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Mathematics · Coordinate Geometry · NCERT Exercises

Exercise 7.2

Complete, independently verified solutions for NCERT Exercise 7.2.

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

section formula · internal division

Find the coordinates of the point which divides the line segment joining A(-1, 7) and B(4, -3) internally in the ratio 2 : 3.

Solution

Board-exam working:

Using the section formula for the ratio 2 : 3:

P=(2(4)+3(−1)2+3,2(−3)+3(7)2+3)P=\left(\frac{2(4)+3(-1)}{2+3},\frac{2(-3)+3(7)}{2+3}\right)

Calculate the x-coordinate:

x=8−35=55=1x=\frac{8-3}{5}=\frac55=1

Calculate the y-coordinate:

y=−6+215=155=3y=\frac{-6+21}{5}=\frac{15}{5}=3

Final answer

The required point is (1, 3).

Question

Question 2

section formula · trisection

Find the coordinates of the points of trisection of the line segment joining A(4, -1) and B(-2, -3).

Solution

Board-exam working:

Let P and Q be the trisection points, so AP = PQ = QB.

Point P divides AB internally in the ratio 1 : 2.

P=(1(−2)+2(4)1+2,1(−3)+2(−1)1+2)P=\left(\frac{1(-2)+2(4)}{1+2},\frac{1(-3)+2(-1)}{1+2}\right)
P=(−2+83,−3−23)=(2,−53)P=\left(\frac{-2+8}{3},\frac{-3-2}{3}\right)=\left(2,-\frac53\right)

Point Q divides AB internally in the ratio 2 : 1.

Q=(2(−2)+1(4)2+1,2(−3)+1(−1)2+1)Q=\left(\frac{2(-2)+1(4)}{2+1},\frac{2(-3)+1(-1)}{2+1}\right)
Q=(−4+43,−6−13)=(0,−73)Q=\left(\frac{-4+4}{3},\frac{-6-1}{3}\right)=\left(0,-\frac73\right)

Final answer

The trisection points are:

(2,−53)and(0,−73)\left(2,-\frac53\right)\quad\text{and}\quad\left(0,-\frac73\right)

Question

Question 3

section formula · distance formula · midpoint · sports day

In the rectangular school ground shown in Figure 7.12, 100 flower pots are placed 1 m apart along AD. Niharika runs one-fourth of AD on the second line and places a green flag. Preet runs one-fifth of AD on the eighth line and places a red flag. Find the distance between the flags and the point where Rashmi should place a blue flag halfway between them.

Rectangular school ground coordinate grid with numbered running lines, flower pots along AD, and green and red flags on lines 2 and 8.
Figure 7-12 from the supplied NCERT chapter.

Solution

Board-exam working:

Take A as the origin, the running-line number as the x-coordinate, and the distance along AD as the y-coordinate.

Niharika is on the second line and covers one-fourth of 100 m:

14×100=25 m\frac14\times100=25\text{ m}

Therefore, the green flag is at G(2, 25).

Preet is on the eighth line and covers one-fifth of 100 m:

15×100=20 m\frac15\times100=20\text{ m}

Therefore, the red flag is at R(8, 20).

Distance between the flags:

GR=(8−2)2+(20−25)2GR=\sqrt{(8-2)^2+(20-25)^2}
GR=62+(−5)2=36+25=61 mGR=\sqrt{6^2+(-5)^2}=\sqrt{36+25}=\sqrt{61}\text{ m}

Rashmi must place the blue flag at the midpoint of G and R:

M=(2+82,25+202)M=\left(\frac{2+8}{2},\frac{25+20}{2}\right)
M=(5,452)=(5,22.5)M=\left(5,\frac{45}{2}\right)=(5,22.5)

Final answer

The flags are √61 m apart. Rashmi should place the blue flag on the fifth line, 22.5 m from A along AD.

Question

Question 4

section formula · ratio of division

Find the ratio in which P(-1, 6) divides the line segment joining A(-3, 10) and B(6, -8).

Solution

Board-exam working:

Let P divide AB internally in the ratio m : n.

Using the x-coordinate in the section formula:

−1=6m−3nm+n-1=\frac{6m-3n}{m+n}

Cross multiplying:

−m−n=6m−3n-m-n=6m-3n
2n=7m2n=7m
mn=27\frac{m}{n}=\frac27
m:n=2:7m:n=2:7

Verification using the y-coordinate:

y=2(−8)+7(10)2+7=−16+709=6y=\frac{2(-8)+7(10)}{2+7}=\frac{-16+70}{9}=6

Final answer

P divides AB internally in the ratio 2 : 7.

Question

Question 5

section formula · x-axis · ratio of division

Find the ratio in which the x-axis divides the line segment joining A(1, -5) and B(-4, 5). Also find the point of division.

Solution

Board-exam working:

Let the x-axis meet AB at P and let AP : PB = m : n.

Since P lies on the x-axis, its y-coordinate is 0.

0=m(5)+n(−5)m+n0=\frac{m(5)+n(-5)}{m+n}
5m−5n=05m-5n=0
m=nm=n
m:n=1:1m:n=1:1

Thus P is the midpoint of AB.

P=(1+(−4)2,−5+52)P=\left(\frac{1+(-4)}{2},\frac{-5+5}{2}\right)
P=(−32,0)P=\left(-\frac32,0\right)

Final answer

The ratio is 1 : 1, and the point of division is:

(−32,0)\left(-\frac32,0\right)

Question

Question 6

midpoint formula · parallelogram · diagonals

If A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram in order, find x and y.

Solution

Board-exam working:

The diagonals of a parallelogram bisect each other. Therefore, the midpoint of AC equals the midpoint of BD.

Midpoint of AC:

(1+x2,2+62)=(1+x2,4)\left(\frac{1+x}{2},\frac{2+6}{2}\right)=\left(\frac{1+x}{2},4\right)

Midpoint of BD:

(4+32,y+52)=(72,y+52)\left(\frac{4+3}{2},\frac{y+5}{2}\right)=\left(\frac72,\frac{y+5}{2}\right)

Equating x-coordinates:

1+x2=72\frac{1+x}{2}=\frac72
1+x=71+x=7
x=6x=6

Equating y-coordinates:

4=y+524=\frac{y+5}{2}
8=y+58=y+5
y=3y=3

Final answer

x = 6 and y = 3.

Question

Question 7

midpoint formula · diameter · circle

Find the coordinates of A if AB is a diameter of a circle whose centre is O(2, -3) and B is (1, 4).

Solution

Board-exam working:

Let A = (x, y). Since O is the centre of the circle, O is the midpoint of diameter AB.

Using the midpoint formula:

(x+12,y+42)=(2,−3)\left(\frac{x+1}{2},\frac{y+4}{2}\right)=(2,-3)

Equating x-coordinates:

x+12=2\frac{x+1}{2}=2
x+1=4x+1=4
x=3x=3

Equating y-coordinates:

y+42=−3\frac{y+4}{2}=-3
y+4=−6y+4=-6
y=−10y=-10

Final answer

A = (3, -10).

Question

Question 8

section formula · fraction of a segment

A and B are (-2, -2) and (2, -4). Find P on AB such that:

AP=37ABAP=\frac37 AB

Solution

Board-exam working:

Since AP is three-sevenths of AB, the remaining part PB is four-sevenths of AB.

AP:PB=3:4AP:PB=3:4

Using the section formula:

P=(3(2)+4(−2)3+4,3(−4)+4(−2)3+4)P=\left(\frac{3(2)+4(-2)}{3+4},\frac{3(-4)+4(-2)}{3+4}\right)

Calculate the x-coordinate:

x=6−87=−27x=\frac{6-8}{7}=-\frac27

Calculate the y-coordinate:

y=−12−87=−207y=\frac{-12-8}{7}=-\frac{20}{7}

Final answer

P=(−27,−207)P=\left(-\frac27,-\frac{20}{7}\right)

Question

Question 9

section formula · four equal parts

Find the coordinates of the three points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.

Solution

Board-exam working:

Let P, Q and R divide AB into four equal parts in that order.

Point P divides AB in the ratio 1 : 3.

P=(1(2)+3(−2)4,1(8)+3(2)4)P=\left(\frac{1(2)+3(-2)}{4},\frac{1(8)+3(2)}{4}\right)
P=(−1,72)P=\left(-1,\frac72\right)

Point Q is the midpoint and divides AB in the ratio 1 : 1.

Q=(−2+22,2+82)=(0,5)Q=\left(\frac{-2+2}{2},\frac{2+8}{2}\right)=(0,5)

Point R divides AB in the ratio 3 : 1.

R=(3(2)+1(−2)4,3(8)+1(2)4)R=\left(\frac{3(2)+1(-2)}{4},\frac{3(8)+1(2)}{4}\right)
R=(1,132)R=\left(1,\frac{13}{2}\right)

Final answer

The three points are:

(−1,72),(0,5),(1,132)\left(-1,\frac72\right),\quad(0,5),\quad\left(1,\frac{13}{2}\right)

Question

Question 10

distance formula · rhombus · area

Find the area of the rhombus whose vertices, in order, are A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1).

Solution

Board-exam working:

The diagonals are AC and BD.

Length of AC:

AC=(−1−3)2+(4−0)2AC=\sqrt{(-1-3)^2+(4-0)^2}
AC=(−4)2+42=32=42AC=\sqrt{(-4)^2+4^2}=\sqrt{32}=4\sqrt2

Length of BD:

BD=(−2−4)2+(−1−5)2BD=\sqrt{(-2-4)^2+(-1-5)^2}
BD=(−6)2+(−6)2=72=62BD=\sqrt{(-6)^2+(-6)^2}=\sqrt{72}=6\sqrt2

Area of a rhombus:

Area=12×d1×d2\text{Area}=\frac12\times d_1\times d_2
Area=12×42×62\text{Area}=\frac12\times4\sqrt2\times6\sqrt2
Area=12×24×2=24 square units\text{Area}=\frac12\times24\times2=24\text{ square units}

Final answer

The area of the rhombus is 24 square units.