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
3 Mark Questions
1. If 
and 
are the vertices of a isosceles triangle, show that 
if AB = BC.
Ans. 
(Given)
2. Find the value of P if the point 
is equidistant from 
and 
Ans. Let 
and 
AB = AC (Given)
3. Find the centroid of the triangle whose vertices are 
and 
Ans. Let 
be the coordinate of centroid
Coordinate of centroid is 
4. Prove that in a right-angled triangle, the mid-point of the hypotenuse is equidistant from the vertices.
Ans. Let 
and
are the vertices of right-angled triangle
Coordinate of 
i.e. (a, b)
Hence, C is Equidistant from the vertices.
5. Prove that diagonals of a rectangle bisect each other and are equal.
Ans. Let ABCD be a rectangle take A as origin the vertices of a rectangle are 
Mid-point of AC = 
Mid-point of BD =
Mid-point of AC = Mid-point of BC
Hence proved.
6. The line joining the points 
and 
is bisected at P. If P lies on the line 
find the value of k.
Ans. Coordinate of 
P lies on equation 
7. Show that the points 
and 
are collinear.
Ans. For collinear
8. The length of a line segment is 10. If one end point is 
and the abscissa of the second end point is 10, show that its ordinate is either 3 or -9.
Ans. Let A (2, -3) be the first end point and B (10, y) be the second end point.
9. Using section formula, show that the points 
and 
are collinear.
Ans. If points 
and are collinear, then one point divides the join of other two in the same ratio. Let 
divides the join of 
and 
in the ratio K:1
and 
and 
Hence Proved.
10. Find the relation between 
and y such that the point is equidistant from the points 
and 
Ans. Let 
be equidistant from the points 
and 
(Given)
11. Determine the ratio in which the line 
divides the line segment joining the points 
and 
Ans. Let the ratio be K: 1
Coordinate of P are 
P lies on the line
12. Show that the points 
and 
are the vertices of a square.
Ans. 
Diagonal 
Diagonal 
Hence proved.
13. If the point is equidistant from the points 
and 
, prove that 
.
Ans. 
(Given)
14. Find the point on the x–axis which is equidistant from (2, –5) and (–2, 9).
Ans. 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:
⇒ 
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)
15. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Ans. We want to find coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
We are given AC = CD = DB
We want to find coordinates of point C and D.
Let coordinates of point C be (x1, y1) and let coordinates of point D be (x2, y2).
Clearly, point C divides line segment AB in 1:2 and point D divides line segment AB in 2:1.
Using Section Formula to find coordinates of point C which divides join of (4, –1) and (–2, –3) in the ratio 1:2, we get
Using Section Formula to find coordinates of point D which divides join of (4, –1) and (–2, –3) in the ratio 2:1, we get
Therefore, coordinates of point C are (2, 
) and coordinates of point D are (0, 
).
16. 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. Niharika runs 14th of the distance AD on the 2nd line and posts a green flag. Preet runs 15th of 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?
Ans. Niharika runs 14th of the distance AD on the 2nd line and posts a green flag.
There are 100 flower pots. It means, she stops at 25th flower pot.
Therefore, the coordinates of point where she stops are (2 m, 25 m).
Preet runs 15th of the distance AD on the eighth line and posts a red flag. There are 100 flower pots. It means, she stops at 20th flower pot.
Therefore, the coordinates of point where she stops are (8, 20).
Using Distance Formula to find distance between points (2 m, 25 m) and (8 m, 20 m), we get
Rashmi posts a blue flag exactly halfway the line segment joining the two flags.
Using section formula to find the coordinates of this point, we get
Therefore, coordinates of point, where Rashmi posts her flag are (5, 
).
It means she posts her flag in 5th line after covering = 22.5 m of distance.
17. If A and B are (–2, –2) and (2, –4) respectively, find the coordinates of P such that AP = 
AB and P lies on the line segment AB.
Ans. A = (–2, –2) and B=(2, –4)
It is given that AP=AB
PB=AB – AP=AB−AB=
AB
So, we have AP:PB = 3:4
Let coordinates of P be (x, y)
Using Section formula to find coordinates of P, we get
Therefore, Coordinates of point P are
.
18. 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)
Ans. (i) (7, –2), (5, 1), (3, k)
Since, the given points are collinear, it means the area of triangle formed by them is equal to zero.
Area of Triangle = 
⇒ ½ [7(1−k)+5 {k−(−2)} +3(−2−1)] = ½ (7−7k+5k+10−9)=0
⇒ ½ (7−7k+5k+1)=0
⇒ ½ (8−2k)=0
⇒ 8−2k=0s
⇒ 2k=8
⇒ k=4
(ii) (8, 1), (k, –4), (2, –5)
Since, the given points are collinear, it means the area of triangle formed by them is equal to zero.
Area of Triangle = 
⇒ ½ [8 {−4−(−5)} +k(−5−1)+2 {1−(−4)}] = ½ (8−6k+10)=0
⇒ ½ (18−6k)=0
⇒ 18−6k=0
⇒ 18=6k
⇒ k=3
19. 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
Ans. Let A = (0, –1) = (x1, y1), B = (2, 1) = (x2, y2) and
C = (0, 3) = (x3, y3)
Area of △ABC = 
⇒ Area of △ABC
= ½ [0(1−3)+2 {3−(−1)} +0(−1−1)] = ½ ×8
=4 sq. units
P, Q and R are the mid–points of sides AB, AC and BC respectively.
Applying Section Formula to find the vertices of P, Q and R, we get
Applying same formula, Area of △PQR= ½ [1(1−2)+0(2−0)+1(0−1)] = ½ 
=1 sq. units (numerically)
Now, 
20. Find the area of the quadrilateral whose vertices taken in order are (–4, –2), (–3, –5), (3, –2) and (2, 3).
Ans. Area of Quadrilateral ABCD
= Area of Triangle ABD + Area of Triangle BCD … (1)
Using formula to find area of triangle:
Area of 
ABD
=
= ½ [−4(−5−3) – 3 {3−(−2)} +2 {−2−(−5)}]
= ½ (32 – 15+6)= ½ (23)=11.5 sq units … (2)
Again using formula to find area of triangle:
Area of △BCD =
= ½ [−3(−2−3)+3 {3−(−5)} +2 {−5−(−2)}]
= ½ (15+24−6)= ½ (33)=16.5 sq units … (3)
Putting (2) and (3) in (1), we get
Area of Quadrilateral ABCD = 11.5 + 16.5 = 28 sq units.
21. We know that 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).
Ans. We have △ABC whose vertices are given.
We need to show thatar(△ABD) = ar(△ACD).
Let coordinates of point D are (x, y)
Using section formula to find coordinates of D, we get
Therefore, coordinates of point D are (4, 0)
Using formula to find area of triangle:
Area of △ABD =
= ½ [4(−2−0)+3 {0−(−6)} +4 {−6−(−2)}]
= ½ (−8+18−16)= ½ (−6)=−3 sq units
Area cannot be in negative.
Therefore, we just consider its numerical value.
Therefore, area of △ABD = 3 sq units … (1)
Again using formula to find area of triangle:
Area of △ACD =
= ½ [4(2−0)+5 {0−(−6)} +4 {−6−2)}]
= ½ (8+30−32)= ½ (6)=3 sq units … (2)
From (1) and (2), we getar(△ABD) = ar(△ACD)
Hence Proved.
22. Find the centre of a circle passing through the points 
and 
Ans. Let P
be the centre of the circle passing through the points A
B
and C (3, 3). Then AP = BP = CP.
Taking AP = BP
AP2 = BP2
……….(i)
Again, taking BP = CP
BP2 = CP2
Putting the value of 
in eq. (i),
Hence, the centre of the circle is 
23. The vertices of aABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively such that 
Calculate the area of the ADE and compare it with the area of ABC.
Ans. Since, 
DE 
BC [By Thales theorem]
ADE 
ABC
……….(i)
Now, Area (ABC) = 
= 
sq. units ……….(ii)
From eq. (i) and (ii),
Area (ADE) = 
Area (ABC) = 
sq. units
Area (ADE) : Area (ABC) = 1 : 16