Coordinate Geometry Class 10 Extra Questions Maths Chapter 7

Coordinate Geometry Class 10 Extra Questions Maths Chapter 7

Extra Questions for Class 10 Maths Chapter 7 Coordinate Geometry. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

You can also download Class 10 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

Coordinate Geometry Class 10 Extra Questions Very Short Answer Type

Question 1.
If the distance between the points (4, k) and (1,0) is 5, then what can be the possible values of it? [CBSE 2017]
Answer:
Let A (4, k), B (1, 0)
AB = 5 given
⇒ \(\sqrt{(4-1)^2+(k-0)^2}\) = 5
⇒ \(\sqrt{3^2+k^2}\) = 5
⇒ k² + 9 = 25
⇒ k² = 25 – 9 = 16
∴ k = ± 4

Question 2.
Find the distance of a point P(x, y) from the origin.
Answer:
Using distance formula for distance between A (x₁, y₁) and B (x₂, y₂)
AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
∴ Reqd. distance = \(\sqrt{(x-0)^2+(y-0)^2}\)
= \(\sqrt{x^2+y^2}\)

Question 3.
Find distance between A(10 cos θ, 0) and B(0, 10 sin θ).
Answer:

Question 4.
Find the coordinates of reflection of Q(- 1, -3) in x-axis.
Answer:
Reflection of Q(- 1, – 3) is (- 1, 3)

Question 5.
Find the coordinates of the point on y-axis which is nearest to the point (- 2, 5). [CBSE Sample Paper 2017]
Answer:
To get the coordinates of the point on y-axis which is nearest to the point P(- 2, 5) drop a perpendicular PM from on y-axis.

Coordinates of M are (0, 5) which is the nearest point to P(-2, 5).

Question 6.
Find the fourth vertex of parallelogram ABCD whose three vertices are A(- 2, 3), B(6, 7) and C(8, 3).
Answer:
Diagonals of parallelogram bisect each other, mid-point of AC = mid-point of BD

Question 7.
If the point (0, 2) is equidistant from the points (3, k) and (k, 5), find the value of k.
Answer:
Let the points be P(0, 2), A(3, k) and B(k, 5).
Now, PA = PB
or, PA² = PB²
⇒ (3 – 0)² + (k – 2)² = (k – 0)² + (5 – 2)²
⇒ 9 + (k – 2)² = k² + 9
⇒ k² = (k – 2)²
⇒ k = ± (k – 2)
k = (k – 2) (impossible)
∴ k = – (k – 2) = – k + 2
or 2k = 2 or k = 1.

Coordinate Geometry Class 10 Extra Questions Short Answer Type-1

Question 1.
Find the value of x for which the distance between the point P (2, – 3) and Q (x, 5) is 10 unit.
Answer:
According as given PQ = 10 units
⇒ \(\sqrt{(2-x)^2+(-3-5)^2}\) = 10
Squaring both sides, we have
4 + x² – 4x + 64 = 100
⇒ x² – 4x – 32 = 0
⇒ x² – 8x + 4x – 32 = 0
⇒ x (x – 8) + 4 (x – 8) = 0
⇒ (x – 8) (x + 4) = 0
⇒ x = 8 or x = – 4.

Question 2.
The coordinates of the vertices of AABC are A (4,1), B (- 3, 2) and C (0, k). Given that the area of ∆ABC is 12 unit², find value of k.
Answer:
Area of ∆ABC = \(\frac{1}{2}\) {4 × 2 + (- 3) × k + 0 × (1)} – {( – 3) × (1) + (0) × 2 + 4 (k)}

⇒ 12 = \(\frac{1}{2}\) |{ 8 – 3k} – { – 3 + 4k}|
= \(\frac{1}{2}\) |11 – 7k|
⇒ |11 – 7k | = 24
⇒ 11 – 7k = ± 24
⇒11 – 7k = 24 or 11 – 7k = – 24
⇒ 7k = – 13 or 7k = 35
k = \(\frac{-13}{7}\) or 5.
[∵ | x | = l ⇒ x = ± 1]

Question 3.
Find the ratio in which y-axis divides the line segment joining the points A(5, -6) and B (-1, -4). Also find the coordinates of the point of division. [CBSE Delhi 2016]
Answer:
Let P be a point on the y-axis dividing the line segment AB in the ratio k : 1 using the section formula. We get

Thus, the y-axis divides the line segment in the ratio 5:1
Also, the coordinates of the point of division are \(\left(0,-\frac{13}{3}\right)\)

Question 4.
Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q. [CBSE Outside Delhi 2016]
Answer:

Question 5.
For what value of k are the points (1, 1), (3, k) and (- 1, 4) collinear?
Answer:
Let the coordinates of the vertices of ∆ABC be A
(1, 1), B (3, k), C(-1, 4).

Since, the points are collinear.
∴ Area of ∆ABC = 0
⇒ \(\frac{1}{2}\)|{k × 1 + 3 × 4 + (- 1) × (1)} – {3 × 1 + (- 1) × k + 1 × 4} | = 0
⇒ \(\frac{1}{2}\)|(11 + k) – (7 – k)| = 0
⇒ |4 + 2k| = 0
⇒ 4 + 2k = 0
⇒ k = – 2.

Question 6.
Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3). Hence find m. [CBSE 2018]
Answer:
Let P divides AB in the ratio k : 1
By section formula

⇒ \(\frac{6 k+2}{k+1}\) = 4
⇒ 6k + 2 = 4k + 4
⇒ 2k = 2
⇒ k = 1
∴ Ratio is 1 : 1
Hence m = \(\frac{-3 k+3}{k+1}=\frac{-3(1)+3}{1+1}\) = 0
⇒ m = 0

Question 7.
Point A(- 1, y) and B(5, 7) lie on a circle with centre 0(2, -3y). Find the values of y. Hence find the radius of the circle.

Answer:
By mid-point formula
⇒ \(\frac{y+7}{2}\) = – 3y
Now, A(- 1, -1) and 0(2, 3)
∴ Radius = AO = \(\sqrt{(2+1)^2+(3+1)^2}\) = 5 units

Question 8.
The x-coordinate of a point P is twice its ^-coordinate. If P is equidistant from Q(2, – 5) and R(-3,6), find the coordinates of P. [CBSE Delhi 2016]
Answer:
Let point P be (2 a, a)
PQ = PR (given)

Squaring both sides, we get
5a² + 2a + 29 = 5a² + 45
⇒ 5a² + 2a – 5a² = 45 – 29
⇒ 2a = 16
⇒ a = 8
Thus, the coordinates of the point P are (16, 8)

Question 9.
Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle. [CBSE Outside Delhi 2016]
Answer:
Let A, B and C be the points (3, 0), (6, 4) and (-1, 3), respectively.
Using Distance formula:

AB = AC = 5
⇒ AABC is isosceles triangle.
Also, (5)² + (5)² = 50 = (5√2)²
⇒ AC² + AB² = BC²
⇒ ∆ABC is right angled at A.

Question 10.
If (- 2, 3), (4, – 3) and (4, 5) are the mid-points of the sides of a triangle, find the coordinates of the centroid.
Answer:
Let the given triangle be ABC and D, E, F are mid-points of sides AB, BC, CA respectively, where D (- 2,3), E (4, – 3), F (4,5). Since centroid of a triangle is same as centroid of triangle obtained by joining mid-points of sides of main triangle.

Question 11.
If \(\left(1, \frac{p}{3}\right)\) is the mid point of the line segment joining the points (2, 0) and \(\left(0, \frac{2}{9}\right)\), then show that the line 5x + 3y + 2 = 0 passes through the point (- 1, 3p).
Answer:
Using Mid point-formula

∴ (- 1, 3p) = (- 1, 3 × \frac{1}{3}) (- 1, 1)
Plug in x = – 1, y = 1 in 5x + 3y + 2 = 0
5(- 1) + 3(1) + 2 = 0
– 5 + 5 = 0 (satisfied)
⇒ (- 1, 3p) i.e. (- 1, 1) lies on line 5x + 3y + 2 = 0

Coordinate Geometry Class 10 Extra Questions Short Answer Type-2

Question 1.
Two friends Seema and Aditya work in the same office at Delhi. In the Christmas vacations, both decided to go to their hometowns represented by Town A and Town B respectively in the figure given below. Town A and Town B are connected by trains from the same station C (in the given figure) in Delhi. Based on the given situation, answer the following questions.

(i) Who will travel more distance, Seema or Aditya, to reach to their hometown?
(ii) Seema and Aditya planned to meet at a location D situated at a point D represented by the mid-point of the line joining the points represented by Town A and Town B. Find the coordinates of the point represented by the point D.
(iii) Find the area of the triangle formed by joining the points represented by A, B and C. [CBSE SQP 2019 (Standard)]
Answer:
(i) Reading the coordinates of A, B, C as given below A(1, 7), B(4,2), C(- 4, 4).
Distance travelled by Seema = AB
= \(\sqrt{(1-4)^2+(7-2)^2}\) = \(\sqrt{9+25}\) = √34
Distance travelled by Aditya
= BC = \(\sqrt{(4+4)^2+(2-4)^2}\) = \(\sqrt{64+4}\) = √68 CE
= √68 > √34
∴ Aditya travels more distance.

(ii) Using mid-point formula,
The coordinates of D are
= \(\left(\frac{1+4}{2}, \frac{7+2}{2}\right)\) = \(\left(\frac{5}{2}, \frac{9}{2}\right)\)

(iii) Area of ∆ ABC = \(\frac{1}{2}\) |1(2 – 4) + 4(4 – 7) – 4(7 – 2)|
= \(\frac{1}{2}\) |- 2 – 12 – 20| = \(\frac{1}{2}\) |- 34|
= 17 sq. units

Question 2.
Find the point on y-axis which is equidistant from the points (5, -2) and (-3, 2).
Answer:
Let die required point on y-axis be P(0, b) and given points be A(5, – 2) and B(- 3,2)

According to question,
PA = PB
⇒ PA² = PB²
⇒ (5 – 0)² + (- 2 – b)² = (- 3 – 0)² + (2 – b)²
⇒ 29 + 4b + b² = 13 + b² – 4b
⇒ b = – 2
∴ Required point is (0, – 2)

Question 3.
The line segment joining the points A(2, 1) and B(5, -8) is trisected by the points P and Q, where P is nearer to A. If the point P also lies on the line 2r -y+A=0, find the value of k.
Or
Show that (a, a), (- a, – a) and (-√3a, √3a) are vertices of an equilateral triangle. [CBSE 2019(C)]
Answer:
Since, P lies nearer to A and is one of the point of trisection.

∴ P divides AB in the ratio 1:2 So, by section formula
P\(\left(\frac{1 \times 5+2 \times 2}{1+2}, \frac{1 \times(-8)+2 \times 1}{1+2}\right)\)
= P\(\left(\frac{9}{3}, \frac{-6}{3}\right)\) = P(3, – 2)
Since, P lies on 2x – y + k = 0
∴ Coordinates of P must satisfy.
⇒ 2(3) – (-2) + k = 0
⇒ 10 + k = 0
⇒ k = – 10.
Or
Let given points be A (a, a) B(-a, -a) and C(-√3a, √3a)
Now using Distance Formula,

Since, AB = BC = CA(= 2√2 a each)
∴ ∆ ABC is an equilateral triangle.

Question 4.
Show that AABC, where A(-2, 0), B(2, 0), C(0, 2) and ∆ PQR where P(- 4, 0), Q(4, 0), R(0, 4) are similar triangles. [CBSE Delhi 2017]
Answer:
Using distance formula:

i.e., sides of As are proportional
⇒ ∆ ABC ~ ∆ PQR

Question 5.
The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is \(\left(\frac{7}{2}, y\right)\), find the value of y. [CBSE 2017]
Answer:

∵ ar ∆ABC = \(\frac{1}{2}\) |x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)
∴ ar (∆ABC)
= \(\frac{1}{2}\) |2(- 2 – y) + 3(y – 1)+ \(\frac{7}{2}\)(1 + 2)| = 5
= \(\frac{1}{2}\) |- 4 – 2y + 3y – 3 + \(\frac{21}{2}\)| = 5
⇒ |y + \(\frac{7}{2}\)| = 10
⇒ y + \(\frac{7}{2}\) = ± 10 [∵ |x| = l ⇒ x = ± l]
either y = 10 – \(\frac{7}{2}\) = \(\frac{13}{2}\) = or y = – 10 – \(\frac{7}{2}\) = \(\frac{-27}{2}\)

Question 6.
If a ≠ b ≠ 0, prove that the point (a, a²), (b, b²) (0, 0) will not be collinear. [CBSE 2017]
Answer:
Let A(a, a²), B(b, b²), O(0, 0) be given points
⇒ ar ∆ABO = \(\frac{1}{2}\)|x₁ (y₂ – y₃) + x₂(y₃ – y₁) +x₃ (y₁ – y₂)|
⇒ = \(\frac{1}{2}\)|a(b²) + b(- a²) + 0|
= \(\frac{1}{2}\) ab(b – a) ≠ 0 as a ≠ b ≠ 0
∴ Given points are not collinear

Question 7.
In figure, ABC is a triangle coordinates of whose vertex A are (0, -1). D and E respectively are the mid-points of the sides AB and AC and their coordinates are (1, 0) and (0, 1) respectively. If F is the mid-point of BC, find the areas of ∆ABC and ∆DEF. [CBSE 2016]

Answer:
Let the coordinates of B and C be (x₂, y₂) and (x₃, y₃) respectively. D is the midpoint of AB. So,

∴ The coordinates of B are (2, 1).
Similarly, E is the midpoint of AC.
So, (0, 1) = \(\left(\frac{x_3+0}{2}, \frac{y_3-1}{2}\right)\)
⇒ 0 = \(\frac{x_3}{2}\) and 1 = \(\frac{y_3-1}{2}\)
⇒ x₃ = 0 and y₃ = 3
∴ The coordinates of C are (0, 3).
Also, F is the midpoint of BC. So, coordinates of F are
\(\left(\frac{2+0}{2}, \frac{1+3}{2}\right)\) = (1, 2)
Now, Area of a triangle
= \(\frac{1}{2}\) |x₁ (y₂ – y₃) + x₂ (y₃ – y₂) + x₃ (y₁ – y₂)|
Thus, the area of ∆ABC is
= \(\frac{1}{2}\) |0(1 – 3) + 2(3 +1) + 0(- 1 – 1)|
= \(\frac{1}{2}\) × 8 = 4 square units
And the area of ∆DEF is
= \(\frac{1}{2}\) |1(1 – 2) + 0(2 – 0) + 1(0 – 1)|
= \(\frac{1}{2}\) |(- 2)| = 1 square unit

Question 8.
If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b). Prove that bx = ay. [CBSE Outside Delhi 2016]
Answer:
Answer:

Question 9.
The vertices of a triangle are (- 2, 0), (2, 3) and (1, – 3). Is the triangle equilateral, isosceles or scalene?
Answer:

Clearly 5 ≠ √37 ≠ 3√2
i.e., AB ≠ BC ≠ CA
⇒ ∆ABC is a scalene triangle.

Question 10.
Find the area of a rhombus if its vertices are (3, 0), (4, 5), ( – 1, 4) and ( – 2, – 1) taken in order.
Answer:
Let ABCD be given rhombus with A (3, 0), B (4, 5), C ( – 1, 4) and D (- 2, – 1).

Thus, AC and BD are its diagonals.
Using distance formula:

Since, area of rhombus = \(\frac{1}{2}\) (Product of its diagonals)
= \(\frac{1}{2}\) (4√2) (6√2) = 24 sq. units

Question 11.
Given two fixed points P(- 3, 4) and Q (5, – 2).; Find the coordinates of points A and B in PQ; such that 5 PA = 3 PQ and 3PB = 2 PQ.

Answer:

Question 12.
Find the coordinates of the points of trisection of the line segment joining the points A (2, – 2) and B (- 7, 4).
Answer:
Let P and Q be the points of trisection of AB.
∴ P divides AB in the ratio 1 : 2 and Q divides AB in the ratio 2 : 1 internally.

Question 13.
If A(-2, 1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides. [CBSE 2018]
Answer:
ABCD is a parallelogram
∴ diagonals AC and BD bisect each other

Therefore
Mid point of BD is same as mid point of AC.

∴ AB = CD = AD = BC = √10 units.
Therefore length of sides are √10 units each.

Question 14.
If A(- 5, 7), B(-4, -5), C(-1, -6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD. [CBSE 2018]
Answer:
Area of quad ABCD = Ar ∆ABD + Ar ∆BCD

Area of ∆ABD
= \(\frac{1}{2}\)|(- 5) (-5 -5) + (- 4) (5 – 7) + (4) (7 + 5)|
= \(\frac{1}{2}\) [50 + 8 + 48]
= \(\frac{1}{2}\) × 106
= 53 sq. units
Area of ∆BCD
= \(\frac{1}{2}\) |(-4) (-6 – 5) + (-1)(5 + 5) + (4)(-5 + 6) |
= \(\frac{1}{2}\) [44 – 10 + 4]
= \(\frac{1}{2}\) × 38
= 19 sq. units
Hence area of quad. ABCD = 53 + 19 = 72 sq. units.

Question 15.
In figure, the vertices of ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line-segment DE is drawn to intersect the sides AB and AC at D and E respectively such that \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{3}\). Calculate the area of ∆ADE and compare it with area of ∆ABC. [CBSE Outside Delhi 2016]

Answer:

So, area of ∆ABC is more than that of ∆ADB by 4.34 Sq. units.

Question 16.
Prove that the area of a triangle with vertices (t, t – 2), (t+2, t+2) and (t+3, t) is independent oft [CBSE Delhi 2016]
Answer:
Let A(t, t – 2), B(t + 2, t + 2) and C(t + 3, t) be the vertices of the given triangle.
We know that the area of the triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is
\(\frac{1}{2}\) |x₁(y₂ – y₃) + x₃ (y₃ – y₁) + x₃(y₁ – y₂)|
∴ Area of ∆ABC
= |\(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃ (y₁ – y₂)l
= |\(\frac{1}{2}\) [t(t + 2 – t) + (t + 2) (t – t + 2) + (t + 3) (t – 2 – t – 2)]|
= |\(\frac{1}{2}\) (2t + 2t + 4 – 4t – 12)
= |- 4|
= 4 square units
Hence, the area of the triangle with given vertices is independent of t.

Question 17.
Determine the ratio in which the line 3x + y -9 = 0 divides the segment joining the points (1, 3) and (2,7).
Answer:
Let line 3x + y – 9 = 0 divides the segment joining the points A (1, 3) and B (2, 7) in the ratio k : 1 at point L.
Using section formula,
Co-ordinates of L are
x = \(\frac{k \times 2+1 \times 1}{k+1}\)
x = \(\frac{2 k+1}{k+1}\)

Coordinate Geometry Class 10 Extra Questions HOTS

Question 1.
The area of triangle is 5 sq. units. Two of its vertices are (2, 1) and (3, -2). The third vertex lies on y = x + 3. Find the coordinates of the third vertex of the triangle.
Answer:
Let C(h, k) be the third vertex and A(2, 1) and B(3, – 2) are other two vertices.

∵ C(h, k) lies on line y = x + 3
⇒ ∴ (h, k) satisfy (i) i.e. k = h + 3.
∴ Coordinates of C are (h, h + 3)
Now ar (∆ABC)
= \(\frac{1}{2}\) |{2 × (- 2) – 3 × 1} + {(3 × (h + 3) – h × (- 2)} + {h × 1 – 2 × (h + 3)}
= \(\frac{1}{2}\) |(- 4 – 3) + (3ft + 9 + 2h) + (h – 2h – 6)|
= \(\frac{1}{2}\) |- 7 + 5h + 9 – h – 6| = \(\frac{1}{2}\) |4h – 4|
But ar (∆ABC) = 5 sq. units [Given]
⇒ \(\frac{1}{2}\) |4h – 4| = 5
⇒ |4h – 4| = 10
⇒ 4h – 4 = ±10
⇒ 4 h = ±10 + 4
⇒ 4h = 14, – 6

Question 2.
The vertices of a AABC are A(4, 6), B{1, 5), C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{4}\) Calculate the area of the ∆ADE and compare it with the area of ∆ABC.
Answer:

Question 3.
The coordinates of mid-point of the line joining the points (3p, 4) and (- 2, 2q) are (5, p). Find the values of p and q.
Answer:
Let A (3p, 4) and B (- 2, 2q) be two given points.
Coordinates of mid-point of AB are
\(\left(\frac{3 p-2}{2}, \frac{2 q+4}{2}\right)\)
According to question,

Question 4.
The line-segment joining the points (3, -4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, -2), \(\left(\frac{5}{3}, q\right)\), respectively, then find the values of p and q.
Answer:
Case I: When P divides AB in the ratio 1 : 2 then coordinates of P are

Multiple Choice Questions

Choose the correct option out of four given in each of the following:

Question 1.
The ratio of the distances of point P(3, 4) from origin to that from y-axis is
(a) 3:5
(b) 5:3
(c) 5:4
(d) 3:4
Answer:
(c) 5:4

Question 2.
AOBC is a rectangle whose three vertices are A(0, 4), O(0, 0) and B(3, 0). The length of its diagonal is
(a) 25 units
(b) 5 units
(c) 3 units
(d) 4 units
Answer:
(b) 5 units

Question 3.
The distance between A(10 cos θ, 0) and B(0, 10 sin θ) is
(a) √10 units
(b) 10 units
(c) 5 units
(d) 10 sin θ cos θ
Answer:
(b) 10 units

Question 4.
Perimeter of the triangle formed by the points O(0, 0), A(A, 0) and B(0, b) is:
(a) a + b
(b) ab
(c) a + b + 2√ab
(d) a + b + \(\sqrt{a^2+b^2}\)
Answer:
(d) a + b + \(\sqrt{a^2+b^2}\)

Question 5.
The points (-8, 0), (8, 0), (0, 5) are the vertices of a
(a) right triangle
(b) isosceles triangle
(c) equilateral triangle
(d) scalene triangle
Answer:
(b) isosceles triangle

Question 6.
The points (-2, 2), (8, -2) and (-4, -3) are the vertices of a
(a) equilateral triangle
(b) isosceles triangle
(c) right triangle
(d) none of these
Answer:
(c) right triangle

Question 7.
The points (1, 7), (4, 2), (-1, 1) and (-4, 4) are the vertices of a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) square
Answer:
(d) square

Question 8.
The line segment joining the points (2, -3) and (5, 6) is divided by x-axis in the ratio
(a) 2:1
(b) 3:1
(c) 1:2
(d) 1:3
Answer:
(a) 2:1

Question 9.
The line segment joining the points (3, 5) and (-4,2) is divided by y-axis in the ratio
(a) 5 : 3
(b) 3 : 5
(c) 4 : 3
(d) 3 : 4
Answer:
(d) 3 : 4

Question 10.
If (3, 2), (4, k) and (5, 3) are collinear then k is equal to
(a) \(\frac{3}{2}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{5}{2}\)
(d) \(\frac{3}{5}\)
Answer:
(c) \(\frac{5}{2}\)

Question 11.
If the points (p, 0), (0, q) and (1, 1) are collinear then \(\frac{1}{p}+\frac{1}{q}\) is equal to
(a) – 1
(b) 1
(c) 2
(d) 0
Answer:
(b) 1

Question 12.
The coordinates of reflection of Q(- 1, -3) in x- axis are
(a) (1, 3)
(b) (-1,3)
(c) (1, -3)
(d) none of these
Answer:
(b) (-1,3)

Question 13.
The distance between the points (-3, 0) and (3, 0) is
(a) 3 units
(b) 3√2 units
(c) 2√3 units
(d) 6 units
Answer:
(d) 6 units

Question 14.
If P\(\left(\frac{a}{3}, 4\right)\) is the mid point of the line segment joining the points A(-6, 5) and B(-2, 3), then value of a is
(a) – 4
(b) – 12
(c) 12
(d) – 6
Answer:
(b) – 12

Question 15.
The fourth vertex D of a parallelogram ABCD whose three vertices are A(-2, 3), B(6, 7) and C(8, 3) is:
(a) (0, 1)
(b) (0, -1)
(c) (1, 0)
(d) (-1, 0)
Answer:
(b) (0, -1)

Fill in the Blanks

Question 1.
Distance of point P(a, b) from origin is __________ .
Answer:
\(\sqrt{a^2+b^2}\)

Question 2.
Coordinates of mid point joining P(α, β) and Q(γ, δ) are __________.
Answer:
\(\left(\frac{\alpha+\gamma}{2}, \frac{\beta+\delta}{2}\right)\)

Question 3.
For given three points A, B, C if out of three possible distances AB, BC and CA the length of the greatest distance is equal to sum of other two distances then the points A, B, C are said to be __________ .
Answer:
collinear

Question 4.
For given four points A, B, C, D if lengths AB, BC, CD and DA are all equal then ABCD is necessarily a __________ .
Answer:
rhombus

Question 5.
For given four points A, B, C, D if lengths AB = BC = CD = DA and AC ≠ BD then ABCD is a __________ but not __________ .
Answer:
rhombus, square

Question 6.
For proving ABCD to be a __________ . It is sufficient to show that opposite sides are equal.
Answer:
parallelogram

Question 7.
For given four points A, B, C, D if AB = CD, BC = DA and AC ≠ BD then ABCD is a __________ but not __________ .
Answer:
parallelogram, rectangle

Question 8.
If area of a triangle is zero square units then its vertices are __________ .
Answer:
collinear

Question 9.
The distance between (α, β) and (- α, – β) is __________ .
Answer:
2\(\sqrt{\alpha^2+\beta^2}\)

Question 10.
If the point P(x, y) divides the line segment joining A(x₁, y₁) and B(x₂, y₂) in the ratio m: n the value of y – x = __________ .
Answer:
\(\frac{m\left(y_2-x_2\right)+n\left(y_1+x_1\right)}{m+n}\)

Extra Questions for Class 10 Maths

NCERT Solutions for Class 10 Maths

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