Coordinate Geometry — CBSE Class 10 Maths Textbook Solutions

Use the distance formula $PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$PQ=√(x_2-x_1)^2+(y_2-y_1)^2 in each case.

(i) $(2,3)$(2,3) and $(4,1)$(4,1):
$d=\sqrt{(4-2)^2+(1-3)^2}=\sqrt{2^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt2\text{ units}$d=√(4-2)^2+(1-3)^2=√2^2+(-2)^2=√4+4=√8=22 units

(ii) $(-5,7)$(-5,7) and $(-1,3)$(-1,3):
$d=\sqrt{(-1+5)^2+(3-7)^2}=\sqrt{4^2+(-4)^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt2\text{ units}$d=√(-1+5)^2+(3-7)^2=√4^2+(-4)^2=√16+16=√32=42 units

(iii) $(a,b)$(a,b) and $(-a,-b)$(-a,-b):
$d=\sqrt{(-a-a)^2+(-b-b)^2}=\sqrt{(-2a)^2+(-2b)^2}=\sqrt{4a^2+4b^2}=2\sqrt{a^2+b^2}\text{ units}$d=√(-a-a)^2+(-b-b)^2=√(-2a)^2+(-2b)^2=√4a^2+4b^2=2√a^2+b^2 units

Using the distance formula with $(x_1,y_1)=(0,0)$(x_1,y_1)=(0,0) and $(x_2,y_2)=(36,15)$(x_2,y_2)=(36,15):
$d=\sqrt{(36-0)^2+(15-0)^2}=\sqrt{36^2+15^2}=\sqrt{1296+225}=\sqrt{1521}=39\text{ units}$d=√(36-0)^2+(15-0)^2=√36^2+15^2=√1296+225=√1521=39 units

In Section 7.2, town A is taken as the origin $(0,0)$(0,0) and town B, which is $36$36 km east and $15$15 km north of A, is taken as $(36,15)$(36,15) (1 km = 1 unit on each axis). So the distance between the towns is the same calculation: $39$39 km.

Let $A(1,5)$A(1,5), $B(2,3)$B(2,3), $C(-2,-11)$C(-2,-11). Three points are collinear only if the sum of two of the pairwise distances equals the third.

$AB=\sqrt{(2-1)^2+(3-5)^2}=\sqrt{1+4}=\sqrt5$AB=√(2-1)^2+(3-5)^2=√1+4=5
$BC=\sqrt{(-2-2)^2+(-11-3)^2}=\sqrt{16+196}=\sqrt{212}$BC=√(-2-2)^2+(-11-3)^2=√16+196=√212
$AC=\sqrt{(-2-1)^2+(-11-5)^2}=\sqrt{9+256}=\sqrt{265}$AC=√(-2-1)^2+(-11-5)^2=√9+256=√265

Here $AB+BC=\sqrt5+\sqrt{212}\approx 2.24+14.56=16.80$AB+BC=5+√212 2.24+14.56=16.80, while $AC=\sqrt{265}\approx16.28$AC=√26516.28. Since $AB+BC\neq AC$AB+BC≠ AC (and no other pair sums to the third either), the three points are not collinear.

Let $A(5,-2)$A(5,-2), $B(6,4)$B(6,4), $C(7,-2)$C(7,-2).

$AB=\sqrt{(6-5)^2+(4+2)^2}=\sqrt{1+36}=\sqrt{37}$AB=√(6-5)^2+(4+2)^2=√1+36=√37
$BC=\sqrt{(7-6)^2+(-2-4)^2}=\sqrt{1+36}=\sqrt{37}$BC=√(7-6)^2+(-2-4)^2=√1+36=√37
$AC=\sqrt{(7-5)^2+(-2+2)^2}=\sqrt{4+0}=2$AC=√(7-5)^2+(-2+2)^2=√4+0=2

Since $AB=BC=\sqrt{37}$AB=BC=√37 (two sides equal), the triangle is isosceles.

Find all four sides and both diagonals of $ABCD$ABCD using the distance formula.

Sides:
$AB=\sqrt{(6-3)^2+(7-4)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt2$AB=√(6-3)^2+(7-4)^2=√9+9=√18=32
$BC=\sqrt{(9-6)^2+(4-7)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt2$BC=√(9-6)^2+(4-7)^2=√9+9=√18=32
$CD=\sqrt{(6-9)^2+(1-4)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt2$CD=√(6-9)^2+(1-4)^2=√9+9=√18=32
$DA=\sqrt{(3-6)^2+(4-1)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt2$DA=√(3-6)^2+(4-1)^2=√9+9=√18=32

All four sides are equal.

Diagonals:
$AC=\sqrt{(9-3)^2+(4-4)^2}=\sqrt{36}=6$AC=√(9-3)^2+(4-4)^2=√36=6
$BD=\sqrt{(6-6)^2+(1-7)^2}=\sqrt{36}=6$BD=√(6-6)^2+(1-7)^2=√36=6

Both diagonals are equal too. Since all four sides are equal and both diagonals are equal, $ABCD$ABCD is indeed a square. Champa is correct (Chameli is wrong).

(i) Let $A(-1,-2)$A(-1,-2), $B(1,0)$B(1,0), $C(-1,2)$C(-1,2), $D(-3,0)$D(-3,0).
$AB=\sqrt{(1+1)^2+(0+2)^2}=\sqrt{4+4}=2\sqrt2,\quad BC=\sqrt{(-1-1)^2+(2-0)^2}=\sqrt{4+4}=2\sqrt2$AB=√(1+1)^2+(0+2)^2=√4+4=22, BC=√(-1-1)^2+(2-0)^2=√4+4=22
$CD=\sqrt{(-3+1)^2+(0-2)^2}=\sqrt{4+4}=2\sqrt2,\quad DA=\sqrt{(-1+3)^2+(-2-0)^2}=\sqrt{4+4}=2\sqrt2$CD=√(-3+1)^2+(0-2)^2=√4+4=22, DA=√(-1+3)^2+(-2-0)^2=√4+4=22
All sides equal. Diagonals: $AC=\sqrt{(-1+1)^2+(2+2)^2}=\sqrt{16}=4$AC=√(-1+1)^2+(2+2)^2=√16=4 and $BD=\sqrt{(-3-1)^2+(0-0)^2}=\sqrt{16}=4$BD=√(-3-1)^2+(0-0)^2=√16=4. All sides and both diagonals are equal, so $ABCD$ABCD is a square.

(ii) Let $A(-3,5)$A(-3,5), $B(3,1)$B(3,1), $C(0,3)$C(0,3), $D(-1,-4)$D(-1,-4).
$AC=\sqrt{(0+3)^2+(3-5)^2}=\sqrt{9+4}=\sqrt{13},\quad CB=\sqrt{(3-0)^2+(1-3)^2}=\sqrt{9+4}=\sqrt{13}$AC=√(0+3)^2+(3-5)^2=√9+4=√13, CB=√(3-0)^2+(1-3)^2=√9+4=√13
$AB=\sqrt{(3+3)^2+(1-5)^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}$AB=√(3+3)^2+(1-5)^2=√36+16=√52=2√13
Since $AC+CB=\sqrt{13}+\sqrt{13}=2\sqrt{13}=AB$AC+CB=√13+√13=2√13=AB, points $A$A, $C$C, $B$B are collinear — so $C$C lies on segment $AB$AB itself. These four points do not form a quadrilateral.

(iii) Let $A(4,5)$A(4,5), $B(7,6)$B(7,6), $C(4,3)$C(4,3), $D(1,2)$D(1,2).
$AB=\sqrt{(7-4)^2+(6-5)^2}=\sqrt{9+1}=\sqrt{10},\quad CD=\sqrt{(1-4)^2+(2-3)^2}=\sqrt{9+1}=\sqrt{10}$AB=√(7-4)^2+(6-5)^2=√9+1=√10, CD=√(1-4)^2+(2-3)^2=√9+1=√10
$BC=\sqrt{(4-7)^2+(3-6)^2}=\sqrt{9+9}=3\sqrt2,\quad DA=\sqrt{(4-1)^2+(5-2)^2}=\sqrt{9+9}=3\sqrt2$BC=√(4-7)^2+(3-6)^2=√9+9=32, DA=√(4-1)^2+(5-2)^2=√9+9=32
Opposite sides are equal ($AB=CD$AB=CD, $BC=DA$BC=DA). Checking the diagonals: $AC=\sqrt{(4-4)^2+(3-5)^2}=2$AC=√(4-4)^2+(3-5)^2=2 and $BD=\sqrt{(1-7)^2+(2-6)^2}=\sqrt{36+16}=\sqrt{52}$BD=√(1-7)^2+(2-6)^2=√36+16=√52 — the diagonals are unequal, so it is not a rectangle. Since opposite sides are equal, $ABCD$ABCD is a parallelogram.

A point on the $x$x-axis has the form $P(x, 0)$P(x, 0). Let $A(2,-5)$A(2,-5) and $B(-2,9)$B(-2,9). Since $P$P is equidistant from $A$A and $B$B, $PA^2=PB^2$PA^2=PB^2:
$(x-2)^2+(0+5)^2=(x+2)^2+(0-9)^2$(x-2)^2+(0+5)^2=(x+2)^2+(0-9)^2
$x^2-4x+4+25=x^2+4x+4+81$x^2-4x+4+25=x^2+4x+4+81
$-4x+29=4x+85$-4x+29=4x+85
$-8x=56 \implies x=-7$-8x=56 x=-7
The required point is $(-7, 0)$(-7, 0).

$PQ^2=(10-2)^2+(y+3)^2=10^2$PQ^2=(10-2)^2+(y+3)^2=10^2
$64+(y+3)^2=100$64+(y+3)^2=100
$(y+3)^2=36$(y+3)^2=36
$y+3=\pm 6$y+3=± 6
So $y=3$y=3 or $y=-9$y=-9. $y = 3$y = 3 or $y = -9$y = -9.

Since $Q$Q is equidistant from $P$P and $R$R, $QP^2=QR^2$QP^2=QR^2.

First, $QP=\sqrt{(5-0)^2+(-3-1)^2}=\sqrt{25+16}=\sqrt{41}$QP=√(5-0)^2+(-3-1)^2=√25+16=√41, so $QP^2=41$QP^2=41.

Also $QR^2=(x-0)^2+(6-1)^2=x^2+25$QR^2=(x-0)^2+(6-1)^2=x^2+25. Setting $QP^2=QR^2$QP^2=QR^2:
$41=x^2+25 \implies x^2=16 \implies x=\pm4$41=x^2+25 x^2=16 x=±4

Since $QR^2=QP^2=41$QR^2=QP^2=41 regardless of the sign of $x$x, $QR=\sqrt{41}$QR=√41 in both cases.

For $PR$PR, use $PR^2=(5-x)^2+(-3-6)^2=(5-x)^2+81$PR^2=(5-x)^2+(-3-6)^2=(5-x)^2+81:

• If $x=4$x=4: $PR=\sqrt{(5-4)^2+81}=\sqrt{1+81}=\sqrt{82}$PR=√(5-4)^2+81=√1+81=√82.
• If $x=-4$x=-4: $PR=\sqrt{(5+4)^2+81}=\sqrt{81+81}=\sqrt{162}=9\sqrt2$PR=√(5+4)^2+81=√81+81=√162=92.

$x=4$x=4 or $x=-4$x=-4; $QR=\sqrt{41}$QR=√41; $PR=\sqrt{82}$PR=√82 (when $x=4$x=4) or $9\sqrt2$92 (when $x=-4$x=-4).

Let $P(x,y)$P(x,y) be equidistant from $A(3,6)$A(3,6) and $B(-3,4)$B(-3,4), so $PA^2=PB^2$PA^2=PB^2:
$(x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2$(x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2
$x^2-6x+9+y^2-12y+36=x^2+6x+9+y^2-8y+16$x^2-6x+9+y^2-12y+36=x^2+6x+9+y^2-8y+16
$-6x-12y+45=6x-8y+25$-6x-12y+45=6x-8y+25
$-12x-4y+20=0$-12x-4y+20=0
Dividing by $-4$-4:
$3x+y-5=0$3x+y-5=0
which is the required relation.

Using the section formula with $A(-1,7)$A(-1,7), $B(4,-3)$B(4,-3), $m_1:m_2=2:3$m_1:m_2=2:3:
$x=\frac{m_1x_2+m_2x_1}{m_1+m_2}=\frac{2(4)+3(-1)}{2+3}=\frac{8-3}{5}=1$x=m_1x_2+m_2x_1/m_1+m_2=2(4)+3(-1)/2+3=8-3/5=1
$y=\frac{m_1y_2+m_2y_1}{m_1+m_2}=\frac{2(-3)+3(7)}{5}=\frac{-6+21}{5}=3$y=m_1y_2+m_2y_1/m_1+m_2=2(-3)+3(7)/5=-6+21/5=3
The required point is $(1, 3)$(1, 3).

Let $A(4,-1)$A(4,-1), $B(-2,-3)$B(-2,-3), and let $P, Q$P, Q be the points of trisection, so $AP=PQ=QB$AP=PQ=QB. $P$P divides $AB$AB in the ratio $1:2$1:2, and $Q$Q divides $AB$AB in the ratio $2:1$2:1.

$P$P (ratio $1:2$1:2):
$x=\frac{1(-2)+2(4)}{1+2}=\frac{-2+8}{3}=2,\quad y=\frac{1(-3)+2(-1)}{3}=\frac{-5}{3}$x=1(-2)+2(4)/1+2=-2+8/3=2, y=1(-3)+2(-1)/3=-5/3
So $P=\left(2, -\frac{5}{3}\right)$P=(2, -5/3).

$Q$Q (ratio $2:1$2:1):
$x=\frac{2(-2)+1(4)}{2+1}=\frac{-4+4}{3}=0,\quad y=\frac{2(-3)+1(-1)}{3}=\frac{-7}{3}$x=2(-2)+1(4)/2+1=-4+4/3=0, y=2(-3)+1(-1)/3=-7/3
So $Q=\left(0, -\frac{7}{3}\right)$Q=(0, -7/3).

The points of trisection are $\left(2, -\frac{5}{3}\right)$(2, -5/3) and $\left(0, -\frac{7}{3}\right)$(0, -7/3).

Take $A$A as the origin, with the 1 m-spaced parallel lines along the $x$x-axis and the direction of $AD$AD along the $y$y-axis. Since $100$100 flower pots are placed $1$1 m apart along $AD$AD, $AD = 100$AD = 100 m.

Green flag (Niharika): on the 2nd line, at $\frac{1}{4}\times100=25$1/4×100=25 m along $AD$AD. Position: $(2, 25)$(2, 25).

Red flag (Preet): on the 8th line, at $\frac{1}{5}\times100=20$1/5×100=20 m along $AD$AD. Position: $(8, 20)$(8, 20).

Distance between flags:
$d=\sqrt{(8-2)^2+(20-25)^2}=\sqrt{36+25}=\sqrt{61}\text{ m}$d=√(8-2)^2+(20-25)^2=√36+25=√61 m

Blue flag (Rashmi): exactly halfway, i.e. the midpoint of $(2,25)$(2,25) and $(8,20)$(8,20):
$\left(\frac{2+8}{2}, \frac{25+20}{2}\right)=(5,\ 22.5)$(2+8/2, 25+20/2)=(5, 22.5)

So Rashmi should post her flag on the 5th line, at a distance of $22.5$22.5 m from $AB$AB (and the distance between the green and red flags is $\sqrt{61}$√61 m).

Let the point $P(-1,6)$P(-1,6) divide $A(-3,10)$A(-3,10) and $B(6,-8)$B(6,-8) in the ratio $k:1$k:1. By the section formula:
$-1=\frac{k(6)+1(-3)}{k+1}$-1=k(6)+1(-3)/k+1
$-1(k+1)=6k-3$-1(k+1)=6k-3
$-k-1=6k-3$-k-1=6k-3
$2=7k \implies k=\frac{2}{7}$2=7k k=2/7
So the ratio is $k:1=2:7$k:1=2:7.

Check with $y$y: $y=\dfrac{k(-8)+1(10)}{k+1}=\dfrac{\frac{2}{7}(-8)+10}{\frac{2}{7}+1}=\dfrac{-\frac{16}{7}+\frac{70}{7}}{\frac{9}{7}}=\dfrac{\frac{54}{7}}{\frac{9}{7}}=6$y=k(-8)+1(10)/k+1=2/7(-8)+102/7+1=-16/7+70/79/7=54/79/7=6 ✓

The required ratio is $2 : 7$2 : 7.

On the $x$x-axis, the $y$y-coordinate is $0$0. Let the point divide $AB$AB in the ratio $k:1$k:1. Using the $y$y-coordinate of the section formula:
$0=\frac{k(5)+1(-5)}{k+1}$0=k(5)+1(-5)/k+1
$5k-5=0 \implies k=1$5k-5=0 k=1
So the ratio is $1:1$1:1.

Using the $x$x-coordinate with $k=1$k=1:
$x=\frac{1(-4)+1(1)}{1+1}=\frac{-3}{2}$x=1(-4)+1(1)/1+1=-3/2

The ratio is $1:1$1:1, and the point of division is $\left(-\frac{3}{2}, 0\right)$(-3/2, 0).

Let the vertices, in order, be $A(1,2)$A(1,2), $B(4,y)$B(4,y), $C(x,6)$C(x,6), $D(3,5)$D(3,5). In a parallelogram, the diagonals bisect each other, so the midpoint of diagonal $AC$AC equals the midpoint of diagonal $BD$BD.

Midpoint of $AC=\left(\dfrac{1+x}{2}, \dfrac{2+6}{2}\right)=\left(\dfrac{1+x}{2}, 4\right)$AC=(1+x/2, 2+6/2)=(1+x/2, 4)

Midpoint of $BD=\left(\dfrac{4+3}{2}, \dfrac{y+5}{2}\right)=\left(\dfrac{7}{2}, \dfrac{y+5}{2}\right)$BD=(4+3/2, y+5/2)=(7/2, y+5/2)

Equating the coordinates:
$\frac{1+x}{2}=\frac{7}{2}\implies 1+x=7\implies x=6$1+x/2=7/2 1+x=7 x=6
$4=\frac{y+5}{2}\implies y+5=8\implies y=3$4=y+5/2 y+5=8 y=3

$x = 6$x = 6, $y = 3$y = 3.

The centre of a circle is the midpoint of any diameter. So the centre $(2,-3)$(2,-3) is the midpoint of $A(x,y)$A(x,y) and $B(1,4)$B(1,4):
$\frac{x+1}{2}=2 \implies x+1=4 \implies x=3$x+1/2=2 x+1=4 x=3
$\frac{y+4}{2}=-3 \implies y+4=-6 \implies y=-10$y+4/2=-3 y+4=-6 y=-10

The point $A$A is $(3, -10)$(3, -10).

Since $AP=\frac{3}{7}AB$AP=3/7AB, the remaining part $PB=AB-AP=\frac{4}{7}AB$PB=AB-AP=4/7AB. So $P$P divides $AB$AB internally in the ratio $AP:PB=3:4$AP:PB=3:4.

Using the section formula with $A(-2,-2)$A(-2,-2), $B(2,-4)$B(2,-4), ratio $3:4$3:4:
$x=\frac{3(2)+4(-2)}{3+4}=\frac{6-8}{7}=-\frac{2}{7}$x=3(2)+4(-2)/3+4=6-8/7=-2/7
$y=\frac{3(-4)+4(-2)}{7}=\frac{-12-8}{7}=-\frac{20}{7}$y=3(-4)+4(-2)/7=-12-8/7=-20/7

$P=\left(-\dfrac{2}{7}, -\dfrac{20}{7}\right)$P=(-2/7, -20/7).

Let $P_1, P_2, P_3$P_1, P_2, P_3 divide $AB$AB into four equal parts, so they correspond to the ratios $1:3$1:3, $1:1$1:1 (midpoint) and $3:1$3:1 respectively.

$P_1$P_1 (ratio $1:3$1:3):
$x=\frac{1(2)+3(-2)}{4}=\frac{2-6}{4}=-1,\quad y=\frac{1(8)+3(2)}{4}=\frac{8+6}{4}=\frac{7}{2}$x=1(2)+3(-2)/4=2-6/4=-1, y=1(8)+3(2)/4=8+6/4=7/2
$P_1=\left(-1, \dfrac{7}{2}\right)$P_1=(-1, 7/2)

$P_2$P_2 (midpoint, ratio $1:1$1:1):
$x=\frac{-2+2}{2}=0,\quad y=\frac{2+8}{2}=5$x=-2+2/2=0, y=2+8/2=5
$P_2=(0, 5)$P_2=(0, 5)

$P_3$P_3 (ratio $3:1$3:1):
$x=\frac{3(2)+1(-2)}{4}=\frac{6-2}{4}=1,\quad y=\frac{3(8)+1(2)}{4}=\frac{24+2}{4}=\frac{13}{2}$x=3(2)+1(-2)/4=6-2/4=1, y=3(8)+1(2)/4=24+2/4=13/2
$P_3=\left(1, \dfrac{13}{2}\right)$P_3=(1, 13/2)

The three points are $\left(-1, \dfrac{7}{2}\right)$(-1, 7/2), $(0, 5)$(0, 5) and $\left(1, \dfrac{13}{2}\right)$(1, 13/2).

Let $A(3,0)$A(3,0), $B(4,5)$B(4,5), $C(-1,4)$C(-1,4), $D(-2,-1)$D(-2,-1). The diagonals of the rhombus are $AC$AC and $BD$BD.

$AC=\sqrt{(-1-3)^2+(4-0)^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt2$AC=√(-1-3)^2+(4-0)^2=√16+16=√32=42
$BD=\sqrt{(-2-4)^2+(-1-5)^2}=\sqrt{36+36}=\sqrt{72}=6\sqrt2$BD=√(-2-4)^2+(-1-5)^2=√36+36=√72=62

Using the hint:
$\text{Area}=\frac12\times AC\times BD=\frac12\times4\sqrt2\times6\sqrt2=\frac12\times24\times2=24$Area=12× AC× BD=12×42×62=12×24×2=24

The area of the rhombus is $24$24 square units.