10th Maths 7.1

Exercise 7.1

Q1. Find the distance between the following points:

$(2, 3), (4,1)$

ii)

$(-5,7), (-1, 3)$

iii)

$(a, b), (-a, -b)$

Sol. :

Given:

Let

$P(2, 3), Q(4, 1)$

Solve:

Distance

$=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$PQ=sqrt((4-2)^2+(1-3)^2)$

$PQ=sqrt((2)^2+(-2)^2)$

$PQ=sqrt(4+4)$

$PQ=sqrt8$

$PQ=2sqrt2$ units.

Answer:

Distance

$2sqrt2$ units.

ii)

Given:

Let

$P(-5, 7), Q(-1, 3)$

Solve:

Distance

$=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$PQ=sqrt((-1+5)^2+(3-7)^2)$

$PQ=sqrt((4)^2+(-4)^2)$

$PQ=sqrt(16+16)$

$PQ=sqrt(32)$

$PQ=16sqrt2$ units

Answer:

Distance

$16sqrt2$ units.

iii)

Given:

Let

$P(a, b), Q(-a, -b)$

Solve:

Distance

$=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$PQ=sqrt((-a-a)^2+(-b-b)^2)$

$PQ=sqrt((-2a)^2+(-2b)^2)$

$PQ=sqrt(4a^2+4b^2)$

$PQ=sqrt(4(a^2+b^2))$

$PQ=2sqrt(a^2+b^2)$ units

Answer:

Distance

$2sqrt(a^2+b^2)$ units.

Q2. 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

$7.2$ .

Given:

Let

$A(0, 0), B(36, 15)$

To find:

Distance between two towns

$A$ and

$B$ .

Solve:

Distance

$=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$AB=sqrt((36-0)^2+(15-0)^2)$

$AB=sqrt((36)^2+(15)^2)$

$AB=sqrt(1296+225)$

$AB=sqrt(1521)$

$AB=39$ units

Answer:

Distance between two towns

$A$ and

$B$ is

$39$ units.

Q3. Determine if the points

$(1, 5), (2, 3)$ and

$(-2, -11)$ are collinear.

Sol. :

Given:

Let

$P(1, 5), Q(2, 3), R(-2, -11)$

To find:

Collinear

Solve:

Distance

$=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$PQ=sqrt((2-1)^2+(3-5)^2)$

$PQ=sqrt((1)^2+(2)^2)$

$PQ=sqrt(1+4)$

$PQ=sqrt5$ units

$QR=sqrt((-2-2)^2+(-11-3)^2)$

$QR=sqrt(-4)^2+(-14)^2)$

$QR=sqrt(16+196)$

$QR=sqrt(212)$

$RP=sqrt((1+2)^2+(5+11)^2)$

$RP=sqrt((3)^2+(16)^2)$

$RP=sqrt(9+256)$

$RP=sqrt(265)$

For collinear

$PQ+QR=RP$

But

$sqrt5+sqrt212nesqrt265$

Answer:

The points are not collinear.

Q4. Check whether

$(5, -2), (6, 4)$ and

$(7, -2)$ are the vertices of an isosceles triangle.

Sol. :

Given:

Let

$P(5, -2), Q(6, 4), R(7, -2)$

To Find:

Points are the vertices of an isosceles triangle.

Solve:

$PQ=sqrt((6-5)^2+(4+2)^2)$

$PQ=sqrt((1)^2+(6)^2)$

$PQ=sqrt(1+36)$

$PQ=sqrt(37)$

$QR=sqrt((7-6)^2+(-2-4)^2)$

$QR=sqrt((1)^2+(-6)^2)$

$QR=sqrt(1+36)$

$QR=sqrt(37)$

$RP=sqrt((7-5)^2+(-2+2)^2)$

$RP=sqrt((2)^2+(0)^2)$

$RP=sqrt4$

$RP=2$

$PQ=QR$

Answer:

Yes, points are the vertices of an isosceles triangle.

Q5. In a classroom,

$4$ friends are seated at the points

$A, B, C$ and

$D$ as shown in figure. 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.

Sol. :

Given:

$A(3, 4), B(6, 7), C(9, 4), D(6, 1)$

To Find:

Which of them (Champa and Chameli) is correct.

Solve:

Distance formula

$=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$AB=sqrt((6-3)^2+(7-4)^2)$

$AB=sqrt((3)^2+(3)^2)$

$AB=sqrt(9+9)$

$AB=sqrt18$

$AB=3sqrt2$

$BC=sqrt((9-6)^2+(4-7)^2)$

$BC=sqrt((3)^2+(-3)^2)$

$BC=sqrt(9+9)$

$BC=sqrt(18)$

$BC=3sqrt2$

$CD=sqrt((6-9)^2+(1-4)^2)$

$CD=sqrt((3)^2+(-3)^2)$

$CD=sqrt(9+9)$

$CD=sqrt(18)$

$CD=3sqrt2$

$DA=sqrt((3-6)^2+(4-1)^2)$

$DA=sqrt((-3)^2+(3)^2)$

$DA=sqrt(9+9)$

$DA=sqrt(18)$

$DA=3sqrt2$

This gives

$AB=BC=CD=DA$

$ABCD$ is rhombus.

Diagonals are

$AC=sqrt((9-3)^2+(4-4)^2)$

$AC=sqrt((6)^2+(0)^2)$

$AC=sqrt(36)$

$AC=6$

$BD=sqrt((6-6)^2+(1-7)^2)$

$BD=sqrt((0)^2+(-6)^2)$

$BD=sqrt(36)$

$BD=6$

Diagonal

$AC=$ Diagonal

$BD$

Answer:

$ABCD$ is a square and hence Champa is correct.

Q6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

$(-1, -2), (1, 0), (-1, 2), (-3, 0)$

ii)

$(-3, 5), (3, 1), (0, 3), (-1, -4)$

iii)

$(4, 5), (7, 6), (4, 3), (1, 2)$

Sol. :

Given:

Let

$P(-1, -2), Q(1, 0), R(-1, 2), S(-3, 0)$

Solve:

$PQ=sqrt((1+1)^2+(0+2)^2)$

$PQ=sqrt((2)^2+(2)^2)$

$PQ=sqrt(4+4)$

$PQ=sqrt8$

$PQ=2sqrt2$

$QR=sqrt(-1-1)^2+(2-0)^2)$

$QR=sqrt(-2)^2+(2)^2$

$QR=sqrt(4+4)$

$QR=sqrt(8)$

$QR=2sqrt2$

$RS=sqrt((-3+1)^2+(0-2)^2)$

$RS=sqrt((-2)^2+(-2)^2)$

$RS=sqrt(4+4)$

$RS=sqrt(8)$

$RS=2sqrt2$

$SP=sqrt((-1+3)^2+(-2-0)^2)$

$SP=sqrt((2)^2+(-2)^2)$

$SP=sqrt(4+4)$

$SP=sqrt8$

$SP=2sqrt2$

Diagonals

$PR=sqrt((-1+1)^2+(2+2)^2)$

$PR=sqrt((0)^2+(4)^2)$

$PR=sqrt(16)$

$PR=4$

$QS=sqrt((-3-1)^2+(0-0)^2)$

$QS=sqrt((-4)^2+(0)^2)$

$QS=sqrt(16)$

$QS=4$

Sides

$PQ=QR=RS=SP$ and diagonals

$PR=QS$ .

Answer:

Quadrilateral is square.

ii)

Given:

Let

$P(-3, 5), Q(3, 1), R(0, 3), S(-1, -4)$

Solve:

$PQ=sqrt((3+3)^2+(1-5)^2)$

$PQ=sqrt((6)^2+(-4)^2)$

$PQ=sqrt(36+16)$

$PQ=sqrt(52)$

$PQ=2sqrt13$

$QR=sqrt((0-3)^2+(3-1)^2)$

$QR=sqrt((-3)^2+(2)^2)$

$QR=sqrt(9+4)$

$QR=sqrt(13)$

$RS=sqrt((-1-0)^2+(-4-3)^2)$

$RS=sqrt((-1)^2+(-7)^2)$

$RS=sqrt(1+49)$

$RS=sqrt(50)$

$RS=5sqrt2$

$SP=sqrt((-3+1)+(5+4)^2)$

$SP=sqrt((-2)^2+(9)^2)$

$SP=sqrt(4+81)$

$SP=sqrt(85)$

$PR=sqrt((0+3)^2+(3-5)^2)$

$PR=sqrt((3)^2+(-2)^2)$

$PR=sqrt(9+4)$

$PR=sqrt13$

$PR+QR=PQ$

$sqrt13+sqrt13=2sqrt13$

Answer:

No quadrilateral is formed.

iii)

Given:

Let

$P(4, 5), Q(7, 6), R(4,3), S(1, 2)$

Solve:

$PQ=sqrt((7-4)^2+(6-5)^2)$

$PQ=sqrt((3)^2+(1)^2)$

$PQ=sqrt(9+1)$

$PQ=sqrt(10)$

$QR=sqrt((4-7)^2+(3-6)^2)$

$QR=sqrt((-3)^2+(-3)^2)$

$QR=sqrt(9+9)$

$QR=sqrt(18)$

$QR=3sqrt2$

$RS=sqrt((1-4)^2+(2-3)^2)$

$RS=sqrt(-3)^2+(-1)^2)$

$RS=sqrt(9+1)$

$RS=sqrt(10)$

$SP=sqrt((4-1)^2+(5-2)^2)$

$SP=sqrt((3)^2+(3)^2)$

$SP=sqrt(9+9)$

$SP=sqrt(18)$

$SP=3sqrt2$

$PQ=RS=sqrt10$ and

$QR=SP=3sqrt2$

Answer:

Parallelogram is formed.

Q7. Find the point on the

$x$ -axis which is equidistance from

$(2, -5)$ and

$(-2, 9)$ .

Sol. :

Given:

Let

$P(x, 0), Q(2, -5)$ and

$R(-2,9)$

To Find:

The point on the

$x-$ axis.

Solve:

$sqrt((2-x)^2+(-5-0))=sqrt((-2-x)^2+(9-0)^2)$

$sqrt((2-x)^2+(-5)^2)=sqrt((-2-x)^2+(9)^2)$

$(2-x)^2+25=(-2-x)^2+81$

$4-4x+x^2+25=4+4x+x^2+81$

$4x+4x=4+25-4-81$

$8x=-56$

$x=-56/8$

$x=-7$

Answer:

The point on the

$x-$ axis is

$(-7, 0)$ .

Q8. Find the values of

$y$ for which the distance between the points

$P(2, -3)$ and

$Q(10, y)$ is

$10$ units.

Sol. :

Given:

$P(2, -3), Q(10, y)$ and

$PQ=10$ units

To Find:

The value of

$y$ .

Solve:

$sqrt((10-2)^2+(y+3)^2)=(10)$

Squaring both the sides.

$(8)^2+(y+3)^2=(10)^2$

$64+y^2+6y+9=100$

$y^2+6y73=100$

$y^2+6y-27=0$

$y^2+9y-3y-27=0$

$y(y+9)-3(y+9)=0$

$(y+9)(y-3)=0$

$y=-9, 3$

Answer:

The value of

$y$ are

$-9, 3.$

Q9. 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

$PR$ .

Sol. :

Given:

$Q(0, 1), P(5, -3), R(x,6)$ and

$QP=QR$ .

To Find:

Value of

$x$ and the distance

$QR$ and

$PR$ .

Solve:

$sqrt((x-0)^2+(6-1)^2)=sqrt((5-0)^2+(-3-1)^2)$

$(x)^2+(5)^2=(5)^2+(-4)^2$

$x^2+25=25+16$

$x^2=16$

$x=+-sqrt16$

$x=+-4$

$QR=sqrt((x-0)^2+(6-1)^2)$

$QR=sqrt((x)^2+(5)^2)$

$QR=sqrt((x)^2+25)$

$QR=sqrt(+-4)^2+25)$

$QR=sqrt(16+25)$

$QR=sqrt(41)$

$PR=sqrt((x-5)^2+(6+3)^2)$

$PR=sqrt((x)^2-10x+25+(9)^2)$

When

$x=+4$

$PR=sqrt((+4)^2-10(4)+25+81)$

$PR=sqrt(16-40+106)$

$PR=sqrt(82)$

When

$x=-4$

$PR=sqrt((-4)-10(-4)+106)$

$PR=sqrt(16+40+106)$

$PR=sqrt(162)$

$PR=9sqrt2$

Answer:

Value of

$x=+-4$

The distance

$QR=sqrt41$ and

$PR=sqrt82$ or

$9sqrt2$ .

Q10. Find a relation between

$x$ and

$y$ such that the point

$(x, y)$ is equidistant from the point

$(3, 6)$ and

$(-3, 4)$ .

Sol. :

Given:

Let

$P(x, y), Q(3, 6)$ , and

$R(-3, 4)$ and

$PQ=PR$

To find:

The relation between

$x$ and

$y$ .

Solve:

$sqrt((x-3)^2+(y-6)^2)$ =sqrt((x+3)^2+(y-4)^2)`

Squaring both the sides

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

$x^2-6x+9+y^2-12y+36=x^2+6x+9+y^2-8y+16$

$6x+6x-8y+12y+25-45=0$

$12x+4y-20=0$

$3x+y-5=0$

Answer:

The relation between

$x$ and

$y$ is

$3x+y-5=0$ .

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10th Maths 7.1

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Exercise 7.1

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