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Mathematics · Coordinate Geometry · NCERT Exercises

Exercise 7.1

Complete, independently verified solutions for NCERT Exercise 7.1.

Mathematics · Chapter 7 · NCERT Exercises

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Verified NCERT questions and solutionsSolutions are checked independently and follow the supplied NCERT chapter edition.

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Question

Question 1

distance formula · coordinate pairs

Find the distance between each pair of points:

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

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

(iii) (a, b) and (-a, -b)

Solution

Board-exam working:

Use the distance formula:

d=(x2−x1)2+(y2−y1)2d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

(i)

d=(4−2)2+(1−3)2d=\sqrt{(4-2)^2+(1-3)^2}
d=22+(−2)2=8=22d=\sqrt{2^2+(-2)^2}=\sqrt8=2\sqrt2

(ii)

d=(−1−(−5))2+(3−7)2d=\sqrt{(-1-(-5))^2+(3-7)^2}
d=42+(−4)2=32=42d=\sqrt{4^2+(-4)^2}=\sqrt{32}=4\sqrt2

(iii)

d=(−a−a)2+(−b−b)2d=\sqrt{(-a-a)^2+(-b-b)^2}
d=(−2a)2+(−2b)2d=\sqrt{(-2a)^2+(-2b)^2}
d=4a2+4b2=2a2+b2d=\sqrt{4a^2+4b^2}=2\sqrt{a^2+b^2}

Final answer

(i) 2√2 units; (ii) 4√2 units; (iii) 2√(a² + b²) units.

Question

Question 2

distance formula · towns · application

Find the distance between (0, 0) and (36, 15). Hence find the distance between towns A and B discussed in Section 7.2.

Solution

Board-exam working:

Using the distance formula:

AB=(36−0)2+(15−0)2AB=\sqrt{(36-0)^2+(15-0)^2}
AB=1296+225AB=\sqrt{1296+225}
AB=1521=39AB=\sqrt{1521}=39

Since one coordinate unit represents 1 km, the distance between the towns is 39 km.

Final answer

The distance is 39 units, so towns A and B are 39 km apart.

Question

Question 3

distance formula · collinearity

Determine whether A(1, 5), B(2, 3) and C(-2, -11) are collinear.

Solution

Board-exam working:

Calculate the three distances:

AB=(2−1)2+(3−5)2=1+4=5AB=\sqrt{(2-1)^2+(3-5)^2}=\sqrt{1+4}=\sqrt5
BC=(−2−2)2+(−11−3)2=16+196=253BC=\sqrt{(-2-2)^2+(-11-3)^2}=\sqrt{16+196}=2\sqrt{53}
AC=(−2−1)2+(−11−5)2=9+256=265AC=\sqrt{(-2-1)^2+(-11-5)^2}=\sqrt{9+256}=\sqrt{265}

For three collinear points, one distance must equal the sum of the other two.

5+253≠265\sqrt5+2\sqrt{53}\ne\sqrt{265}

No one of the three distances equals the sum of the other two. Therefore, the points are not collinear.

Final answer

The three points are not collinear.

Question

Question 4

distance formula · isosceles triangle

Check whether A(5, -2), B(6, 4) and C(7, -2) are the vertices of an isosceles triangle.

Solution

Board-exam working:

Using the distance formula:

AB=(6−5)2+(4−(−2))2=1+36=37AB=\sqrt{(6-5)^2+(4-(-2))^2}=\sqrt{1+36}=\sqrt{37}
BC=(7−6)2+(−2−4)2=1+36=37BC=\sqrt{(7-6)^2+(-2-4)^2}=\sqrt{1+36}=\sqrt{37}
AC=(7−5)2+(−2−(−2))2=4=2AC=\sqrt{(7-5)^2+(-2-(-2))^2}=\sqrt4=2
AB=BCAB=BC

Two sides are equal. Therefore, △ABC is an isosceles triangle with base AC.

Final answer

Yes. The points form an isosceles triangle with AB = BC = √37 units.

Question

Question 5

distance formula · square · classroom grid

In Figure 7.8, four friends sit at A(3, 4), B(6, 7), C(9, 4) and D(6, 1). Champa says ABCD is a square, while Chameli disagrees. Use the distance formula to decide who is correct.

A ten-by-ten classroom coordinate grid showing A(3, 4), B(6, 7), C(9, 4), and D(6, 1).
Figure 7-8 from the supplied NCERT chapter.

Solution

Board-exam working:

Calculate the four sides:

AB=(6−3)2+(7−4)2=9+9=32AB=\sqrt{(6-3)^2+(7-4)^2}=\sqrt{9+9}=3\sqrt2
BC=(9−6)2+(4−7)2=9+9=32BC=\sqrt{(9-6)^2+(4-7)^2}=\sqrt{9+9}=3\sqrt2
CD=(6−9)2+(1−4)2=9+9=32CD=\sqrt{(6-9)^2+(1-4)^2}=\sqrt{9+9}=3\sqrt2
DA=(3−6)2+(4−1)2=9+9=32DA=\sqrt{(3-6)^2+(4-1)^2}=\sqrt{9+9}=3\sqrt2

Now calculate the diagonals:

AC=(9−3)2+(4−4)2=6AC=\sqrt{(9-3)^2+(4-4)^2}=6
BD=(6−6)2+(1−7)2=6BD=\sqrt{(6-6)^2+(1-7)^2}=6
AB=BC=CD=DAAB=BC=CD=DA
AC=BDAC=BD

All four sides are equal and the diagonals are equal. Therefore, ABCD is a square.

Final answer

Champa is correct: ABCD is a square.

Question

Question 6

distance formula · quadrilateral classification

Name the type of quadrilateral formed, if any, by each ordered set of points. Give reasons:

(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

Board-exam working:

Let the four points in each part be A, B, C and D in the given order.

(i) Calculate the sides:

AB=BC=CD=DA=(2)2+(2)2=22AB=BC=CD=DA=\sqrt{(2)^2+(2)^2}=2\sqrt2

Calculate the diagonals:

AC=4,BD=4AC=4,\qquad BD=4

All four sides are equal and both diagonals are equal. Therefore, ABCD is a square.

(ii) Calculate the sides:

AB=(3−(−3))2+(1−5)2=213AB=\sqrt{(3-(-3))^2+(1-5)^2}=2\sqrt{13}
AC=(0−(−3))2+(3−5)2=13AC=\sqrt{(0-(-3))^2+(3-5)^2}=\sqrt{13}
CB=(3−0)2+(1−3)2=13CB=\sqrt{(3-0)^2+(1-3)^2}=\sqrt{13}
AC+CB=13+13=213=ABAC+CB=\sqrt{13}+\sqrt{13}=2\sqrt{13}=AB

Therefore, A, C and B are collinear. Three of the given points lie on one straight line, so the four points do not form a quadrilateral.

(iii) Calculate the sides:

AB=(7−4)2+(6−5)2=10AB=\sqrt{(7-4)^2+(6-5)^2}=\sqrt{10}
BC=(4−7)2+(3−6)2=32BC=\sqrt{(4-7)^2+(3-6)^2}=3\sqrt2
CD=(1−4)2+(2−3)2=10CD=\sqrt{(1-4)^2+(2-3)^2}=\sqrt{10}
DA=(4−1)2+(5−2)2=32DA=\sqrt{(4-1)^2+(5-2)^2}=3\sqrt2
AB=CD,BC=DAAB=CD,\qquad BC=DA

Both pairs of opposite sides are equal. Therefore, ABCD is a parallelogram.

Final answer

(i) Square; (ii) no quadrilateral; (iii) parallelogram.

Question

Question 7

x-axis · equidistant point · distance formula

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

Solution

Board-exam working:

A point on the x-axis has y-coordinate 0. Let the required point be P(x, 0).

Since P is equidistant from A and B:

PA=PBPA=PB

Squaring both sides:

(x−2)2+(0−(−5))2=(x−(−2))2+(0−9)2(x-2)^2+(0-(-5))^2=(x-(-2))^2+(0-9)^2

Expanding:

x2−4x+4+25=x2+4x+4+81x^2-4x+4+25=x^2+4x+4+81
−4x+29=4x+85-4x+29=4x+85
−8x=56-8x=56
x=−7x=-7

Final answer

The required point is (-7, 0).

Question

Question 8

distance formula · quadratic equation · two values

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

Solution

Board-exam working:

Using the distance formula:

(10−2)2+(y−(−3))2=10\sqrt{(10-2)^2+(y-(-3))^2}=10

Squaring both sides:

82+(y+3)2=1028^2+(y+3)^2=10^2
64+(y+3)2=10064+(y+3)^2=100
(y+3)2=36(y+3)^2=36
y+3=±6y+3=\pm6

Therefore:

y=3ory=−9y=3\quad\text{or}\quad y=-9

Final answer

y = 3 or y = -9.

Question

Question 9

equidistant point · distance formula · two values

Q(0, 1) is equidistant from P(5, -3) and R(x, 6). Find x. Also find QR and PR.

Solution

Board-exam working:

First calculate QP²:

QP2=(5−0)2+(−3−1)2=25+16=41QP^2=(5-0)^2+(-3-1)^2=25+16=41

Since Q is equidistant from P and R:

QR2=QP2QR^2=QP^2
(x−0)2+(6−1)2=41(x-0)^2+(6-1)^2=41
x2+25=41x^2+25=41
x2=16x^2=16
x=4orx=−4x=4\quad\text{or}\quad x=-4

In both cases:

QR=41 unitsQR=\sqrt{41}\text{ units}

When x = 4, R = (4, 6):

PR=(4−5)2+(6−(−3))2=1+81=82PR=\sqrt{(4-5)^2+(6-(-3))^2}=\sqrt{1+81}=\sqrt{82}

When x = -4, R = (-4, 6):

PR=(−4−5)2+(6−(−3))2=81+81=92PR=\sqrt{(-4-5)^2+(6-(-3))^2}=\sqrt{81+81}=9\sqrt2

Final answer

x = 4 or -4; QR = √41 units. PR = √82 units when x = 4, and 9√2 units when x = -4.

Question

Question 10

equidistant point · distance formula · locus

Find a relation between x and y such that P(x, y) is equidistant from A(3, 6) and B(-3, 4).

Solution

Board-exam working:

Since P is equidistant from A and B:

PA=PBPA=PB

Squaring both sides:

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

Expanding:

x2−6x+9+y2−12y+36=x2+6x+9+y2−8y+16x^2-6x+9+y^2-12y+36=x^2+6x+9+y^2-8y+16

Cancelling equal terms and simplifying:

−6x−12y+45=6x−8y+25-6x-12y+45=6x-8y+25
12x+4y−20=012x+4y-20=0
3x+y−5=03x+y-5=0

Final answer

The required relation is 3x + y - 5 = 0.