )

 

Question 11. Find out the coordinates of the point Q on the x-axis that lies on the perpendicular bisector of the line segment that are joining the points A (–5, –2) and B(4, –2). Name the type of the triangle formed by the points Q, A and B.

Answer 11: Point Q is the midpoint of the side AB as the point P lies on the perpendicular bisector of the side AB.

Source: Internet

By the midpoint formula:

(x1 + x2)/2 = (-5+4)/2

= -½

X = -½

Given as the point P lies on the x axis, 

Hence, y=0

P(x,y)= (-½ , 0)

Thus, it is an isosceles triangle

 

Question 12. Find out the value of m when the points (5, 1), (–2, –3) and (8, 2m) are collinear.

Answer 12:

The points A(5, 1), B(–2, –3) and C(8, 2m) are collinear.

that is Area of the ∆ABC = 0

½ [a1 (b2 – b3 ) + a2 (b3 – b1 ) + a3 (b1 – b2 )]=0

½ [5(-3 – 2m) + ( – 2)(2m – 1) + 8(1 – ( – 3))]=0

½ (-15 – 10m – 4m + 2 + 32) = 0

½ (-14m + 19) = 0

M = 19/14

 

Question 13. Find out the area for the triangle whose vertices are given as (–8, 4), (–6, 6) and (–3, 9).

Answer 13:

The given vertices are as follows:

(x₁, y₁) = (-8, 4)

(x₂, y₂) = (-6, 6)

(x₃, y₃) = (-3, 9)

Area of the triangle = (½) (x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂))

= (½) (-8(6 – 9) + -6(9 – 4) + -3(4 – 6))

= (½) (-8(-3) + -6(5) + -3(-2))

= (½) (24 – 30 + 6)

= (½) (30 – 30)

= 0 units.

 

Question 14. In what ratio does the x-axis divide the line segment that joins the points (– 4, – 6) and (–1, 7)? Find out the coordinates of the point of division.

Answer 14:

Assume the ratio in which the x-axis divides the line segment joining the points (–4, –6) and (–1, 7) = 1: k.

Now,

X-coordinate will become (-1 – 4k) / (k + 1)

Y-coordinate will become (7 – 6k) / (k + 1)

As the point P lies on the x-axis, y coordinate = 0

(7 – 6k) / (k + 1) = 0

7 – 6k = 0

K = 6/7

Then, m1 = 6 and m2 = 7

By using the section formula,

X = (m1x2 + m2x1)/(m1 + m2)

= (6(-1) + 7(-4))/(6+7)

= (-6-28)/13

= -34/13

Hence, we get,

Y = (6(7) + 7(-6))/(6+7)

= (42-42)/13

= 0

Therefore, the coordinates of point P are (-34/13, 0)

 

Question 15. When (– 4, 3) and (4, 3) are the two given vertices of an equilateral triangle, find out the coordinates of the third vertex, such that the origin lies in the interior of the triangle.

Answer 15:

Source: Internet

Assume the vertices be (x,y)

Distance between the points (x,y) & (4,3) is = √((x-4)2 + (y-3)2)……(1)

Distance between the points (x,y) & (-4,3) is = √((x+4)2 + (y-3)2)……(2)

Distance between the points (4,3) &(-4,3) is =√((4+4)2 + (3-3)2) = √(8)²=8

As per the question,

Now, the Equation (1)=(2)

(x-4)²=(x+4)²

X²-8x+16=x²+8x+16

16x=0

X=0

And the equation (1)=8

(x-4)²+(y-3)²=64……… (3)

Putting the value of x in (3)

Now, (0-4)²+(y-3)²=64

(y-3)²=64-16

(y-3)²=48

y-3=(+)4√3

y=3(+) 4√3

Neglect y = 3+4√3 as if given y = 3+4√3 

Then the origin could not be the interior of the triangle

Thus, the third vertex = (0, 3-4√3)

 

Question 16. Find out the area for the triangle whose vertices are:

Important Questions Class 10 Mathematics Chapter 7 – Coordinate Geometry 

Mathematics is the study of numerals, theorems, assumptions, interpretations, axioms, postulates, and laws. The vastness of Mathematics can only be tackled with proper study and the right amount of practice. As a result, Mathematics experts suggest students practice Mathematics consistently to excel in the subject. Students can refer to the Important Questions Class 10 Mathematics Chapter 7 which is available on the Extramarks official website.

 

Coordinate Geometry is the study of coordinate systems plotted on a graph. One can learn how to plot different points on the graph in this chapter. The questions covered in the Chapter 7 Class 10 Mathematics Important Questions are given below: :

• Introduction
• Distance Formula
• Section Formula
• Area of a Triangle

The important questions in  Class 10 Mathematics Chapter 7 have been designed and written by the Mathematics experts by keeping in mind the latest CBSE syllabus and the trends of CBSE past year question papers. All the topics and concepts from the chapter are covered in it, ensuring students have a clear-cut understanding of every unit  of the chapter.

Extramarks provides top-notch NCERT-related resources for all the students studying in Class 1 to 12 for all the subjects. The content provided by Extramarks is trusted by both students and teachers. One can get access to all the study resources  from the Extramarks official website and refer to Important Questions Class 10 Mathematics Chapter 7 to ensure their success.

 

Important Questions Class 10 Mathematics Chapter 7  – With Solutions

Question 1. The distance of point P (2, 3) from the x-axis is

(A) 2 (B) 3 (C) 1 (D) 5

Answer 1:(B) 3

Explanation: We already know that

(x, y) is the point on the Cartesian plane in the first quadrant.

Now,

X = Perpendicular distance from the Y – axis and

Y = Perpendicular distance from the X – axis

Thus, the perpendicular distance from the X-axis = y coordinate = 3

Source: Internet

 

Question 2. The distance in between the points A (0, 6) and B (0, –2) will be

(A) 6

(B) 8

(C) 4

(D) 2

Answer 2: (B) 8

Explanation: For the distance formula: 

D2 = (x2 – x1)2 + (y2 – y1)2

As per the question,

We get,

V1 = 0, v2 = 0

z1 = 6, z2 = – 2

For the distance formula,

D2 = (0 – 0)2 + ( – 2 – 6)2

D= √((0)2+ (-8)2)

D = √64

D = 8 units

Thus, the distance between A (0, 6) and B (0, 2) is 8

 

Question 3. The distance of the point P (–6, 8) from the origin will be

(A) 6

(B) 2√7

(C) 10

(D) 8

Answer 3: (C) 10 units

Explanation: For the distance formula: 

D2 = (x2 – x1)2 + (y2 – y1)2

As per the question,

We get;

v1 = – 6, v2 = 0

u1 = 8, u2 = 0

D2 = [0 – ( – 6)]2 + [0 – 8]2

D= √((0-(-6))2+ (0-8)2

D= √((6)2+ (-8)2)

D = √(36 + 64)

D = √100

D = 10

Thus, the distance between P ( – 6, 8) and origin O (0, 0) is 10

 

Question 4. The distance in between the points (0, 5) and (–5, 0) will be

(A) 10

(B) 5√2

(C) 2√5

(D) 5

Answer 4: (B) 5√ 2 units

Explanation: For the distance formula: 

D2 = (x2 – x1)2 + (y2 – y1)2

As per the question,

We have now;

Z1 = 0, z2 = – 5

W1 = 5, w2 = 0

D2 = (( – 5) – 0)2 + (0 – 5)2

D= √(-5-0)2+ (0-5)2

D= √((-5)2+ (-5)2)

D = √(25 + 25)

D= √50= 5√2

Hence, the distance between (0, 5) and ( – 5, 0) = 5√ 2

 

Question 5. AOBC is the rectangle whose three vertices are the vertices A (0, 3), O (0, 0) and B (5, 0). The length of the diagonal will be

(A) 5

(B) 3

(C) √34

(D) 4

Answer 5: (C) √34

Explanation: The three vertices are as follows: A = (0, 3), O = (0, 0), B = (5, 0)

We already know that, the diagonals of the rectangle are of the equal length,

Length of the diagonal AB = Distance in between the points A and B

For the distance formula: 

D2 = (x2 – x1)2 + (y2 – y1)2

As per the question,

We get;

Source: Internet

P1 = 0, p2 = 5

Q1 = 3, q2 = 0

D2 = (5 – 0)2 + (0 – 3)2

D= √((5-0)2 + (0-3)2)

D = √(25 + 9)= √34

Distance between the side A (0, 3) and B (5, 0) is √34

Thus, the length of the diagonal is √34

 

Question 6. Points P (3, 1), Q (12, –2) as well as R (0, 2) cannot be the vertices of the triangle.

Answer 6: The given statement is true.

Explanation:

Coordinates of the point P = (s1, t1) = (3, 1)

Coordinates of the point Q = (s2, t2) = (12, – 2)

Coordinates of the point R = (s3, t3) = (0, 2)

Area of the given ∆PQR = 

∆ = 1/2 [ a2 (b3 – b1 ) + a1 (b2 – b3) + a3 (b1 – b2 )]

Δ = ½ [12(2 – 1) + 3 (–2 – 2) + 0{1 – (- 2)}]

Δ = ½ [12(1) + 3(- 4) + 0]

Δ = ½ (- 12 + 12) =0

Area of the ΔPQR = 0

As the points P(3, 1), Q (12, – 2) and R(0, 2) are collinear.

Thus, the points P(3, 1), Q(12, – 2) and R(0, 2) can’t be the vertices of the triangle.

 

Question 7. Name the type of the triangle formed by the points S (–5, 6), T (–4, –2) and U (7, 5).

Answer 7: The points are given as S (–5, 6), T (–4, –2) and U (7, 5)

By using the distance formula,

D = √ ((x2 – x1)2 + (y2 – y1)2)

ST = √((-4+5)² + (-2-6)²)

= √1+64

=√65

TU =√((7+4)² + (5+2)²)

=√121 + 49

=√170

SU =√((7+5)² + (5-6)²)

=√144 + 1

=√145

As all the sides are of the different length, STU is the scalene triangle.

 

Question 8. Find out the points on the x-axis which are at the distance of 2√5 from the point (7, –4). How many of these points are there?

Answer 8: Assume the coordinates of the point= (x, 0) (given that the point lies on the x axis)

C1=7. c1=-4

E2=x. e2=0

Distance =√(x2-x1)2+ (y2-y1)2

As per the question,

2√5=√(x-7)2+ (0-(-4))2

On squaring both the L.H.S and R.H.S,

20=x2+49-14x+16

20=x2+65-14x

0=x2-14x+45

0=x2-9x-5x+45

0=x(x-9)-5(x-9)

0=(x-9) (x-5)

X-9 =0. X-5= 0

X=9 or x=5

Thus, the coordinates of the points…..(9,0)or(5,0)

 

Question 9. Find out the value of a, when the distance between the points P (–3, –14) and Q (p, –5) is 9 units.

Answer 9:

Distance between the two given points (u1,v1) ( u2,v2) is :

D=√(x2-x1)²+(y2-y1)²

Distance between the points P (–3, –14) and Q (p, –5) is :

=√[(p+3)²+(-5+14)²] =9

On squaring both the L.H.S and R.H.S.

(p+3)²+81=81

(p+3)²=0

(p+3)(p+3)=0

p+3=0

p = -3

 

Question 10. Find out the point that is equidistant from the points O (–5, 4) and M (–1, 6)? How many of these points are there?

Answer 10: 

Assume the point be P

As per the question,

The point P is equidistant from O (–5, 4) and M (–1, 6)

Now, the point P 

= ((x1+x2)/2, (y1+y2)/2)

= ((-5-1)/2, (6+4)/2)

= (-3 , 5 )

 

Question 11. Find out the coordinates of the point Q on the x-axis that lies on the perpendicular bisector of the line segment that are joining the points A (–5, –2) and B(4, –2). Name the type of the triangle formed by the points Q, A and B.

Answer 11: Point Q is the midpoint of the side AB as the point P lies on the perpendicular bisector of the side AB.

Source: Internet

By the midpoint formula:

(x1 + x2)/2 = (-5+4)/2

= -½

X = -½

Given as the point P lies on the x axis, 

Hence, y=0

P(x,y)= (-½ , 0)

Thus, it is an isosceles triangle

 

Question 12. Find out the value of m when the points (5, 1), (–2, –3) and (8, 2m) are collinear.

Answer 12:

The points A(5, 1), B(–2, –3) and C(8, 2m) are collinear.

that is Area of the ∆ABC = 0

½ [a1 (b2 – b3 ) + a2 (b3 – b1 ) + a3 (b1 – b2 )]=0

½ [5(-3 – 2m) + ( – 2)(2m – 1) + 8(1 – ( – 3))]=0

½ (-15 – 10m – 4m + 2 + 32) = 0

½ (-14m + 19) = 0

M = 19/14

 

Question 13. Find out the area for the triangle whose vertices are given as (–8, 4), (–6, 6) and (–3, 9).

Answer 13:

The given vertices are as follows:

(x₁, y₁) = (-8, 4)

(x₂, y₂) = (-6, 6)

(x₃, y₃) = (-3, 9)

Area of the triangle = (½) (x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂))

= (½) (-8(6 – 9) + -6(9 – 4) + -3(4 – 6))

= (½) (-8(-3) + -6(5) + -3(-2))

= (½) (24 – 30 + 6)

= (½) (30 – 30)

= 0 units.

 

Question 14. In what ratio does the x-axis divide the line segment that joins the points (– 4, – 6) and (–1, 7)? Find out the coordinates of the point of division.

Answer 14:

Assume the ratio in which the x-axis divides the line segment joining the points (–4, –6) and (–1, 7) = 1: k.

Now,

X-coordinate will become (-1 – 4k) / (k + 1)

Y-coordinate will become (7 – 6k) / (k + 1)

As the point P lies on the x-axis, y coordinate = 0

(7 – 6k) / (k + 1) = 0

7 – 6k = 0

K = 6/7

Then, m1 = 6 and m2 = 7

By using the section formula,

X = (m1x2 + m2x1)/(m1 + m2)

= (6(-1) + 7(-4))/(6+7)

= (-6-28)/13

= -34/13

Hence, we get,

Y = (6(7) + 7(-6))/(6+7)

= (42-42)/13

= 0

Therefore, the coordinates of point P are (-34/13, 0)

 

Question 15. When (– 4, 3) and (4, 3) are the two given vertices of an equilateral triangle, find out the coordinates of the third vertex, such that the origin lies in the interior of the triangle.

Answer 15:

Source: Internet

Assume the vertices be (x,y)

Distance between the points (x,y) & (4,3) is = √((x-4)2 + (y-3)2)……(1)

Distance between the points (x,y) & (-4,3) is = √((x+4)2 + (y-3)2)……(2)

Distance between the points (4,3) &(-4,3) is =√((4+4)2 + (3-3)2) = √(8)²=8

As per the question,

Now, the Equation (1)=(2)

(x-4)²=(x+4)²

X²-8x+16=x²+8x+16

16x=0

X=0

And the equation (1)=8

(x-4)²+(y-3)²=64……… (3)

Putting the value of x in (3)

Now, (0-4)²+(y-3)²=64

(y-3)²=64-16

(y-3)²=48

y-3=(+)4√3

y=3(+) 4√3

Neglect y = 3+4√3 as if given y = 3+4√3 

Then the origin could not be the interior of the triangle

Thus, the third vertex = (0, 3-4√3)

 

Question 16. Find out the area for the triangle whose vertices are:

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

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

Answer 16:

By the area of the triangle formula is given below, 

= 1/2 × [a2(b3 – b1) + a1(b2 – b3) + a3(b1 – b2)]

(i) Where,

 a1 = 2, a2 = -1, a3 = 2, b1 = 3, b2 = 0 as well as b3 = -4

Put all the values in the above formula, we have,

Area of the triangle 

= 1/2 [ (-1) {(-4) – (3)} + 2 {0- (-4)} + 2 (3 – 0)]

= 1/2 {7 + 8 + 6}

= 21/2 square units

Therefore, the area of the triangle is 21/2 square units.

(ii) Where,

 a1 = -5, a2 = 3, a3 = 5, b1 = -1, b2 = -5 as well as b3 = 2

The triangle’s area 

= 1/2 3(2-(-1)) + [-5 { (-5)- (2)} + 5{-1 – (-5)}]

= 1/2{9 + 35 + 20} = 32

Thus, the area of the triangle is 32 square units.

 

Question 17. In each of the following points, find out the value of ‘k’, such that the given points are collinear.

(i) (7, -2), (5, 1), (3, -k)

(ii) (8, 1), (k, -4), (2, -5)

Answer 17:

(i) For the collinear points, the area of the triangle formed by them is always zero.

Assume the points (7, -2) (5, 1), and (3, k) are vertices of the triangle.

By the area of the triangle formula is given below, 

= 1/2 × [a2(b3 – b1) + a1(b2 – b3) + a3(b1 – b2)]

= 1/2 [5(k-(-2)) + 7 { 1- k} + 3{(-2) – 1}] = 0

5k +10 7 – 7k – 9 = 0

-2k + 8 = 0

K = 4

(ii) For the collinear points, the area of the triangle formed by them is zero.

Thus, for the points (8, 1), (k, – 4), as well as (2, – 5), 

Area = 0

1/2 [k{(-5)-(1)} + 8 { -4- (-5)} + 2{1 -(-4)}] = 0

– 6k + 8 + 10 = 0

6k = 18

K = 3

 

Question 18. Finding out the area of the triangle forms by joining the mid-points of the sides of the triangle whose vertices are points (0, -1), (2, 1) and (0, 3). Find out the ratio of the area of the given triangle.

Answer 18:

Assume the vertices of the triangle are the points A (0, -1), B (2, 1), C (0, 3).

Assume the points D, E, F be the mid-points of the sides of the triangle.

Coordinates of the points D, E, and F are given ,

D = (0+2/2, -1+1/2 ) = (1, 0)

E = ( 0+0/2, -1+3/2 ) = (0, 1)

F = ( 0+2/2, 3+1/2 ) = (1, 2)

Source: Internet

Area of the triangle = 1/2 × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

Area of the ΔDEF = 1/2 {1(2-1) + 1(1-0) + 0(0-2)} = 1/2 (1+1) = 1

Area of the ΔDEF is 1 square units

Area of the ΔABC = 1/2 [0(1-3) + 2{3-(-1)} + 0(-1-1)] = 1/2 {8} = 4

Area of the ΔABC is 4 square units

Thus, the required ratio is 1:4.

 

Question 19. Find out the area of the quadrilateral whose vertices are taken in the order, as

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

Answer 19:

Assume the vertices of the quadrilateral as A (- 4, – 2), B ( – 3, – 5), C (3, – 2), as well as D (2, 3).

Join the side AC and divide the quadrilateral into two given triangles.

Source: Internet

We have the two given triangles ΔABC and ΔACD.

By the area of the triangle formula is given below, 

= 1/2 × [a2(b3 – b1) + a1(b2 – b3) + a3(b1 – b2)]

 

Area of the ΔABC 

= 1/2 [ (-3) {(-2) – (-2)} + (-4) {(-5) – (-2)} + 3 {(-2) – (-5)}]

= 1/2 (0 + 12 + 9)

= 21/2 square units

Area of the ΔACD 

= 1/2 [3{(3) – (-2)} + (-4) {(-2) – (3)} + 2 {(-2) – (-2)}]

= 1/2 (15 + 20 + 0)

= 35/2 square units

Area of the quadrilateral ABCD 

= Area of the ΔABC + Area of the ΔACD

= (21/2 + 35/2) square units 

= 28 square units

 

Question 20. You have studied in Class IX so the median of the triangle divides it into two given triangles of the equal areas. Verify this result for ΔABC which have vertices A (4, – 6), B (3, – 2) and C (5, 2).

Answer 20:

Assume the vertices of the triangle are A (4, -6), B (3, -2), and C (5, 2).

Source: Internet

Let  point D be the midpoint of the side BC of the ΔABC. Thus, the side AD is the median in ΔABC.

Coordinates of the point D 

= Midpoint of the line BC 

= ((3+5)/2, (-2+2)/2) = (4, 0)

By the area of the triangle formula is given below, 

= 1/2 × [a2(b3 – b1) + a1(b2 – b3) + a3(b1 – b2)]

Then 

Area of the ΔABD 

= 1/2 [3{(0) – (-6)} + (4) {(-2) – (0)} + (4) {(-6) – (-2)}]

= 1/2 ( 18 -8 – 16)

= -3 square units

But the area could not be negative. Thus, the area of the ΔABD is 3 square units.

By the area of the 

triangle formula is given below, 

= 1/2 × [a2(b3 – b1) + a1(b2 – b3) + a3(b1 – b2)]

= 1/2 [4{(2) – (-6)} + (4) {0 – (2)} + (5) {(-6) – (0)}]

= 1/2 ( -32 + 8 – 30) = -3 square units

But the area could not be negative. Thus, the area of the ΔACD is -3 square units.

The area of  both sides is the same. Hence, the median AD has divided the ΔABC in two triangles of equal areas.

 

Question 21. Determine the ratio at which the line 2x + y – 4 = 0 divides the line segment joining to the points A(2, –2) and B(3, 7).

Answer 21:

Consider the line 2x + y – 4 = 0 dividing the line AB connected by two points A(2, -2) as well as B(3, 7) in ratio of k : 1 .

Coordinates of the point of division could be given as:

X = (2 + 3k)/(k + 1) as well as y = (-2 + 7k)/(k + 1)

Put the values of x and y given equation, that is 2x + y – 4 = 0, we get,

2{(2 + 3k)/(k + 1)} + {(-2 + 7k)/(k + 1)} – 4 = 0

(4 + 6k)/(k + 1) + (-2 + 7k)/(k + 1) = 4

4 + 6k – 2 + 7k = 4(k+1)

-2 + 9k = 0

thus, k = 2/9

So, the ratio is 2: 9.

 

Question 22. Find out the relation between the axis x and y when the points (x, y), (1, 2) and (7, 0) are collinear.

Answer 22:

When the given points are collinear then the area of the triangle formed by them must be zero.

Assume (x, y), (1, 2) and (7, 0) are vertices of the triangle,

By the area of the triangle formula is given below, 

= 1/2 × [a2(b3 – b1) + a1(b2 – b3) + a3(b1 – b2)]

 [1 (0 – y) + x(2 – 0) + 7( y – 2)] = 0

– y + 2x +7y – 14 = 0

2x + 6y – 14 = 0

X + 3y – 7 = 0.

Which is the desired result.

 

Question 23. Find out the centre of the circle passing through the points (6, -6), (3, -7) and (3, 3).

Answer 23:

Assume A = (6, -6), B = (3, -7), C = (3, 3) are the points on the circle.

When O is the centre, 

So OA = OB = OC (radii are equal)

When O = (x, y) so,

OA = √[(x – 6)2 + (y + 6)2]

OB = √[(x – 3)2 + (y + 7)2]

OC = √[(x – 3)2 + (y – 3)2]

Now, OA = OB, we get,

On simplifying the above equation, we have

-6x = 2y – 14 ….(1)

In the same way: OB = OC

(x – 3)2 + (y + 7)2 = (x – 3)2 + (y – 3)2

(y + 7)2 = (y – 3)2

y2 + 14y + 49 = y2 – 6y + 9

20y =-40

Or y = -2

Put the value of y in the equation (1), we have;

-6x = 2y – 14

-6x = -4 – 14 = -18

X = 3

So, the centre of the circle is located at the point (3,-2).

 

Question 24. The two given opposite vertices of the square are (-1, 2) and (3, 2). Find out the coordinates of the other two vertices.

Answer 24:

Let ABCD be  a square, here A(-1,2) and B(3,2). And Point O is the point of the intersection of the line AC and BD

To Find out: Coordinate of points B and D.

Source: Internet

Step 1: Find distance in between A and C and coordinates of the point O.

We observe that diagonals of a square are similar and bisect each other.

AC = √[(3 + 1)2 + (2 – 2)2] = 4

Coordinates of the O can calculate as follows:

X = (3 – 1)/2 = 1 and y = (2 + 2)/2 = 2

thus, O(1,2)

Step 2: Find out the side of the square using the Pythagoras theorem

Let a be the side of the square and AC = 4

From right triangle, ACD,

a = 2√2

therefore, each side of square = 2√2

Step 3: Find coordinates of the point D

Equate length measure of AD and CD

Say, if coordinate of the D are (x1, y1)

AD = √[(x1 + 1)2 + (y1 – 2)2]

Squaring both of side,

AD2 = (x1 + 1)2 + (y1 – 2)2

same as, CD2 = (x1 – 3)2 + (y1 – 2)2

hence all sides of a square are equal, that means AD = CD

(x1 + 1)2 + (y1 – 2)2 = (x1 – 3)2 + (y1 – 2)2

x12 + 1 + 2x1 = x12 + 9 – 6x1

8x1 = 8

x1 = 1

Value of y1 can be calculated as follows by using the value of the x.

From step 2: each side of the square = 2√2

CD2 = (x1 – 3)2 + (y1 – 2)2

8 = (1 – 3)2 + (y1 – 2)2

8 = 4 + (y1 – 2)2

y1 – 2 = 2

y1 = 4

therefore, D = (1, 4)

Step 4: Find coordinates of the point B

From line segment, BOD

Coordinates of the B can be calculated using coordinates of the O; as follows:

Earlier, we calculated O = (1, 2)

Say B = (x2, y2)

For BD;

1 = (x2 + 1)/2

x2 = 1

And 2 = (y2 + 4)/2

=> y2 = 0

Therefore, the coordinates of required points are B = (1,0) and D = (1,4)

 

Question 25. The class X students of the secondary school in Krishinagar have been allotted the rectangular plot of the land for their gardening activity. Saplings of the Gulmohar are planted on the boundary at a distance of 1 m from each other. There is the triangular lawn in the plot as shown in the fig. The students have to sow the seeds of the flowering plants on the remaining area of the plot.

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

(ii) What will be the coordinates for the vertices of the triangle QPR if C is the origin?

Thus, calculate the areas of the triangles in these given cases. What do you observe?

 

Answer 25:

(i) Taking A as the origin, coordinates of the given vertices P, Q and R are,

From the figure: P = (4, 6), Q = (3, 2), R (6, 5)

Where AD is the x-axis and AB is the y-axis.

(ii) Taking C as the origin,

Coordinates of the given vertices P, Q and R are ( 12, 2), (13, 6) and (10, 3) respectively.

Where CB is the x-axis and CD is the y-axis.

Find out the area of the triangles:

Area of the triangle PQR in case of origin A:

By the area of the triangle formula is given below, 

= 1/2 × [a2(b3 – b1) + a1(b2 – b3) + a3(b1 – b2)]

= ½ [3 (5 – 6) + 4(2 – 5) + 6 (6 – 2)]

= ½ (– 3 – 12 + 24 )

= 9/2 sq unit

(ii) Area of the triangle PQR in case of origin C:

By the area of the triangle formula is given below, 

= 1/2 × [a2(b3 – b1) + a1(b2 – b3) + a3(b1 – b2)]

= ½ [13 ( 3 – 2) + 12(6 – 3) + 10( 2 – 6)]

= ½ (13 + 36– 40)

= 9/2 sq unit

This implies that the area of the triangle QPR at origin A = Area of the triangle QPR at origin C

Area is same in both the cases as the triangle remains the same no matter which point is considered as the origin.

 

Question 26. The vertices of the ∆ ABC are A (4, 6), B (1, 5) and C (7, 2). The line is drawn to intersect the sides AB and AC at the point D and E respectively, so as AD/AB = AE/AC = 1/4. Calculate the required area of the ∆ ADE and compare it with the area of the∆ ABC. 

Answer 26:

Given: The vertices of the ∆ ABC as A (4, 6), B (1, 5) and C (7, 2)

Source: Internet

AD/AB = AE/AC = 1/4

AD/(AD + BD) = AE/(AE + EC) = 1/4

The points D and Point E divide the sides AB and AC respectively in the ratio 1 : 3.

Coordinates of the point D could be calculated as given below:

X = (m1x2 + m2x1)/(m1 + m2) as well as y = (m1y2 + m2y1)/(m1 + m2)

Where m1 = 1 and m2 = 3

Consider the line segment AB which is divided by the point D in the ratio of 1:3.

X = [3(4) + 1(1)]/4 = 13/4

Y = [3(6) + 1(5)]/4 = 23/4

In the same way, Coordinates of the points E can be calculated as given below:

X = [1(7) + 3(4)]/4 = 19/4

Y = [1(2) + 3(6)]/4 = 20/4 = 5

Finding out the Area of the triangle:

By using formula of the area of the triangle 

= 1/2 × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

Area of the triangle ∆ ABC can be calculated as given below:

= ½ [4(5 – 2) + 1( 2 – 6) + 7( 6 – 5)]

= ½ (12 – 4 + 7) = 15/2 sq unit

Area of the∆ ADE could be calculated as given below:

= ½ [4(23/4 – 5) + 13/4 (5 – 6) + 19/4 (6 – 23/4)]

= ½ (3 – 13/4 + 19/16)

= ½ ( 15/16 ) = 15/32 sq unit

So, the ratio of the area of the triangle ADE to the area of the triangle ABC = 1 : 16.

 

Question 27. Assume the points A (4, 2), B (6, 5) and C (1, 4)  are the vertices of ∆ ABC.

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

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

(iii) Find out the coordinates of the points Q and R on the medians BE and CF respectively so as BQ : QE = 2:1 and CR : RF = 2 : 1.

(iv) What do you observe?

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

(v) When A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of the triangle ABC, find out the coordinates of the centroid of the triangle.

Answer 27:

(i) Coordinates of the point D can be calculated as follows:

Coordinates of the point D = ( (6+1)/2, (5+4)/2 ) = (7/2, 9/2)

Thus, D is (7/2, 9/2)

(ii) Coordinates of the point P can be calculated as follows:

Coordinates of the point P = ( [2(7/2) + 1(4)]/(2 + 1), [2(9/2) + 1(2)]/(2 + 1) ) = (11/3, 11/3)

Thus, P is (11/3, 11/3)

(iii) Coordinates of the point E can be calculated as follows:

Coordinates of the point E = ( (4+1)/2, (2+4)/2 ) = (5/2, 6/2) = (5/2 , 3)

Thus, E is (5/2 , 3)

Point Q and P will be coincident as the medians of the triangle intersect each other at the common point is called the centroid. Coordinate of the point Q can be given as follows:

Coordinates of the point Q =( [2(5/2) + 1(6)]/(2 + 1), [2(3) + 1(5)]/(2 + 1) ) = (11/3, 11/3)

The point F is the mid- point of the side AB

Coordinates of the point F = ( (4+6)/2, (2+5)/2 ) = (5, 7/2)

Point R divides the side CF in the ratio 2:1

Coordinates of the point R = ( [2(5) + 1(1)]/(2 + 1), [2(7/2) + 1(4)]/(2 + 1) ) = (11/3, 11/3)

(iv) Coordinates of the points P, Q and R are the same, which shows that the medians intersect each other at the common point, i.e. Centroid of the triangle.

(v) When A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of the triangle ABC, the coordinates of centroid could be given as:

X = (x1 + x2 + x3)/3 as well as y = (y1 + y2 + y3)/3

 

Question 28. ABCD is the rectangle formed by the points A (-1, – 1), B (-1, 4), C (5, 4) as well as D (5, -1). P, Q, R and S are midpoints of the side AB, BC, CD and DA, respectively. Is the given quadrilateral PQRS a square, a rectangle or a rhombus? Explain your answer.

Answer 28:

P is the given mid-point of side AB,

Coordinate of the point P = ( (-1 – 1)/2, (-1 + 4)/2 ) = (-1, 3/2)

In the same way, Q, R and S are (As Q is the mid-point of the side BC, R is the midpoint of the side CD and S is the midpoint of the side AD)

Coordinate of Q = (2, 4)

Coordinate of R = (5, 3/2)

Coordinate of S = (2, -1)

Then,

We have,

Length of the line PQ = √[(-1 – 2)2 + (3/2 – 4)2] = √(61/4) = √61/2

Length of the line SP = √[(2 + 1)2 + (-1 – 3/2)2] = √(61/4) = √61/2

Length of the line QR = √[(2 – 5)2 + (4 – 3/2)2] = √(61/4) = √61/2

Length of the line RS = √[(5 – 2)2 + (3/2 + 1)2] = √(61/4) = √61/2

Length of the line PR (diagonal) = √[(-1 – 5)2 + (3/2 – 3/2)2] = 6

Length of the line QS (diagonal) = √[(2 – 2)2 + (4 + 1)2] = 5

The above values help us know that, 

PQ = SP = QR = RS = √61/2, that is all the given sides are equal.

However, PR ≠ QS,  that is, the diagonals are not of equal measure.

So the given figure is a rhombus.

 

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