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Mathematics · Coordinate Geometry

NCERT Examples

All 10 worked examples from Coordinate Geometry, with independently verified solutions and direct navigation.

Mathematics · Chapter 7 · NCERT Examples

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NCERT Example

Example 1

distance formula · right triangle · converse of Pythagoras theorem

Do the points P(3, 2), Q(-2, -3) and R(2, 3) form a triangle? If so, name the type of triangle formed.

Solution

Board-exam working:

Using the distance formula:

PQ=(3−(−2))2+(2−(−3))2PQ=\sqrt{(3-(-2))^2+(2-(-3))^2}
PQ=52+52=50=52PQ=\sqrt{5^2+5^2}=\sqrt{50}=5\sqrt2
QR=(2−(−2))2+(3−(−3))2QR=\sqrt{(2-(-2))^2+(3-(-3))^2}
QR=42+62=52=213QR=\sqrt{4^2+6^2}=\sqrt{52}=2\sqrt{13}
PR=(3−2)2+(2−3)2PR=\sqrt{(3-2)^2+(2-3)^2}
PR=12+(−1)2=2PR=\sqrt{1^2+(-1)^2}=\sqrt2

The sum of any two distances is greater than the third, so the three points form a triangle.

Now:

PQ2+PR2=50+2=52=QR2PQ^2+PR^2=50+2=52=QR^2

By the converse of the Pythagoras theorem, ∠P = 90°. Therefore, △PQR is right-angled at P.

Final answer

The points form a right triangle, right-angled at P.

NCERT Example

Example 2

distance formula · square · diagonals

Show that the points A(1, 7), B(4, 2), C(-1, -1) and D(-4, 4) are the vertices of a square.

Solution

Board-exam working:

Using the distance formula for each side:

AB=(4−1)2+(2−7)2=9+25=34AB=\sqrt{(4-1)^2+(2-7)^2}=\sqrt{9+25}=\sqrt{34}
BC=(−1−4)2+(−1−2)2=25+9=34BC=\sqrt{(-1-4)^2+(-1-2)^2}=\sqrt{25+9}=\sqrt{34}
CD=(−4−(−1))2+(4−(−1))2=9+25=34CD=\sqrt{(-4-(-1))^2+(4-(-1))^2}=\sqrt{9+25}=\sqrt{34}
DA=(1−(−4))2+(7−4)2=25+9=34DA=\sqrt{(1-(-4))^2+(7-4)^2}=\sqrt{25+9}=\sqrt{34}

Now calculate the diagonals:

AC=(−1−1)2+(−1−7)2=4+64=68AC=\sqrt{(-1-1)^2+(-1-7)^2}=\sqrt{4+64}=\sqrt{68}
BD=(−4−4)2+(4−2)2=64+4=68BD=\sqrt{(-4-4)^2+(4-2)^2}=\sqrt{64+4}=\sqrt{68}
AB=BC=CD=DAAB=BC=CD=DA
AC=BDAC=BD

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

Final answer

ABCD is a square.

NCERT Example

Example 3

distance formula · collinear points · classroom grid

Figure 7.6 shows Ashima, Bharti and Camella seated at A(3, 1), B(6, 4) and C(8, 6). Are they seated in a line? Give reasons.

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

Solution

Board-exam working:

Using the distance formula:

AB=(6−3)2+(4−1)2=9+9=32AB=\sqrt{(6-3)^2+(4-1)^2}=\sqrt{9+9}=3\sqrt2
BC=(8−6)2+(6−4)2=4+4=22BC=\sqrt{(8-6)^2+(6-4)^2}=\sqrt{4+4}=2\sqrt2
AC=(8−3)2+(6−1)2=25+25=52AC=\sqrt{(8-3)^2+(6-1)^2}=\sqrt{25+25}=5\sqrt2

Now compare the distances:

AB+BC=32+22=52=ACAB+BC=3\sqrt2+2\sqrt2=5\sqrt2=AC

Since AB + BC = AC, point B lies between A and C. Therefore, A, B and C are collinear.

Final answer

Yes. Ashima, Bharti and Camella are seated in a straight line.

NCERT Example

Example 4

equidistant point · distance formula · perpendicular bisector

Find a relation between x and y such that P(x, y) is equidistant from A(7, 1) and B(3, 5).

Coordinate axes showing A(7, 1), B(3, 5), and their perpendicular bisector x minus y equals 2.
Figure 7-7 from the supplied NCERT chapter.

Solution

Board-exam working:

Since P is equidistant from A and B:

AP=BPAP=BP

Squaring both sides:

AP2=BP2AP^2=BP^2
(x−7)2+(y−1)2=(x−3)2+(y−5)2(x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2

Expanding both sides:

x2−14x+49+y2−2y+1=x2−6x+9+y2−10y+25x^2-14x+49+y^2-2y+1=x^2-6x+9+y^2-10y+25

Cancelling equal terms and simplifying:

−14x−2y+50=−6x−10y+34-14x-2y+50=-6x-10y+34
8y−8x+16=08y-8x+16=0
x−y=2x-y=2

Final answer

The required relation is x - y = 2.

NCERT Example

Example 5

y-axis · equidistant point · distance formula

Find a point on the y-axis which is equidistant from A(6, 5) and B(-4, 3).

Coordinate axes showing A(7, 1), B(3, 5), and their perpendicular bisector x minus y equals 2.
Figure 7-7 from the supplied NCERT chapter.

Solution

Board-exam working:

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

Since P is equidistant from A and B:

AP=BPAP=BP

Squaring both sides:

(6−0)2+(5−y)2=(−4−0)2+(3−y)2(6-0)^2+(5-y)^2=(-4-0)^2+(3-y)^2

Expanding:

36+25+y2−10y=16+9+y2−6y36+25+y^2-10y=16+9+y^2-6y

Cancelling y² and simplifying:

61−10y=25−6y61-10y=25-6y
36=4y36=4y
y=9y=9

Check:

AP=(6−0)2+(5−9)2=36+16=52AP=\sqrt{(6-0)^2+(5-9)^2}=\sqrt{36+16}=\sqrt{52}
BP=(−4−0)2+(3−9)2=16+36=52BP=\sqrt{(-4-0)^2+(3-9)^2}=\sqrt{16+36}=\sqrt{52}

Thus AP = BP.

Final answer

The required point is (0, 9).

NCERT Example

Example 6

section formula · internal division · ratio

Find the coordinates of the point which divides the line segment joining (4, -3) and (8, 5) internally in the ratio 3 : 1.

Solution

Board-exam working:

Let A(4, -3), B(8, 5), and let P(x, y) divide AB in the ratio 3 : 1.

Using the section formula:

x=3(8)+1(4)3+1x=\frac{3(8)+1(4)}{3+1}
x=24+44=284=7x=\frac{24+4}{4}=\frac{28}{4}=7
y=3(5)+1(−3)3+1y=\frac{3(5)+1(-3)}{3+1}
y=15−34=124=3y=\frac{15-3}{4}=\frac{12}{4}=3

Final answer

The required point is (7, 3).

NCERT Example

Example 7

section formula · ratio of division · coordinate comparison

In what ratio does P(-4, 6) divide the line segment joining A(-6, 10) and B(3, -8)?

Solution

Board-exam working:

Let P divide AB internally in the ratio m₁ : m₂.

Using the x-coordinate in the section formula:

−4=3m1−6m2m1+m2-4=\frac{3m_1-6m_2}{m_1+m_2}

Cross multiplying:

−4m1−4m2=3m1−6m2-4m_1-4m_2=3m_1-6m_2
2m2=7m12m_2=7m_1
m1m2=27\frac{m_1}{m_2}=\frac27
m1:m2=2:7m_1:m_2=2:7

Verification using the y-coordinate:

y=2(−8)+7(10)2+7y=\frac{2(-8)+7(10)}{2+7}
y=−16+709=549=6y=\frac{-16+70}{9}=\frac{54}{9}=6

The y-coordinate agrees with P(-4, 6).

Final answer

The point divides the line segment internally in the ratio 2 : 7.

NCERT Example

Example 8

section formula · trisection · internal division

Find the coordinates of the points of trisection of the line segment joining A(2, -2) and B(-7, 4).

Line segment AB from A(2, -2) to B(-7, 4), divided into three equal parts by P and Q.
Figure 7-11 from the supplied NCERT chapter.

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(−7)+2(2)1+2,1(4)+2(−2)1+2)P=\left(\frac{1(-7)+2(2)}{1+2},\frac{1(4)+2(-2)}{1+2}\right)
P=(−7+43,4−43)=(−1,0)P=\left(\frac{-7+4}{3},\frac{4-4}{3}\right)=(-1,0)

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

Q=(2(−7)+1(2)2+1,2(4)+1(−2)2+1)Q=\left(\frac{2(-7)+1(2)}{2+1},\frac{2(4)+1(-2)}{2+1}\right)
Q=(−14+23,8−23)=(−4,2)Q=\left(\frac{-14+2}{3},\frac{8-2}{3}\right)=(-4,2)

Final answer

The points of trisection are (-1, 0) and (-4, 2).

NCERT Example

Example 9

section formula · y-axis · ratio of division

Find the ratio in which the y-axis divides the line segment joining A(5, -6) and B(-1, -4). Also find the point of intersection.

Solution

Board-exam working:

Let the y-axis divide AB at P in the ratio k : 1.

By the section formula:

P=(k(−1)+1(5)k+1,k(−4)+1(−6)k+1)P=\left(\frac{k(-1)+1(5)}{k+1},\frac{k(-4)+1(-6)}{k+1}\right)

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

−k+5k+1=0\frac{-k+5}{k+1}=0
−k+5=0-k+5=0
k=5k=5

Therefore, the ratio is 5 : 1.

Substituting k = 5 in the y-coordinate:

y=5(−4)+1(−6)5+1y=\frac{5(-4)+1(-6)}{5+1}
y=−20−66=−266=−133y=\frac{-20-6}{6}=\frac{-26}{6}=-\frac{13}{3}

Final answer

The ratio is 5 : 1, and the point of intersection is:

(0,−133)\left(0,-\frac{13}{3}\right)

NCERT Example

Example 10

midpoint formula · parallelogram · diagonals

If A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram in order, find p.

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:

(6+92,1+42)=(152,52)\left(\frac{6+9}{2},\frac{1+4}{2}\right)=\left(\frac{15}{2},\frac52\right)

Midpoint of BD:

(8+p2,2+32)=(8+p2,52)\left(\frac{8+p}{2},\frac{2+3}{2}\right)=\left(\frac{8+p}{2},\frac52\right)

Equating the x-coordinates:

152=8+p2\frac{15}{2}=\frac{8+p}{2}
15=8+p15=8+p
p=7p=7

Final answer

p = 7.