10th Maths 7.1

NCERT Class 10th solution of Exercise 7.2

NCERT Class 10th solution of Exercise 7.3

Exercise 7.1

Q1. Find the distance between the following points:
i) (2,3), 
ii) (-5,7), (-1, 3)
iii) (a, b), (-a, -b)
Sol. :
i)
Given:
Let P(2, 3), Q(4, 1)
Solve:
 Distance =sqrt((x_2-x_1)^2+(y_2-y_1)^2)
PQ=sqrt((4-2)^2+(1-3)^2)
PQ=sqrt((2)^2+(-2)^2)
PQ=sqrt(4+4)
PQ=sqrt8
PQ=2sqrt2 units.
Answer:
Distance 2sqrt2 units.
ii)
Given: 
Let P(-5, 7), Q(-1, 3)
Solve:
Distance =sqrt((x_2-x_1)^2+(y_2-y_1)^2)
PQ=sqrt((-1+5)^2+(3-7)^2)
PQ=sqrt((4)^2+(-4)^2)
PQ=sqrt(16+16)
PQ=sqrt(32)
PQ=16sqrt2 units
Answer:
Distance 16sqrt2 units.
iii)
Given:
Let P(a, b), Q(-a, -b)
Solve:
Distance =sqrt((x_2-x_1)^2+(y_2-y_1)^2)
PQ=sqrt((-a-a)^2+(-b-b)^2)
PQ=sqrt((-2a)^2+(-2b)^2)
PQ=sqrt(4a^2+4b^2)
PQ=sqrt(4(a^2+b^2))
PQ=2sqrt(a^2+b^2) units
Answer:
Distance 2sqrt(a^2+b^2) units.
Q2. 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.
Given:
Let A(0, 0), B(36, 15)
To find:
Distance between two towns A and B.
Solve:
Distance =sqrt((x_2-x_1)^2+(y_2-y_1)^2)
AB=sqrt((36-0)^2+(15-0)^2)
AB=sqrt((36)^2+(15)^2)
AB=sqrt(1296+225)
AB=sqrt(1521)
AB=39 units
Answer:
Distance between two towns A and B is 39 units.
Q3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Sol. :
Given:
Let P(1, 5), Q(2, 3), R(-2, -11)
To find:
Collinear
Solve:
Distance =sqrt((x_2-x_1)^2+(y_2-y_1)^2)
PQ=sqrt((2-1)^2+(3-5)^2)
PQ=sqrt((1)^2+(2)^2)
PQ=sqrt(1+4)
PQ=sqrt5 units
QR=sqrt((-2-2)^2+(-11-3)^2)
QR=sqrt(-4)^2+(-14)^2)
QR=sqrt(16+196)
QR=sqrt(212)
RP=sqrt((1+2)^2+(5+11)^2)
RP=sqrt((3)^2+(16)^2)
RP=sqrt(9+256)
RP=sqrt(265)
For collinear
PQ+QR=RP
But
sqrt5+sqrt212nesqrt265
Answer:
The points are not collinear.
Q4. Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Sol. :
Given:
Let P(5, -2), Q(6, 4), R(7, -2)
To Find:
Points are the vertices of an isosceles triangle.
Solve:
PQ=sqrt((6-5)^2+(4+2)^2)
PQ=sqrt((1)^2+(6)^2)
PQ=sqrt(1+36)
PQ=sqrt(37)
QR=sqrt((7-6)^2+(-2-4)^2)
QR=sqrt((1)^2+(-6)^2)
QR=sqrt(1+36)
QR=sqrt(37)
RP=sqrt((7-5)^2+(-2+2)^2)
RP=sqrt((2)^2+(0)^2)
RP=sqrt4
RP=2
PQ=QR
Answer:
Yes, points are the vertices of an isosceles triangle.
Q5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in figure. 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.
manysolution12.blogspot.com
Sol. :
Given:
A(3, 4), B(6, 7), C(9, 4), D(6, 1)
To Find:
Which of them (Champa and Chameli) is correct.
Solve:
Distance formula =sqrt((x_2-x_1)^2+(y_2-y_1)^2)
AB=sqrt((6-3)^2+(7-4)^2)
AB=sqrt((3)^2+(3)^2)
AB=sqrt(9+9)
AB=sqrt18
AB=3sqrt2
BC=sqrt((9-6)^2+(4-7)^2)
BC=sqrt((3)^2+(-3)^2)
BC=sqrt(9+9)
BC=sqrt(18)
BC=3sqrt2
CD=sqrt((6-9)^2+(1-4)^2)
CD=sqrt((3)^2+(-3)^2)
CD=sqrt(9+9)
CD=sqrt(18)
CD=3sqrt2
DA=sqrt((3-6)^2+(4-1)^2)
DA=sqrt((-3)^2+(3)^2)
DA=sqrt(9+9)
DA=sqrt(18)
DA=3sqrt2
This gives
AB=BC=CD=DA
So ABCD is rhombus.
Diagonals are
AC=sqrt((9-3)^2+(4-4)^2)
AC=sqrt((6)^2+(0)^2)
AC=sqrt(36)
AC=6
BD=sqrt((6-6)^2+(1-7)^2)
BD=sqrt((0)^2+(-6)^2)
BD=sqrt(36)
BD=6
DiagonalAC=DiagonalBD
Answer:
ABCD is a square and hence Champa is correct.
Q6. 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)
Sol. :
i)
Given:
Let P(-1, -2), Q(1, 0), R(-1, 2), S(-3, 0)
Solve:
PQ=sqrt((1+1)^2+(0+2)^2)
PQ=sqrt((2)^2+(2)^2)
PQ=sqrt(4+4)
PQ=sqrt8
PQ=2sqrt2
QR=sqrt(-1-1)^2+(2-0)^2)
QR=sqrt(-2)^2+(2)^2
QR=sqrt(4+4)
QR=sqrt(8)
QR=2sqrt2
RS=sqrt((-3+1)^2+(0-2)^2)
RS=sqrt((-2)^2+(-2)^2)
RS=sqrt(4+4)
RS=sqrt(8)
RS=2sqrt2
SP=sqrt((-1+3)^2+(-2-0)^2)
SP=sqrt((2)^2+(-2)^2)
SP=sqrt(4+4)
SP=sqrt8
SP=2sqrt2
Diagonals
PR=sqrt((-1+1)^2+(2+2)^2)
PR=sqrt((0)^2+(4)^2)
PR=sqrt(16)
PR=4
QS=sqrt((-3-1)^2+(0-0)^2)
QS=sqrt((-4)^2+(0)^2)
QS=sqrt(16)
QS=4
Sides PQ=QR=RS=SP and diagonals PR=QS.
Answer:
Quadrilateral is square.
ii)
Given:
Let P(-3, 5), Q(3, 1), R(0, 3), S(-1, -4)
Solve:
PQ=sqrt((3+3)^2+(1-5)^2)
PQ=sqrt((6)^2+(-4)^2)
PQ=sqrt(36+16)
PQ=sqrt(52)
PQ=2sqrt13
QR=sqrt((0-3)^2+(3-1)^2)
QR=sqrt((-3)^2+(2)^2)
QR=sqrt(9+4)
QR=sqrt(13)
RS=sqrt((-1-0)^2+(-4-3)^2)
RS=sqrt((-1)^2+(-7)^2)
RS=sqrt(1+49)
RS=sqrt(50)
RS=5sqrt2
SP=sqrt((-3+1)+(5+4)^2)
SP=sqrt((-2)^2+(9)^2)
SP=sqrt(4+81)
SP=sqrt(85)
PR=sqrt((0+3)^2+(3-5)^2)
PR=sqrt((3)^2+(-2)^2)
PR=sqrt(9+4)
PR=sqrt13
PR+QR=PQ
sqrt13+sqrt13=2sqrt13
Answer:
No quadrilateral is formed.
iii)
Given:
Let P(4, 5), Q(7, 6), R(4,3), S(1, 2)
Solve:
PQ=sqrt((7-4)^2+(6-5)^2)
PQ=sqrt((3)^2+(1)^2)
PQ=sqrt(9+1)
PQ=sqrt(10)
QR=sqrt((4-7)^2+(3-6)^2)
QR=sqrt((-3)^2+(-3)^2)
QR=sqrt(9+9)
QR=sqrt(18)
QR=3sqrt2
RS=sqrt((1-4)^2+(2-3)^2)
RS=sqrt(-3)^2+(-1)^2)
RS=sqrt(9+1)
RS=sqrt(10)
SP=sqrt((4-1)^2+(5-2)^2)
SP=sqrt((3)^2+(3)^2)
SP=sqrt(9+9)
SP=sqrt(18)
SP=3sqrt2
PQ=RS=sqrt10 and QR=SP=3sqrt2
Answer:
Parallelogram is formed.
Q7. Find the point on the x-axis which is equidistance from (2, -5) and (-2, 9).
Sol. :
Given:
Let P(x, 0), Q(2, -5) and R(-2,9)
To Find:
The point on the x-axis.
Solve:
sqrt((2-x)^2+(-5-0))=sqrt((-2-x)^2+(9-0)^2)
sqrt((2-x)^2+(-5)^2)=sqrt((-2-x)^2+(9)^2)
(2-x)^2+25=(-2-x)^2+81
4-4x+x^2+25=4+4x+x^2+81
4x+4x=4+25-4-81
8x=-56
x=-56/8
x=-7
Answer:
The point on the x-axis is (-7, 0).
Q8. Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Sol. :
Given:
P(2, -3), Q(10, y) and PQ=10 units
To Find:
The value of y.
Solve:
sqrt((10-2)^2+(y+3)^2)=(10)
Squaring both the sides.
(8)^2+(y+3)^2=(10)^2
64+y^2+6y+9=100
y^2+6y73=100
y^2+6y-27=0
y^2+9y-3y-27=0
y(y+9)-3(y+9)=0
(y+9)(y-3)=0
y=-9, 3
Answer:
The value of y are -9, 3.
Q9. If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also find the distance QR and PR.
Sol. :
Given:
Q(0, 1), P(5, -3), R(x,6) and QP=QR.
To Find:
Value of x and the distance QR and PR.
Solve:
sqrt((x-0)^2+(6-1)^2)=sqrt((5-0)^2+(-3-1)^2)
(x)^2+(5)^2=(5)^2+(-4)^2
x^2+25=25+16
x^2=16
x=+-sqrt16
x=+-4
QR=sqrt((x-0)^2+(6-1)^2)
QR=sqrt((x)^2+(5)^2)
QR=sqrt((x)^2+25)
QR=sqrt(+-4)^2+25)
QR=sqrt(16+25)
QR=sqrt(41)
PR=sqrt((x-5)^2+(6+3)^2)
PR=sqrt((x)^2-10x+25+(9)^2)
When x=+4
PR=sqrt((+4)^2-10(4)+25+81)
PR=sqrt(16-40+106)
PR=sqrt(82)
When x=-4
PR=sqrt((-4)-10(-4)+106)
PR=sqrt(16+40+106)
PR=sqrt(162)
PR=9sqrt2
Answer:
Value of x=+-4
The distance QR=sqrt41 and PR=sqrt82 or 9sqrt2.
Q10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4)
Sol. :
Given:
Let P(x, y), Q(3, 6), and R(-3, 4) and PQ=PR
To find:
The relation between x and y.
Solve:
sqrt((x-3)^2+(y-6)^2)=sqrt((x+3)^2+(y-4)^2)`
Squaring both the sides
(x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2
x^2-6x+9+y^2-12y+36=x^2+6x+9+y^2-8y+16
6x+6x-8y+12y+25-45=0
12x+4y-20=0
3x+y-5=0
Answer:
The relation between x and y is 3x+y-5=0.
  

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