## CBSE Class 10 Maths Chapter-7 Coordinate Geometry – Free PDF Download

Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 7 – Coordinate Geometry prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.

CBSE Class 10 Maths Chapter-7 Coordinate Geometry Important Questions

**CBSE Class 10 Maths Important Questions Chapter 7 – Coordinate Geometry**

**4 Mark Questions**

**1. If the points is equidistant from the points and prove that **

**Ans.** Let

**2. andare three concyclic points whose centre is . Find the possible value of and y.**

**Ans.** Radius of circle

**3. Find the vertices of the triangle, the mid-points of whose sides are and **

**Ans.** Let vertices of be and

By mid-points formula

Adding (i), (iii) and (v)

Adding (ii), (iv)and (vi)

Subtracting (i), (iii) and (v) from (vii)

We get,

Subtracting (ii), (iv) and (vi) from eq. (viii)

We get,

**4. The two opposite vertices of a square are and. Find the coordinates of the other two vertices.**

**Ans. **

In right

Put the value of y in eq. (i)

Now

And

**5. Find the coordinates of the circumcentre of a triangle whose vertices are A(4,6), B(0,4) and C(6,2). Also find its circum-radius.**

**Ans.** Let P be the circum-centre of then PA = PB = PC

On solving equations (i) and (ii),

Circum-radius (PA) =

**6. If two vertices of an equilateral triangle are Find the third vertex.**

**Ans.**

Put the value of y in eq. (i),

**7. If P and Q are two points whose coordinates are and respectively and S is the point show that is independent of .**

**Ans. **

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

**Ans. **

= 10.5 sq. units

area of quadrilateral = 10.5 + 17.5 = 28 sq. units.

**9. The vertices of are and . A line is drawn to intersect sides AB and AC at D and E respectively such that . Calculate the area of the and compare it with the area of .**

**Ans. **

Now coordinate of D and E are

and

**10. Prove that the points and are the vertices of an equilateraltriangle. Calculate the area of this triangle.**

**Ans.** Let

**11. are the vertices of a is the mid-point of BC and P is a point on AD joined such that find the coordinates of P.**

**Ans.** Let and are the vertices of is the mid- point of BC

Coordinate of D

i.e.,

Coordinate of P are

**12. The coordinates of the vertices of are. Given that the area of is 12, find the value of K.**

**Ans. **

Value of

**13. Find the lengths of the medians of the triangle whose vertices are and .**

**Ans.** Coordinates of points D, E and F are

Length of the median AD

Length of the median BE

And length of the median CF

**14. The area of a triangle is 5. Two of its vertices are and. The third vertex lies on. Find the third vertex.**

**Ans.** Let the third vertex be. Other two vertices of the are and

lies on eq.

On solving eq.

We get

Similarly, on solving eq.

We get

**15. Prove that the point and are collinear, if **

**Ans.** Since are collinear

Area = 0

Dividing by ,

**16. 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.**

**Ans.** It is given that Q is equidistant from P and R. UsingDistance Formula, we get

PQ = RQ

PQ^{2} = RQ^{2}

⇒

⇒

Squaring both sides, we get

⇒ 25+16=x^{2}+25

⇒ x^{2}=16

⇒ x=4,−4

Thus, Q is (4, 6) or (–4, 6).

Using Distance Formula to find QR, we get

Using value of x = 4

QR=

Using value of x = –4

QR=

Therefore, QR=

Using Distance Formula to find PR, we get

Using value of x = 4

PR=

Using value of x =–4

PR=

Therefore, x = 4, –4

QR=, PR=

**17. Find the coordinates of the points which divides the line segment joining A(–2, 2) and B(2, 8) into four equal parts.**

**Ans.** A = (–2, 2) and B = (2, 8)

Let P, Q and R are the points which divide line segment AB into 4 equal parts.

Let coordinates of point P =(x_{1}, y_{1}), Q =(x_{2}, y_{2}) and R =(x_{3}, y_{3})

We know AP = PQ = QR = RS.

It means, point P divides line segment AB in 1:3.

Using Section formula to find coordinates of point P, we get

Since, AP = PQ = QR = RS.

It means, point Q is the mid-point of AB.

Using Section formula to find coordinates of point Q, we get

Because, AP = PQ = QR = RS.

It means, point R divides line segment AB in 3:1

Using Section formula to find coordinates of point P, we get

Therefore, P=(–1, ), Q= (0, )and R =(1, )

**18. The two opposite vertices of a square are and Find the coordinates of the other two vertices.**

**Ans.** Let ABCD be a square and B be the unknown vertex.

AB = BC

AB^{2} = BC^{2}

……….(i)

In ABC, AB^{2} + BC^{2} = AC^{2}

……….(ii)

Putting the value of in eq. (ii),

= 0 or 4

**19. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmoharare 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 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 PQR if C is the origin? Also calculate the area of the triangle in these cases. What do you observe?**

**Ans. (i)** Taking A as the origin, AD and AB as the coordinate axes. Clearly, the points P, Q and

R are (4, 6), (3, 2) and (6, 5) respectively.

**(ii)** Taking C as the origin, CB and CD as the coordinate axes. Clearly, the points P, Q and R are given by (12, 2), (13, 6) and (10, 3) respectively.

We know that the area of the triangle =

Area of PQR (First case) =

=

= = sq. units

And Area of PQR (Second case) =

=

= = sq. units

Hence, the areas are same in both the cases.

**20. ABCD is a rectangle formed by joining points A B C and D P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? Or a rhombus? Justify your answer.**

**Ans.** Using distance formula, PQ =

= =

QR =

= =

RS =

= =

SP =

= =

PQ = QR = RS = SP

Now, PR = = = 6

And SQ = = = 5

PR SQ

Since all the sides are equal but the diagonals are not equal.

PQRS is a rhombus.

**21. 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.**

**Ans.** 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 =

BC =

CD =

DA =

Therefore, All the sides of ABCD are equal here. … (1)

Now, we will check the length of its diagonals.

AC =

BD =

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.

**22. 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)**

**Ans. (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 =

BC=

CD=

DA=

Therefore, all four sides of quadrilateral are equal. … (1)

Now, we will check the length of diagonals.

AC=

BD=

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=

BC=

CD=

DA=

We cannot find any relation between the lengths of different sides.

Therefore, we cannot give any name to the quadrilateral 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=

BC=

CD=

DA=

Here opposite sides of quadrilateral ABCD are equal. … (1)

We can now find out the lengths of diagonals.

AC=

BD=

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.

**23. Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ABC.**

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

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

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

**(iv) What do you observe?**

**(Note: The point which is common to all the three medians is called centroid and this point divides each median in the ratio 2 : 1)**

**(v) If A B and C are the vertices of ABC, find the coordinates of the centroid of the triangle.**

**Ans.**Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ABC.

**(i)**Since AD is the median of ABC.

D is the mid-point of BC.

Its coordinates are =

**(ii)**Since P divides AD in the ratio 2 : 1

Its coordinates are =

**(iii)**Since BE is the median of ABC.

E is the mid-point of AD.

Its coordinates are =

Since Q divides BE in the ratio 2 : 1.

Its coordinates are =

Since CF is the median of ABC.

F is the mid-point of AB.

Its coordinates are =

Since R divides CF in the ratio 2 : 1.

Its coordinates are =

**(iv)**We observe that the points P, Q and R coincide, i.e., the medians AD, BE and CF are concurrent at the point . This point is known as the centroid of the triangle.

**(v)**According to the question, D, E, and F are the mid-points of BC, CA and AB respectively.

Coordinates of D are

Coordinates of a point dividing AD in the ratio 2 : 1 are

=

The coordinates of E are .

The coordinates of a point dividing BE in the ratio 2 : 1 are

=

Similarly the coordinates of a point dividing CF in the ratio 2 : 1 are

Thus, the point is common to AD, BE and CF and divides them in the ratio 2 : 1.

The median of a triangle are concurrent and the coordinates of the centroid are .