Coordinate Geometry Class 10

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Ex7.1,1
Find the distance between the following pairs of points :
(i) (2, 3), (4, 1)
Let the two points be P(2, 3) & Q(4, 1)
We need to find distance PQ

PQ= V(x — “1? + O2 - 91)"
Here, x, = x coordinate of P = 2

y, = y coordinate of P =3

X, = x coordinate of Q=4

Y> = y coordinate of Q=1

Find the distance between the following pairs of points :
(−5, 7), (−1, 3)

Find the distance between the following pairs of points :
(a, b), (−a, −b)

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Ex 7.1, 2 CAEROOCOM
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.
Let the two points be P(0, 0) & Q(36, 15)
We need to find distance PQ
PQ= V(x, — X41)? + 2 — yi)?

Here, x, = x coordinate of P = 0

y, = y coordinate of P = 0

xX, = x coordinate of Q = 36

Y2 = y coordinate of Q=15

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Ex 7.1, 3
Determine if the points (1, 5), (2, 3) and (-2, — 11) are collinear.
Let the 3 points be A (1, 5), B (2, 3) & C (-2, -11)
Collinear points are points which fall on the same line
There are three cases possible
Case 1

—+>—___+—__ +—

A B c

A,B & C are collinear if

AB+BC =AC

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Ex 7.1, 4
Check whether (5, —2), (6, 4) and (7, -2) are the vertices of an
isosceles triangle.
In an isosceles triangle, any 2 of the 3 sides are equal.
Let the three points be P(5, -2), Q(6, 4) & R(7, -2)
In order to be isosceles,
Either P (5, -2)
PQ=PR
or
PQ=QR
or
PR=QR Q(6, 4) R(7, -2}

Ex 7.1,5 teachoo.com
Ina classroom, 4 friends are seated at the points A, B, Cand Das
shown in Fig. 7.8. 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.
As seen from the figure, four points are
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ina square ‘hier
wares ‘Se TTT 64)
¢ All sides are equal 2 A+ es
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Columns

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Ex 7.1, 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) D{-3, 0) c(-1, 2)
Let the points be

A(-1, -2), B(1, 0),

C(-1, 2), D(-3, 0)

A(-1, -2) B(1, 0)

We find the distances AB, BC, CD & AD

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(ii) (–3, 5), (3, 1), (0, 3), (–1, – 4)

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(iii) (4, 5), (7, 6), (4, 3), (1, 2)

EX7.1,7 teachoo.com

Find the point on the x-axis which is equidistant from

(2, -5) and (-2, 9).

Let the given points be

P (2,-5) , Q(-2, 9)

And the point required be

R (a, 0) Note: Since the point is on the x-axis, y = 0
And assuming x =a
Hence the point is (a, 0)

As per question,

point R is equidistant from P& Q

Hence, PR=QR

Ex 7.1, 8 feachoo.com
Find the values of y for which the distance between the points
P (2, —3) and Q(10, y) is 10 units.
Let the points be
Given that PQ = 10 units
By distance formula
PQ= V(x, —%)*+ 2 - yy)’
X,=2, y,=-3
x,=10, y. =y
PQ=/(10 —2)*+ (y-(-3))?

Ex 7.1, 9 teachoo.com
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.
Since Q is equidistant from P & R
QP=QR

Finding QP
x,=0, y,=1
%=5, y,=-3
OP= VO) — m+ Or — VM

=V¥(5 —0)?+(-3 —1)?

= (6) + (4

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EX 7.1, 10 Acnoo-com
Find a relation between x and y such that the point (x, y) is
equidistant from the point (3, 6) and (-3, 4).
Let the points be

A(x, y), B(3, 6), C{-3, 4)
According to question, point A is equidistant from B & C
Hence AB = AC
So, we will find AB & AC using distance formula

Ex 7.2,1 teachoo.com
Find the coordinates of the point which divides the join of (-1, 7)
and (4, —3) in the ratio 2 : 3.
Let the given points be A(-1, 7) & B(4, -3)
Let the point be P(x, y) which divides AB in ratio 2:3
Here,
_ ™yX2 + m4 xy
x= m, +m,
a
A P B
Where, (-1, 7) (x, y) (4, -3)
m,=2, m,=3
X,=-1, %=4

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Ex 7.2, 2 Cachoo.com
Find the coordinates of the points of trisection of the line segment
joining (4, —1) and (-—2, —3).
Given points are A(4, -1) & B(-2, -3)
1 1 1

T_T SE?

A p Q B
Let P & Q be the points of trisection
" AP = PQ = QB

Ex 7.2, 3 teachoo.com
To conduct Sports Day activities, in your rectangular shaped school
ground ABCD, lines have been drawn with chalk powder at a
distance of 1m each. 100 flower pots have been placed at a distance
of 1m from each other along AD, as shown in Fig. 7.12. Niharika
runs =i the distance AD on the 2nd line and posts a green flag.
Preet runs! 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? sg c

i
Here let us draw a simpler figure of sl

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what is given sil A

iwi Wer. FY

ol .

Ml |” RY |

wil ma

iwi BY

iwi

Al B

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Ex 7.2, 4 teachoo.com
Find the ratio in which the line segment joining the points (- 3, 10)
and (6, — 8) is divided by (-1, 6).
Let the points be
A(-3, 10) , B(6, -8) , C(-1, 6)
k 1
a
A Cc B
(-3, 10) (-1, 6) (6, 8)
We need to find ratio between AC & CB
Let the ratio be k:1
Hence,
m,=k,m,=1

EX 7.2, 5 teachoo.com
Find the ratio in which the line segment joining A(1, -5) & B (-4, 5)
is divided by the x-axis. Also find the coordinates of the point of
division.
Note: Point P is on x-axis,
hence its y coordinate is 0.
So, it is of the form P(x, 0)
Now, we have to find ratio k 1
.
Let ratio be k: 1 A P B
(1, -5) (x, 0) (-4, 5)
Hence,
m,=k, m,=1
x=1, yy=-5

Ex 7.2,6 teachoo.com
If (1, 2), (4, y), (% 6) and (3, 5) are the vertices of a parallelogram
taken in order, find x and y.

A(1, 2) B(4, y}

D(3, 5) C(x, 6)

Let the points be
A(1, 2), B(4, y), C(x, 6), D(3, 5)
We know that diagonals of parallelogram bisect each other
So, O is the mid-pint of AC & BD

Ex 7.2,7 teachoo.com
Find the coordinates of a point A, where AB is the diameter of a
circle whose centre is (2, — 3) and Bis (1, 4).
Let the circle be as shown with centre C (2, -3)
Let AB be the diameter of the circle
Since AB is the diameter,
Centre C must be the mid-point of AB (™
Let A(x, y) A B
xX, 1,4
Since C is the mid-point of AB ( er] (1,4)
+
x-coordinate of C= a
+
y-coordinate of C= a

Ex 7.2, 8 teachoo.com
If A and Bare (—2, —2) and (2, — 4), respectively, find the
coordinates of P such that AP = = AB & P lies on the line segment AB.
Let the co-ordinates of point P be P(x, y)
It is given that
tT
3 A P B
AP =7 (AB) (-2, -2) (x, y) (2, -4)
3
AP =~ (AP + PB)
TAP = 3AP + 3PB
7AP — 3AP = 3PB
AAP = 3PB
AP _3
PB 4
Hence the point P divides AB in the ratio of 3:4

Ex 7.2,9 teachoo.com
Find the coordinates of the points which divide the line segment
joining A(—2, 2} and B(2, 8) into four equal parts.
Let the points that divide AB into 4 equal
Parts be P,, P, and P;
—$__+-—___+—___+—.
A P P P B
1 2 3
We know that {-2, 2) (2, 8)
AP, = P,P, = P,P; = P3B
Assuming
AP, = P,P, = P,P; = P3B=k
Hence
AP, AP, + P,P,
PB P,P, +P,B

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Ex 7.2, 10
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4)
and (- 2, — 1) takenin order.
[Hint : Area of a rhombus = ; (product of its diagonals)]
A(3, 0) B (4,5)
Let the vertices be
A (3,0), B(4, 5)
C(-1, 4), D(-2, -1) D c
(-2, -1) (-1, 4)
We know that
Area of Rhombus = ; (Product of diagonals)
=5*ACx BD

Example 1 teachoo.com
Do the points (3, 2), (-2, -3) and (2, 3) form a triangle? If so,
name the type of triangle formed.
Let the three points be P(3, 2), Q(-2, -3) & R(2, 3)
We find the distances PQ, QR, and PR
Calculating PQ

x, =3, y, =2

%)=-2, y. =-3

PQ=¥ (2 — 14)? + On — 1)?

=/(-2 -3)?+(-3 - 2)

Example 2 teachoo.com
Show that the points (1, 7), (4, 2), (-1, -1) and (—4, 4) are the
vertices of a square.
A(1,7) B (4, 2)

Let the points be
A(-1, -2), B(1, 0), C(-1, 2), D{-3, 0)
To prove that ABCD is a square, D (-4, 4) c(-1, -1)
We have to prove all sides equal,

ie. AB=BC=CD=AD
& Diagonals equal

i.e. AC=BD
We find the distances AB, BC, CD & AD and AC & BD

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Example 3
Given figure shows the arrangement of desks in a classroom. Ashima,
Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6)
respectively. Do you think they are seated in a line? Give reasons for
your answer. “OTE
9
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6 <)
are seated in a line, then points Rows 5 Lt tT TTT ese |
ae TTT hilal |
A, B & C must be collinear ; 1} | i) @t tt
i.e. AB + BC = AC 2 Lt tt tt tt TT
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12345 678 9 10
Columns
We Find AB, BC & AC using distance formula

Example 4 teachoo.com
Find a relation between x and y such that the point (x, y) is
equidistant from the points (7, 1) and (3, 5).
Let the points be
A(x, y}, B(7, 1), €(3, 5)
Given that point A is equidistant from B & C
Hence,
AB=AC
Finding AB & AC using distance formula

Example 5 teachoo.com
Find a point on the y-axis which is equidistant from the points A(6, 5)
and B(-4, 3).
Given A(6, 5) & B(-4, 3) Y
Cc (0, a)

Since the required point is in y-axis,
its x -coordinate will be zero x x
Let Required point = C (0, a}

y’
As per question,
point C is equidistant from A & B
Hence, AC = BC
Finding AC & BC separately

Example 6 teachoo.com
Find the coordinates of the point which divides the line segment
joining the points (4, — 3) and (8, 5) in the ratio 3 : 1 internally.
Let the given points be A(4, -3) & B(8, 5)
Let the point be P(x, y) which divides AB in ratio 3: 1
3 1
— pt
A P B
(4, -3) (x y) (8, 5)

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Example 7
In what ratio does the point (— 4, 6) divide the line segment joining
the points A(—6, 10) and B(3, — 8)?
Given points A(-6, 10) & B(3, -8)
Let point C(—4, 6) k 1
es
A Cc B
(-6, 10) (-4, 6) (3, -8)
We need to find ratio between AC & CB
Let the ratio be k:1
Hence,
m,=k,m,=1

Example 8 teachoo.com
Find the coordinates of the points of trisection {i.e., points dividing
in three equal parts) of the line segment joining the points A(2, — 2)
and B(-7, 4).
Se [a Se
A Pp Q B
(2, -2) (-7, 4)
Let the given points be
A(2, -2) & B(-7, 4)
P & Qare two points on AB such that
AP =PQ=QB
Let k = AP = PQ= QB

Example 9 teachoo.com
Find the ratio in which the y-axis divides the line segment joining
the points (5, —6) and (-1, — 4). Also find the point of intersection.
Let the point be A(5, —6) & B(-1, -4)
ee
A P B
(5, -6) (0, y) (-1, -4)
Let Point P the required required point
Since Point P is on y-axis,
hence its x coordinate is 0.
So, it is of the form P(0, y)
Now, we have to find ratio
Let ratio be k: 1

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Example 10
If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of
a parallelogram, takenin order, find the value of p.
A(6, 1) B(8, 2)

Let the points be NU]

A(6,1), B(8, 2) </

C(9, 4), D(p, 3) pO N

D (p, 3} Cc (9, 4)

We know that diagonals of parallelogram bisect each other
So, Ois the mid-pint of AC & BD
«. We find x co-ordinate of O from both AC & BD

Example 11 teachoo.com
Find the area of a triangle whose vertices are (1, -1), (-4, 6) and (-
3, -5).
Let the two points be A (1, -1) , B(-4, 6} & C(-3, -5)
+ 1

Area of triangle ABC = z [ X1(¥o — V3) + Xo(¥3— Va) + X3(¥a — Vo) J
Here

x,=1, y,=-1

xX) =-4, y2=6

X,=-3, Y3=7-5
Putting values
Area of triangle ABC = = [ 1(6 - (-5)) + (-4)(-5- (-1) ) + (-3)(-1- 6) ]

Example 12 teachoo.com
Find the area of a triangle formed by the points A(5, 2), B(4, 7) and
C(7, -4).
. 1
Area of triangle ABC = > [ x:{¥2— Ys) + Xol¥3 — Ya) + Xa(¥i — Ya) ]
Here
X=5, yy, =2
%=4, y=7
%=7, Y3=-4
Putting values
Area of triangle ABC = ; [5(7 —(-4)) + 4(-4-2) + 7(2- 7) ]

Example 13 teachoo.com
Find the area of the triangle formed by the points P(—-1.5, 3),
Q(6, —2) and R(-3, 4).
: 1

Area of triangle PQR = z [ X1(¥2— Ya) + Xoly3— Va} + X3(¥1— Yo) J
Here

X,=—-1L5, y,=3

X,=6, y,=-2

xX,=-3, y,=4
Putting values
Area of triangle POR == [ (-1.5)(-2-4) + 64-3) + (-3)(3 - (-2))]

Example 14 teachoo.com
Find the value of k if the points A(2, 3), B(4, k) and C(6, —3) are
collinear.
If points A, B, C are collinear,
they will lie on the same line,
i.e. they will not form triangle
Therefore,
Area of AABC = 0
1
z [ X1{¥2 — Ya) + Xa(¥3— Va) + Xe¥1 - Yo) ] = 0

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Example 15
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.
A(-5, 7) B(- 4, —5)
Let the vertices of quadrilateral be
A(-5, 7), B(-4, -5)
C(-1, -6), D(4, 5)
D(4, 5) C(-1, —6)
Joining AC
There are 2 triangles formed ABC & ACD
Hence,
Area of quadrilateral ABCD = Area of A ABC + Area of A ADC

In order to conduct Sports Day activities in your School, lines have been drawn with chalk powder at a distance of 1 m each, in a rectangular shaped ground ABCD, 100 flowerpots have been placed at a distance of 1 m from each other along AD, as shown in given figure below. Niharika runs 1/4 th the distance AD on the 2nd line
and posts a green flag. Preet runs 1/5 th distance AD on the eighth line and posts a red flag
This question is
inspired from
Ex 7.2, 3 - Chapter 7 Class 10 - Coordinate Geometry
Question 1
Find the position of green flag
(a) (2, 25)
(b) (2, 0.25)
(c) (25, 2)
(d) (0, –25)
Question 2
Find the position of red flag
(a) (8, 0)
(b) (20, 8)
(c) (8, 20)
(d) (8, 0.2)
Question 3
What is the distance between both the flags?
(a) √41
(b) √11
(c) √61
(d) √51
Question 4
If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
(a) (5, 22.5)
(b) (10, 22)
(c) (2, 8.5)
(d) (2.5, 20)
Question 5
If Joy has to post a flag at one-fourth distance from green flag, in the line segment joining the green and red flags, then where should he post his flag?
(a) (3.5, 24)
(b) (0.5, 12.5)
(c) (2.25, 8.5)
(d) (25, 20)

The class X students school in krishnagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is 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.
This question is
inspired from
Ex 7.4, 5 (Optional) - Chapter 7 Class 10 - Coordinate Geometry
Question 18 - CBSE Class 10 - Sample Paper for 2021 Board
Question 1
Taking A as origin, find the coordinates of P
(a) (4, 6)
(b) (6, 4)
(c) (0, 6)
(d) (4, 0)
Question 2
What will be the coordinates of R, if C is the origin?
(a) (8, 6)
(b) (3, 10)
(c) (10, 3)
(d) (0, 6)
Question 3
What will be the coordinates of Q, if C is the origin?
(a) (6, 13)
(b) (–6, 13)
(c) (–13, 6)
(d) (13, 6)
Question 4
Calculate the area of the triangles if A is the origin
(a) 4.5
(b) 6
(c) 8
(d) 6.25
Question 5
Calculate the area of the triangles if C is the origin
(a) 8
(b) 5
(c) 6.25
(d) 4.5

Question Ayush Starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter's school and then reaches the office. (Assume that all distances covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in km.
Question 1 What is the distance between house and bank? (a) 5 km (b) 10 km (c) 12 km (d) 27 km
Given that
House (2, 4)
& Bank (5, 8)

Question In a room, 4 friends are seated at the points A, B, C and D as shown in figure. Reeta and Meeta walk into the room and after observing for a few minutes, Reeta asks Meeta

If the distance between the points (2, –2) and (–1, x) is 5, one of
the values of x is
(A) –2 (B) 2 (C) –1 (D) 1
Given
Distance between (2, −2) and (−1, x) = 5
√(( −1 −2)2+(𝑥−(−2))2) = 5
√(( −3)2+(𝑥+2)2) = 5
√(9+𝑥^2+2^2+4𝑥) = 5
√(9+𝑥^2+4+4𝑥) = 5
√(𝑥^2+4𝑥+13) = 5
Squaring both sides
(√(𝑥^2+4𝑥+13))^2 = 52
x2 + 4x + 13 = 25
x2 + 4x + 13 − 25 = 0
x2 + 4x − 12 = 0
Solving by splitting the middle term
x2 + 6x − 2x − 12 = 0
x(x + 6) − 2(x + 6) = 0
(x − 2) (x + 6) = 0

The mid-point of the line segment joining the points A (–2, 8) and
B (– 6, – 4) is
(A) (– 4, – 6) (B) (2, 6) (C) (– 4, 2) (D) (4, 2)

The points A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0) are the vertices of a
(A) square (B) rectangle (C) rhombus (D) trapezium
Since this is an MCQ,
we can plot these points on a graph
and check from there
Plotting A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0)

The distance of the point P (2, 3) from the x-axis is
(A) 2 (B) 3 (C) 1 (D) 5
Plotting P(2, 3) in the graph

The distance between the points A (0, 6) and B (0, –2) is
(A) 6 (B) 8 (C) 4 (D) 2
Distance between A (0, 6) and B (0, –2)
= √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2)
= √(( 0−0)2+(−2 −6)2)
= √(02+(−8)2)
= √(0+64)
= √64
= √(8^2 )
= 8 units
So, the correct answer is (B)

The distance of the point P (–6, 8) from the origin is
(A) 8 (B) 2 √7 (C) 10 (D) 6
Distance between P (−6, 8) and Origin O (0, 0)
= √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2)
= √(( 0−(−6))2+(0 −8)2)
= √(62+(−8)2)
= √(36+64)
= √100
= √(10^2 )
= 10 units
So, the correct answer is (C)

The distance between the points (0, 5) and (–5, 0) is
(A) 5 (B) 5 √2 (C) 2 √5 (D) 10
Distance between (0, 5) and (−5, 0)
= √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2)
= √(( −5−0)2+(0 −5)2)
= √((−5)2+(−5)2)
= √(25+25)
= √50
= √(2 × 25) = √(2 × 5^2 )
= 5√𝟐 units
So, the correct answer is (B)

AOBC is a rectangle whose three vertices are vertices A (0, 3), O (0, 0) and B (5, 0). The length of its diagonal is
(A) 5 (B) 3 (C) √34 (D) 4
Length of Diagonal
= Distance b/w A (0, 3) and B (5, 0)
= √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2)
= √(( 5−0)2+(0 −3)2)
= √((5)2+(−3)2)
= √(25+9)
= √𝟑𝟒
So, the correct answer is (C)

The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(A) 5 (B) 12 (C) 11 (D) 7 + √5
Let A (0, 4), B (0, 0) and C (3, 0)

The area of a triangle with vertices A (3, 0), B (7, 0) and C (8, 4) is
(A) 14 (B) 28 (C) 8 (D) 6
Since points are A (3, 0), B (7, 0) and C (8, 4)
Here
x1 = 3 , y1 = 0
x2 = 7 , y2 = 0
x3 = 8 , y3 = 4

The points (–4, 0), (4, 0), (0, 3) are the vertices of a
(A) right triangle (B) isosceles triangle
(C) equilateral triangle (D) scalene triangle
Let A (−4, 0), B (4, 0) and C (0, 3)

The point which divides the line segment joining the points (7, –6) and (3, 4) in ratio 1 : 2 internally lies in the
(A) I quadrant (B) II quadrant
(C) III quadrant (D) IV quadrant

The point which lies on the perpendicular bisector of the line segment joining the points A (–2, –5) and B (2, 5) is
(A) (0, 0) (B) (0, 2) (C) (2, 0) (D) (–2, 0)
Let Line l be the perpendicular bisector of AB
And let it intersect AB at point P

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)
Since ABCD is a parallelogram
It’s Diagonals bisect each other

If the point P (2, 1) lies on the line segment joining points A (4, 2) and B (8, 4), then
(A) AP = 1/3 AB (B) AP = PB (C) PB = 1/3 AB (D) AP = 1/2 AB

If P = (𝑎/3, 4) is the mid-point of the line segment joining the points Q (–6, 5) and R (–2, 3), then the value of a is
(A) – 4 (B) – 12 (C) 12 (D) – 6

The perpendicular bisector of the line segment joining the points A (1, 5) and B (4, 6) cuts the y-axis at
(A) (0, 13) (B) (0, –13) (C) (0, 12) (D) (13, 0)
Let Line l be the perpendicular bisector of AB

The coordinates of the point which is equidistant from the three vertices of the ∆ AOB as shown in the Fig. 7.1 is
(x, y) (B) (y, x)
(C) (𝑥/2,𝑦/2) (D) (𝑥/2,𝑦/2)

A circle drawn with origin as the centre passes through (13/2, 0). The point which does not lie in the interior of the circle is
(A) ((−3)/4,1) (B) (2, 7/3) (C) (5,(−1)/2 ) (D) (−6, 5/2)
Since this is an MCQ

A line intersects the y-axis and x-axis at the points P and Q, respectively. If (2, –5) is the mid-point of PQ, then the coordinates of P and Q are, respectively
(A) (0, – 5) and (2, 0) (B) (0, 10) and (– 4, 0)
(C) (0, 4) and (– 10, 0) (D) (0, – 10) and (4, 0)

The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is
(A) (a + b + c)2 (B) 0 (C) a + b + c (D) abc
Given vertices (a, b + c), (b, c + a) and (c, a + b)

If the distance between the points (4, p) and (1, 0) is 5, then the value of p is
(A) 4 only (B) ± 4 (C) – 4 only (D) 0
Given
Distance between (4, p) and (1, 0) = 5
√(( 𝟏 −𝟒)𝟐+(𝟎−𝒑)𝟐) = 5
√(( −3)2+(−𝑝)2) = 5
√(9+𝑝^2 ) = 5
Squaring both sides
𝟗+𝒑^𝟐=𝟐𝟓
𝑝^2=25−9

If the points A (1, 2), O (0, 0) and C (a, b) are collinear, then
(A) a = b (B) a = 2b (C) 2a = b (D) a = –b
Since points A, O, C are collinear,
they will lie on the same line,
i.e. they will not form triangle

If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (−2, 5), then the coordinates of the other end of the diameter are (A) (−6, 7) (B) (6, −7) (C) (6, 7) (D) (−6, −7)
Let co-ordinates of A be (x, y)

In fig 6.52, the area of ∆ ABC (in square units) is (A) 15 (B) 10 (C) 7.5 (D) 2.5
Vertices of triangle are A(1, 3), B(-1, 0) and C(4, 0)
Here
x1 = 1 , y1 = 3
x2 = −1 , y2 = 0
x3 = 4 , y3 = 0

The point on the x – axis which is equidistant from points (−1, 0) and (5, 0) is (A) (0, 2) (B) (2, 0) (C) (3, 0) (D) (0, 3)

If A (4, 9), B (2, 3) and C (6, 5) are the vertices of ∆ ABC, then the length of median through C is (A) 5 units (B) √10 units (C) 25 units (D) 10
Joining vertex C to line AB such that
CD is the median

If P (2, 4), Q (0, 3), R (3, 6) and S (5, y) are the vertices of a parallelogram PQRS, then the value of y is (A) 7 (B) 5 (C) −7 (D) −8
Since PQRS is a parallelogram
It’s Diagonals bisect each other

If A (x, 2), B (−3, −4), and C (7, −5) are collinear, then the value of x is (A) −63 (B) 63 (C) 60 (D) −60
Since points A, B, C are collinear,
they will lie on the same line,
i.e. they will not form triangle

ABCD is a rectangle whose three vertices are B (4, 0), C (4, 3) and D (0, 3). The length of one of its diagonals is (A) 5 (B) 4 (C) 3 (D) 25
Length of Diagonal
= Distance b/w B (4, 0) and D (0, 3)
= √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2)
= √((0−4)2+(3 −0)2)
= √((−4)2+(3)2)
= √(16+9)
= √25
= 5 units
So, the correct answer is (A)

teachoo.co
Ex7.3,1 CAEROOCOM
Find the area of the triangle whose vertices are :
(i) (2, 3), (1, 0), (2, — 4)
Let the two points be A(2, 3) , B(-1, 0) & C(2, -4)
, 1

Area of triangle ABC = 2 [ X1(¥2 — Ya) + Xal¥3 — Va) + X3(¥a — Yo) J
Here

X,=2, y,=3

X= -1, y,=0

X,=2, y3=-4
Putting values

Area of triangle ABC = ; [2(0 — (-4)) + (-1)(-4- 3 ) + 2(3-0)]

Find the area of the triangle whose vertices are :
(ii) (–5, –1), (3, –5), (5, 2)

teachoo.com
Ex7.3,2
In each of the following find the value of ‘k’, for which the points
are collinear.
(i) (7, -2), (5, 1), (3, k)
Let the given points be A (7, -2}, B (5, 1), C (3, k)
If the above points are collinear,
they will lie on the same line,
i.e. the will not form triangle
or We can say that

In each of the following find the value of ‘k’, for which the points are collinear.
(ii) (8, 1), (k, – 4), (2, –5)

Ex 7.3, 3 teachoo.com
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.
Let the vertices of triangle be
A(O, -1
A(0,-1), (2, 1), C(O, 3) ON
Let the mid-point of
P
AB be P R
BC beQ
ACbeR
€ B(2, 1) Q C(0, 3)
Joining points P, Q,R
We get APQR

teachoo.co
Ex7.3,4 CAEROOCOM
Find the area of the quadrilateral whose vertices, taken in order,
are (-4, —2), (-3, —5), (3, —2) and (2, 3).
Let the vertices of quadrilateral be
A(-4, -2
A(-A, -2}, B(-3, -5) 4, =) B(-3, -5)
os INQ
D(2, 3) C(3, -2)
Joining AC,
There are 2 triangles formed ABC & ACD
Hence,
Area of quadrilateral ABCD = Area of A ABC + Area of AADC

Ex7.3,5 teachoo.com
You have studied in Class IX, (Chapter 9, Example 3), that a median
of a triangle divides it into two triangles of equal areas. Verify this
result for AABC whose vertices are A(4, — 6), B(3, —2) and C(5, 2)
Let A ABC be as shown in the figure
Let AD be the median which divides BC into two equal parts,
BD &CD
A(4, -6)

Hence,
Coordinates of D

- Ge vy +2)

“77 2 B(3,-2) D4, 0) C(5, 2)

345 -242
~ ( 2’ 2 )

Ex 7.4, 1 teachoo.com
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).
AB is the line segment joining the Points A (2, -2) and B (3, 7)
Let line 2x + y - 4 = 0 divide AB in the ratio k : 1 at point P
kG) +1(2) k(7)+1(-2
Coordinates of Point P = Fer. ol
k+1 k+1 roy
4
[Ro ~| Ry
= |——— , ——_
k+4°k+1 a
A kivf1 B
(2, -2) P (3, 7)

Ex 7.4, 2 teachoo.com
Find a relation between x and y if the points (x, y}, (1, 2} and (7, 0)
are collinear.
Let A(x, y), B (1, 2) and c (7, 0} be the 3 collinear Points
If the above Points are collinear,
they will lie on the same line,
ie. they will not form a triangle .
So,

Area of A ABC =0

1

3 [x10¥2 —¥3) + X23 —YD +%301 — y2)] =0
Here, X, =X, Yi =¥

%=L y.=2

Ex 7.4, 3 teachoo.com
Find the centre of a circle passing through the points (6, — 6), (3, - 7)
and (3, 3).

Let the circle pass through points (6, -6)

Points A (6, -6), B (3, -7) and C (3, 3) A

Let O(x, y) be the centre of the circle

C B
(3, 3) (3, -7)

Since the centre is equidistant from all the Points Lying on the
circle.

Ex 7.4, 4 teachoo.com
The two opposite vertices of a square are (—1, 2) and (3, 2). Find the
coordinates of the other two vertices.
The two given vertices of the square ABCD are A(-1, 2) and C(3, 2).
A (-1, 2) B (x, y)
Let B(x, y) be the unknown vertex.
We know, all the sides of square are equal.
AB=CB
D C (3, 2)
Using Distance Formula,
AB=,/(x + 1)2 + (y — 2)"
CB= (x — 3)? +(~y—- 2)"

Ex 7.4,5 teachoo.com
The Class X students of a secondary school in Krishinagar have been
allotted a rectangular plot of land for their gardening activity.
Sapling of Gulmohar are planted on the boundary at a distance of
1m from each other. There is a triangular grassy lawn in the plot as
shown in the Fig. 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. as. c

5 Ries R

:

rv: |

A12345678 910 D

If A is the origin, AD and AB are coordinate axes.
our graph will look like this

Ex 7.4, 6 (Method 1) teachoo.com
The vertices of a A ABC 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 =
AE 1 ra
ae Calculate the area of the A ADE and compare it with the
area of A ABC. (Recall Theorem 6.2 and Theorem 6.6).
A (4,6)
Given
AD _AE_1 D E
AB AC 4
B Cc
(1, 5) (7, 2)
AD A 4
AB 4 AC 4

EX 7.4, 7 teachoo.com
Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of A ABC.
(i) The median from A meets BC at D. Find the coordinates of the
point D. Al4, 2)
Since Median is AD
D is the mid-point of BC
B C
(6, 5) D (1,4)
Coordinates of D = Coordinates of mid-point of BC
- (@ + XQ Vit 2)
2 7’ 2

Here x, =6 y,=5

%M=1 yw=4

EX7.4, 8 teachoo.com
ABCD is a rectangle formed by the points A(—1, —1), B(—-1, 4), C(5, 4)
and D(5, —1). P,Q, R and S are the mid-points of AB, BC, CD and DA
respectively. Is the quadrilateral PQRS a square? a rectangle? or a
rhombus? Justify your answer
D(5,-1) R c (5, 4)
1 | :
A(-1,-1) P B(-1, 4)
Point P Point Q
P is the mid point of AB. Qis the mid point of BC.
Applying mid point Formula, Applying mid point Formula,
Coordinates of P Coordinates of Q
Xt X2 va t+ Y. _ (%14 *2 ¥1+ Ya
“A ) od (oor ~)