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Coordinate Geometry Maths-class-10-ex-7-1

Table of contents:

NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1

The Distance formula in Coordinate Geometry

If there are two points A(x1,y1) and B(x2,y2) on a cartesian plane, then their distance can be calculated with the following formula :

Distance between two points (x2x1)2+(y2y1)2


Table of contents:

This is known as distance formula in the world of Coordinate Geometry. It has been derived from the famous Pythagorean theorem. All problems in this page are related to finding distance between two points.

To find the distance of a point, say (x,y), from the origin (0,0), the formula becomes:

Distance between origin and one point x2+y2

which is the simplified form of (x0)2+(y0)2




Solved problems

Class 10 Maths Chapter 7 Exercise 7.1

Q. 1(i): Find the distance between the pairs of points (2,3) and (4,1).

Solution

For points (2,3) and (4,1), x1=2,y1=3,x2=4,y2=1 So, the distance between those two points =(x2x1)2+(y2y1)2=(42)2+(13)2=22+(2)2=4+4=8=22×2=22 unit Ans.

Maths Class 10 Exercise 7.1 Q1 (i) X′XYY′A(2, 3)B(4, 1)

Q. 1(ii): Find the distance between the pairs of points (5,7) and (1,3).

Solution

For points (5,7) and (1,3), x1=5,y1=7,x2=1,y2=3 So, the distance between those two points =(1(5))2+(37)2=(1+5)2+(4)2=42+16=16+16=32=42×2=42 unit Ans. Following is the graph of the straight line with some additional information:


Maths Class 10 Exercise 7.1 Q1 (ii) X′XYY′A(-5, 7)B(-1, 3)

Q. 1(iii) Find the distance between the pairs of points (a,b) and (a,b).

Solution

Here, x1=a,y1=b,x2=a,y2=b So, the distance between those two points =(aa)2+(bb)2=(2a)2+(2b)2=4a2+4b2=4(a2+b2)=2(a2+b2) unit Ans.



Q.2. Find the distance between the pairs of points (0,0) and (36,15).

Solution

Here, x1=0,y1=0,x2=36,y2=15 So, the distance between those two points =(360)2+(150)2=362+152=1296+225=1521=32×132=3×13=39 unit Ans. Following is the graph of the straight line:


Maths Class 10 Exercise 7.1 Q2 X′XYY′A(0, 0)B(36, 15)

Q.3. Determine if the points (1,5),(2,3) and (2,11) are collinear.

Solution

Let us name the given points as P(1,5),Q(2,3) and R(2,11). Distance between P & Q, Q & R and P & R are as follows:
For PQ, x1=1,y1=5,x2=2,y2=3 PQ=(21)2+(35)2=12+(2)2=1+4=5 unit  For QR, x1=2,y1=3,x2=2,y2=11 QR=(22)2+(113)2=(4)2+(14)2=16+196=212=22×53=253 unit  For PR, x1=1,y1=5,x2=2,y2=11 PR=(21)2+(115)2=(3)2+(16)2=9+256=265 unit  If the 3 points P, Q and R are collinear, then PR(the whole line) must be equal to sum of PQ and QR. But in the given case, PQ+QRPR. So, the three points are not collinear. Ans.

Following is the graph of the three points P, Q and R.


Maths Class 10 Exercise 7.1 Q3 X′XYY′P(1, 5)Q(2, 3)R(-2, -11)

Q. 4.Check whether (5,2),(6,4) and (7,2) are the vertices of an isosceles triangle.

Solution

Let us name the given points as P(5,2),Q(6,4) and R(7,2).
We calculate the distances between P and Q, Q and R, P and R.
For PQ, x1=5,y1=2,x2=6,y2=4 PQ=(65)2+(4(2))2=12+(6)2=1+36=37 unit  For QR, x1=6,y1=4,x2=7,y2=2 QR=(76)2+(24)2=12+(6)2=1+36=37 unit  For PR, x1=5,y1=2,x2=7,y2=2 PR=(75)2+(2(2))2=22+(2+2)2=4+02=4=2 unit 

Since side PQ = side QR, PQR is an isosceles triangle. So, points (5,2),(6,4) and (7,2) are the vertices of an isosceles triangle. Ans.

An isosceles triangle has two equal sides. Following is the graph of the 3 lines formed by the given coordinates.



Maths Class 10 Exercise 7.1 Q4 X′XYY′-3-2-101234567891054321-1-2-3P(5, -2)Q(6, 4)R(7, -2)

It is clear that PQ=QR. So the three given points form an ISOSCELES triangle.



Q.5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in the given 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.



Exercise 7.1 Q-5 🤹 A 🤷 B 🤵 C 🤴 D (3, 4)(6, 7)(9, 4)(6, 1)

Solution

According to the box, let us name the given points as A, B, C and D. So the coordinates are A(3,4),B(6,7),C(9,4) and D(6,1). Let us now find the distances AB, BC, CD and DA. AB=(63)2+(74)2=32+32=9+9=18=32×2=32 unit BC=(96)2+(47)2=32+(3)2=9+9=18=32×2=32 unit CD=(69)2+(14)2=(3)2+(3)2=9+9=18=32×2=32 unit AD=(63)2+(14)2=32+(3)2=9+9=18=32×2=32 unit  Till now we find that AB=BC=CD=AD=32.
So, this quadrilateral is either a square or a rhombus.


Note: The diagonals of a square are always equal to each other. But diagonals of a rhombus are unequal.


To be sure, let us compare the lengths of diagonals AC and BD ... AC=(93)2+(44)2=62+02=36=6 unit BD=(66)2+(17)2=02+(6)2=0+36=36=6 unit  Since diagonals are equal to each other, and the four sides are also equal, ABCD is a square, and Champa was correct. Ans.



Q.6(i). Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(1,2),(1,0),(1,2),(3,0)

Solution

Let us name the given points as P(1,2),Q(1,0),R(1,2) and S(3,0). PQ=(1(1))2+(0(2))2=22+22=4+4=8=22×2=22 unit QR=(11)2+(20)2=(2)2+22=4+4=8=22×2=22 unit RS=(3(1))2+(02)2=(2)2+(2)2=4+4=8=22×2=22 unit PS=(3(1))2+(0(2))2=(2)2+(0+2)2=4+4=8=22×2=22 unit  As all the 4 sides are equal, the quadrilateral should either be a square, or a rhombus. The two diagonals of a square are equal to each other. But a rhombus has unequal diagonals.

Let us measure the diagonals ... Diagonal PR=(1(1))2+(2(2))2=(1+1)2+(2+2)2=02+42=16=4 unit Diagonal QS=(31)2+(00)2=(4)2+02=16+0=16=4 unit  Since all sides are equal, and the two diagonals are also equal to each other, the shape is a square.  Ans.

Please check the following graph ...


Maths Class 10 Exercise 7.1 Q6(i) P(-1, -2)Q(1, 0)R(-1, 2)S(-3, 0)All sides are equal.Two diagonals are also equal.Conclusion: it is a square.

Q.6(ii). Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(3,5),(3,1),(0,3),(1,4)

Solution

Let us name the given points as A(3,5),B(3,1),C(0,3) and D(1,4). AB=(3(3))2+(15)2=(3+3)2+(4)2=62+16=36+16=52=22×13=213 unit BC=(03)2+(31)2=(3)2+22=9+4=13 unit CD=(10)2+(43)2=(1)2+(7)2=1+49=50=52×2=52 unit AD=(1(3))2+(45)2=22+(9)2=4+81=85 unit AC=(0(3))2+(35)2=32+(2)2=9+4=13 unit BD=(13)2+(41)2=(4)2+(5)2=16+25=41 unit  Since AC+BC=13+13=213=AB,
points A,B,C are collinear.
So no quadrilateral is possible with the given four points.  Ans.


Maths Class 10 Exercise 7.1 Q6(ii) X′XYY′A(-3, 5)B(3, 1)C(0, 3)D(-1, -4)Points A, B, C are collinear.So, ABCD cannot form a quadrilateral.

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

Solution

Let us name the given points as A(4,5),B(7,6),C(4,3) and D(1,2).
For distance A to B, x1=4,y1=5,x2=7,y2=6 (74)2+(65)2 =32+12=9+1=10 unit  For distance B to C, x1=7,y1=6,x2=4,y2=3 (47)2+(36)2 =(3)2+(3)2=9+9=18=32×2=32 unit  For distance C to D, x1=4,y1=3,x2=1,y2=2 (14)2+(23)2 =(3)2+(1)2=9+1=10 unit  For distance A to D, x1=4,y1=5,x2=1,y2=2 (14)2+(25)2 =(3)2+(3)2=9+9=18=32×2=32 unit  For diagonal AC, x1=4,y1=5,x2=4,y2=3 (44)2+(35)2 =02+(2)2=0+4=2 unit  For diagonal BD, x1=7,y1=6,x2=1,y2=2 (17)2+(26)2 =(6)2+(4)2=36+16=52=22×13=213 unit  Since AB=CD=10, and BC=AD=32,
the opposite sides of this quadrilateral are equal.
The quadrilateral is either a rectangle or a parallelogram.

Note: The diagonals of a rectangle are always equal to each other. But diagonals of a parallelogram are unequal.


In our case, the diagonals are not equal; ACBD, since 2213,.
So, the given quadrilateral is a parallelogram. Ans.



Maths Class 10 Exercise 7.1 Q6(iii) A(4, 5)B(7, 6)C(4, 3)D(1, 2)All sides are not equal.Only opposite sides are equal.This is either a rectangle, or a parallelogram.Since two diagonals are not equal, this is a parallelogram.

Q.7. Find the point on the x-axis which is equidistant from (2,5) and (2,9).

Solution

Given that an unkonwn point is equidistant from (2,5) and (2,9).
It is also given that the said point lies on the X-axis.
So, from the 2nd condition we know that y coordinate of the unknown point is 0.
Let the unknown point be P(x,0).

Distance between (x,0) and (2,5): =(2x)2+(50)2=(222.2.x+x2)+(5)2=44.x+x2+25=x24x+29 -- (i) Distance between (x,0) and (2,9): =(2x)2+(90)2=((2)22.(2).x+x2)+(9)2=4+4.x+x2+81=x2+4x+85 -- (ii) According to the box, (i) and (ii) are equal: x24x+29=x2+4x+85or,x24x+29=x2+4x+85or,4x+29=4x+85or,4x4x=8529or,8x=56x=56÷(8)=7 Since x=7,
the required point is (7,0)  Ans.



Maths Class 10 Exercise 7.1 Q7 X′XYY′A(2, -5)B(-2, 9)P(-7, 0)Point P is equidistant from A and B.

Q.8. Find the values of y for which the distance between the points P(2,3) and Q(10,y) is 10 unit.

Solution

Distance between P(2,3) and Q(10,y) is : =(102)2+(y(3))2=82+(y+3)2=64+y2+2.y.3+32=64+y2+6y+9=y2+6y+73 Given that the distance between P and Q is 10. y2+6y+73=10y2+6y+73=102=100y2+6y+73100=0y2+6y27=0y2+9y3y27=0y(y+9)3(y+9)=0(y+9)(y3)=0either (y+9)=0y=9or (y3)=0y=3 So, the value of y is 9 or 3.  Ans.

Following is the graph of the two points:

Maths Class 10 Exercise 7.1 Q8 P(2, -3)[y as 3] Q(10, 3)[y as -9] Q(10, -9)Points (10, 3) and (10, -9) are equidistant from P(2, -3).

Q.9. If Q(0,1) is equidistant from P(5,3), and R(x,6), find the value of x. Also, find the distances QR and PR.

Solution

For distance between Q(0,1) and P(5,3), x1=0,y1=1,x2=5,y2=3

For distance between Q(0,1) and R(x,6), x1=0,y1=1,x2=x,y2=6

Given that distance
Q(0,1) to P(5,3)=Q(0,1) to R(x,6)(50)2+(31)2=(x0)2+(61)2(50)2+(31)2=(x0)2+(61)252+(4)2=x2+5225+16=x2+25x2+25=25+16x2=16x=±4 The value of x=+4or4 When x=4,R(x,6) becomes R(4,6) Distance QR when x=4, =(40)2+(61)2=(42+52)=16+25=4141 unit When x=4,R(x,6) becomes R(4,6) Distance QR when x=4, =(40)2+(61)2=(4)2+52=16+25=4141 unit Distance PR when x=4, =(45)2+(6(3))2=(12+92)=1+81=82 unit Distance PR when x=4, =(45)2+(6(3))2=(9)2+92=81+81=81×2=92×2=92 unit So, x=4,4
Distance QR=41
Distance PR=82,92 Ans.

Following is the graph of the given 3 points with additional 1 point:

Maths Class 10 Exercise 7.1 Q9 X′XYY′Q(0, 1)P(5, -3)R(4, 6)[when x = -4], R′(-4, 6)

Q.10. Find a relation between x and y such that the point ;(x,y) is equidistant from the points (3,6) and (3,4).

Solution

Distance (x,y) to (3,6)=(x,y) to (3,4) (3x)2+(6y)2=(3x)2+(4y)2(3x)2+(6y)2=(3x)2+(4y)2322.3.x+x2+622.6.y+y2=(3)22.(3).x+x2+422.4.y+y296x+x2+3612y+y2=9+6x+x2+168y+y2x2+y26x12y+45=x2+y2+6x8y+256x12y+45=6x8y+256x12y+456x+8y25=012x4y+20=03xy+5=03x+y5=0y=3x+5 Here, the relation between x and y is y=3x+5 Ans.



Maths Class 10 Exercise 7.1 Q10 X′XYY′P(3, 6)Q(-3, 4)A(0, 5)B(1, 2)C(2, -1)

In the diagram, points A, B and C are equidistant from points P and Q. Their coordinates (0,5),(1,2) and (2,1) have been calculated from the equation y=3x+5 When x=0,y=5, when x=1,y=2, when x=2,y=1 ...




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