# Class 10 NCERT Solutions- Chapter 7 Coordinate Geometry – Exercise 7.1

**Question 1) Find the distance between the following pairs of points:**

**(i) (2,3), (4,1)**

**(ii) (-5, 7), (-1, 3)**

**(iii) (a, b), (-a, -b)**

**Solution:**

Formula used in the above question is : √(x

_{2 }– x_{1})^{2 }+ (y_{2 }– y_{1})^{2 }(i.e., Distance Formula)

(i)Here, x_{1}= 2, y_{1}= 3, x_{2}= 4, y_{2}= 1Now, applying the distance formula :

= √(4-2)

^{2}+ (1-3)^{2}=√(2)

^{2}+ (-2)^{2}= √8

= 2√2 units

(ii)Here, x_{1}= -5, y_{1}= 7, x_{2}= -1, y_{2}= 3Now, applying the distance formula :

= √(-1 – (-5))

^{2 }+ (3 – 7)^{2}= √(4)

^{2 }+ (-4)^{2}= 4√2

(iii)Here, x_{1}= a, y_{1}= b . x_{2}= -a, y_{2}= -bNow, applying the distance formula :

= √(-a – a)

^{2}+(-b – b)^{2}= √(-2a)

^{2}+ (-2b)^{2}= √4a

^{2}+ 4b^{2}= 2√a

^{2}+ b^{2}

**Question 2) 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 and shown in following figure.**

**Solution:**

Formula used in the above question is : √(x

_{2}– x_{1})^{2}+ (y_{2}– y_{1})^{2}(i.e, Distance Formula)Considering, Point A as (0, 0) and Point B as (36, 15) and applying the distance formula we get :

Distance between the two points : √(36 – 0)

^{2}+ (15 – 0)^{2}= √(36)

^{2}+ (15)^{2}= √1296 + 225

= √1521

= 39 units

Hence, the distance between two towns A and B is 39 units.

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

**Solution:**

Let (1, 5), (2, 3) and (-2, -11) be points A, B and C respectively.

Collinear term means that these 3 points lie in the same line. So, to we’ ll check it.

Using distance formula we will find the distance between these points.

AB = √(2 – 1)

^{2 }+ (3 – 5)^{2}=√(1)

^{2}+ (-2)^{2}=√1 + 4 =√5BC = √(-2 – 2)

^{2}+ (-11 – 3)^{2}= √(-4)

^{2 }+ (-14)^{2 }= √16 + 196 = √212CA = √(-2 – 1)

^{2}+ (-11 – 5)^{2}= √(-3)

^{2}+ (-16)^{2 }= √9 + 256 =√265As, AB + BC ≠ AC (Since, one distance is not equal to sum of other two distances, we can say that they do not lie in the same line.)

Hence, points A, B and C are not collinear.

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

**Solution:**

Let (5, – 2), (6, 4) and (7, – 2) be the points A, B and C respectively.

Using distance formula :

AB = √(6 – 5)

^{2}+ (4 – (-2))^{2}= √(1 + 36) = √37

BC = √(7 – 6)

^{2}+ (-2 – 4)^{2}= √(1 + 36) = √37

AC = √(7 – 5)

^{2}+ (-2 – (-2))^{2}= √(4 + 0) = 2

As, AB = BC ≠ AC (Two distances equal and one distance is not equal to sum of other two)

So, we can say that they are vertices of an isosceles triangle.

**Question 5) In a classroom, 4 friends are seated at the points A, B, C and D as 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.**

**Solution:**

From the given fig, find the coordinates of the points

AB = √(6 – 3) + (7 – 4)

= √9+9 = √18 = 3√2

BC = √(9 – 6) + (4 – 7)

= √9+9 = √18 = 3√2

CD = √(6 – 9) + (1 – 4)

= √9 + 9 = √18 = 3√2

DA = √(6 – 3) + (1 – 4)

= √9+9 =√18 =3√2

AB = BC = CD = DA = 3√2

All sides are of equal length. Therefore, ABCD is a square and hence, Champa was correct.

**Question 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)**

**(ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4)**

**(iii) (4, 5), (7, 6), (4, 3), (1, 2)**

**Solution:**

(i)Here, let the given points are P(-1, -2), Q(1, 0), R(-1, 2) and S(-3, 0) respectively.PQ = √(1 – (-1))² + (0 – (-2))²

= √(1+1)²+(0+2)²

= √8 = 2 √2

QR = √(−1−1)²+(2−0)²

= √(−2)²+(2)²

= √8 = 2 √2

RS = √(−3−(−1))²+(0−2)²

= √8 = 2 √2

PS = √((−3−(−1))²+(0−(−2))²

= √8 = 2 √2

Here, we found that the length of all the sides are equal.

Diagonal PR = √(−1−(−1))²+(2−(−2))²

= √ 0+16

= 4

Diagonal QS = √(−3−1)²+(0−0)²

= √ 16 = 4

Finally, we also found that the length of diagonal are also same.

Here, PQ = QR = RS = PS = 2√2

and QS = PR = 4

This is the property of SQUARE. Hence, the given figure is SQUARE.

(ii)Let the points be P(-3, 5), Q(3, 1), R(0, 3) and S(-1, -4)PQ = √(3−(−3))²+(1−5)²

= √(3+3)²+(−4)²

= √36+16 = √52 = 2 √13

QR = √(0−3)²+(3−1)²

= √(−3)²+(2)²

= √9 + 4 = √13

RS = √(−1−0)²+(−4−3)²

= √(−1)²+(−7)²

= √1+49 = √50 = 5 √2

PS = √(−1−(−3))²+(−4−5)²

= √(−1+3)²+(−9)²

= √4+81 = √85

Here, All the lengths of sides are unequal.

So, The given points will not create any quadrilateral.

(iii)Let the points be P(4, 5), Q(7, 6), R(4, 3) and S(1, 2)PQ = √(7−4)²+(6−5)²

= √(3)²+(1)² = √9+1 = √10

QR = √(4−7)²+(3−6)²

= √(−3)²+(−3)² = √9+9 =3 √2

RS = √(1−4)²+(2−3)²

= √(−3)²+(−1)² = √9+1 = √10

PS = √(1−4)²+(2−5)²

= √(−3)²+(−3)² = √9+9 =3 √2

We see that the opposite sides are equal. Lets find the diagonal now.

Diagonal PR = √(4−4)²+(3−5)²

= √0+4 = 2

Diagonal QS = √(1−7)²+(2−6)²

= √36+16

= √52

Here, PQ = RS = √10

and QR = PS = 3√2

We see that the diagonals are not equal.

Hence, the formed quadrilateral is a PARALLELOGRAM.

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

**Solution:**

Let the point on the X axis be (x, 0)

Given, distance between the points (2, -5), (x, 0) = distance between points (-2, 9), (x, 0)

[ Applying Distance Formula ]

⇒ √(x – 2)²+(0 – (-5))² = √(x – (-2))²+(0 – 9)²

On squaring both the sides, we get

⇒ (x – 2)² + 5² = (x + 2)² + 9²

⇒ x² – 4x + 4 + 25 = x² + 4x + 4 + 81

⇒ -4x -4x = 85 – 29

⇒ -8x = 56

⇒ x = -7

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

**Solution:**

It is given that, the distance b/w two points is 10 units.

So, we’ll find the distance and equate

PQ = √ (10 – 2)

^{2}+ (y – (-3))^{2}= √ (8)

^{2}+ (y +3)^{2}On squaring both the sides, we get :

64 +(y+3)

^{2}= (10)^{2}(y+3)

^{2}= 36y + 3 = ±6

y + 3 = +6 or y + 3 = −6

Hence, y = 3 or -9.

**Question 9) 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.**

**Solution:**

Given, PQ = QR

We will apply distance formula and find the distance between them,

PQ = √(5 – 0)

^{2 }+ (-3 – 1)^{2}= √ (- 5)

^{2 }+ (-4)^{2}= √ 25 + 16 = √41

QR = √ (0 – x)

^{2}+ (1 – 6)^{2}= √ (-x)

^{2}+ (-5)^{2}= √ x

^{2}+ 25As they both are equal so on equating them, x

^{2 }+ 25 = 41x

^{2 }=16, x = ± 4So, putting the value of x and obtaining the value of QR and PR through distance formula

For x = +4, PR = √ (4 – 5)

^{2}+ (6 – (-3))^{2}= √ (-1)

^{2}+ (9)^{2}= √ 82

QR = √ (0 – 4)

^{2}+ (1 – 6)^{2}= √ 41

For x = -4, QR = √ (0 – (-4))

^{2}+ (1 – 6)^{2}= √ 16 + 25 = √ 41

PR = √ (5 + 4)

^{2 }+ (-3 -6)^{2}= √ 81 + 81 = 9 √ 2

**Question 10) Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).**

**Solution:**

Let, (x, y) be point P and (3, 6), (-3, 4) be pt A and B respectively.

It is given that their distance is equal, so we will equate the equations.

PA = PB (given)

⇒ √(x – 3)

^{2 }+(y – 6)^{2}= √(x-(-3))^{2}+ (y – 4)^{2}[ By applying distance formula ]On squaring both sides,

(x-3)

^{2}+(y-6)^{2}= (x +3)^{2}+(y-4)^{2}x

^{2}+9-6x+y^{2}+36-12y = x^{2}+9+6x+y^{2}+16-8y36-16 = 6x+6x+12y-8y

20 = 12x+4y

3x+y = 5

3x+y-5 = 0

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