प्रश्नावली - 1A

Question 1 

रिक्त स्थानों की पूर्ति करें :-

(i) यदि b = ac तो b, a का ................ है ।

(ii) यदि भाग एल्गोरिदम a = bq + r में a = 72, b = 9 तो r = ....................

(iii) यदि भाग एल्गोरिदम a = bq + r में b = 4, q = 5, r = 1 तो a = .................

(iv) यदि भाग एल्गोरिदम a = bq + r में a = 37, b = 4, r = 1 तो q = 0 ...................

(v) 92 = 12×7 + 8 तो म०स० (92,12) = म०स० (12, .....) = ......

(vi) यदि 42 = 35×1 + 7, 35 = 7×5 + 0 तो म०स० (42, 35) = ...................

Sol :

(i) गुणज

(ii)  

a = bq + r 

भागफल की गणना (Quotient Calculation):

$q=\dfrac{a}{b}$

$q=\dfrac{72}{8}$

q = 8

शेष की गणना (Remainder Calculation):

r = a - b × q

r = 72 - 9 × 8

r = 72 - 72 = 0

इस प्रकार,  r = 0

(iii)  

a = b × q + r 

a = 4 × 5 + 1

a = 21

(iv)

37 = 4 × q + 1

4q = 37-1

4q = 36

$q=\dfrac{36}{4}=9$

(v)

92 = 12×7+8

म०स० (92 , 12) = म०स० (12 , 8)

12 = 8×1 + 4

म०स० (12,8)=म०स० (8,4)

8 = 4×2+0

म०स० (8,4)=4

तो , म०स० (92,12)=4 

(vi) 

42 = 35×1+7

35=7×5+0

तो , म०स० (42,35)=7

Question 2

भाग एल्गोरिदम a = bq + r का q और r बताएं, यदि 

(i) a = 10, b = 3 

(ii) a = 4, b = 19 

(iii) a = 81, b = 3 

(iv) a = 27, b = 2 

Sol :

(i)  

a = bq + r 

10 = 3×3 + 1 

q = 3 ,  r = 1

(ii)

4 = 19 × 0 + 4

81 = 3 × 27 + 0

q = 27 , r = 0

(iii)

a = 27 , b = 2

27 = 2 × 13 + 1

q = 13 , r = 1

Question 3

यूक्लिडियन एल्गोरिद्म के प्रयोग से निम्नलिखित के म०स० ज्ञात करें।

(i) 252 और 1540

(ii) 196 और 38220

(iii) 365 और 12450

(iv) 135 और 225

(v) 12576 और 4052

(vi) 455 और 42

Sol :

SELINA Solution Class 9 Chapter 28 Distance Formula Exercise 28 A

Question 1.1

Find the distance between the following pairs of points:
(-3, 6) and (2, -6)

Sol:

(-3, 6) and (2, -6)
Distance between the given points
= ${\displaystyle \sqrt{{(2+3)}^2+{(-6-6)}^2}}$
= ${\displaystyle \sqrt{{\left(5\right)}^2+{(-12)}^2}}$
= ${\displaystyle \sqrt{25+144}}$
= ${\displaystyle \sqrt{169}}$
= 13

Question 1.2

Find the distance between the following pairs of points:
(-a, -b) and (a, b)

Sol:

(-a, -b) and (a, b)
Distance between the given points
= ${\displaystyle \sqrt{{(\text{a}+\text{a})}^2+{(\text{b}+\text{b})}^2}}$
= ${\displaystyle \sqrt{{\left(2\text{a}\right)}^2+{\left(2\text{b}\right)}^2}}$
= ${\displaystyle \sqrt{4{\text{a}}^2+4{\text{b}}^2}}$
= 2${\displaystyle \sqrt{{\text{a}}^2+{\text{b}}^2}}$

Question 1.3

Find the distance between the following pairs of points:
${\displaystyle (\frac{3}{5},2)\text{and}(-\frac{1}{5},1\frac{2}{5})}$

Sol:

${\displaystyle (\frac{3}{5},2)\text{and}(-\frac{1}{5},1\frac{2}{5})}$
Distance between the given points
= ${\displaystyle \sqrt{{(-\frac{1}{5}-\frac{3}{5})}^2+{(1\frac{2}{5}-2)}^2}}$
= ${\displaystyle \sqrt{{(-\frac{4}{5})}^2+{\left(\frac{7-10}{5}\right)}^2}}$
= ${\displaystyle \sqrt{\frac{16}{25}+\frac{9}{25}}}$
= ${\displaystyle \sqrt{\frac{25}{25}}}$
= 1

Question 1.4

Find the distance between the following pairs of points:
${\displaystyle (\sqrt{3}+1,1)}$ and ${\displaystyle (0,\sqrt{3})}$

Sol:

${\displaystyle (\sqrt{3}+1,1)}$ and ${\displaystyle (0,\sqrt{3})}$
Distance between the given points
= ${\displaystyle \sqrt{{(0-\sqrt{3}-1)}^2+{(\sqrt{3}-1)}^2}}$
 = ${\displaystyle \sqrt{3+1+2\sqrt{3}+3+1-2\sqrt{3}}}$
= ${\displaystyle \sqrt{8}}$
= 2${\displaystyle \sqrt{2}}$

Question 2.1

Find the distance between the origin and the point:
(-8, 6) 

Sol:

Coordinates of origin are O (0, 0).
A (-8, 6)
AO = ${\displaystyle \sqrt{{(0+8)}^2+{(0-6)}^2}}$
= ${\displaystyle \sqrt{64+36}}$
= ${\displaystyle \sqrt{100}}$
= 10

Question 2.2

Find the distance between the origin and the point:
(-5, -12)

Sol:

Coordinates of origin are O (0, 0).
B (-5, -12)
BO = ${\displaystyle \sqrt{{(0+5)}^2+{(0+12)}^2}}$
= ${\displaystyle \sqrt{25+144}}$
= ${\displaystyle \sqrt{169}}$
= 13

Question 2.3

Find the distance between the origin and the point:
(8, -15)

Sol;

Coordinates of origin are O (0, 0).
C (8, -15)
CO = ${\displaystyle \sqrt{{(0-8)}^2+{(0+15)}^2}}$
= ${\displaystyle \sqrt{64+225}}$
= ${\displaystyle \sqrt{289}}$
= 17

Question 3

The distance between the points (3, 1) and (0, x) is 5. Find x.

Sol:

It is given that the distance between the points A (3, 1) and B (0, x) is 5.
∴ AB = 5
AB² = 25
(0 - 3)² + (x - 1)² = 25
9 + x² + 1 - 2x = 25
x² - 2x - 15 = 0
x² - 5x + 3x - 15 = 0
x(x - 5) + 3(x - 5)  = 0
(x - 5)(x + 3) = 0
x = 5, -3

Question 4

Find the co-ordinates of points on the x-axis which are at a distance of 17 units from the point (11, -8).

Sol;

Let the coordinates of the point on x-axis be (x, 0).
From the given information, we have:
${\displaystyle \sqrt{{(x-11)}^2+{(0+8)}^2}}$ = 17
(x - 11)² + (0 + 8)² = 289
x² + 121 - 22x + 64 = 289
x² - 22x - 104 = 0
x² - 26x + 4x - 104 = 0
x(x - 26) + 4(x - 26) = 0
(x - 26)(x + 4) = 0
x = 26, -4
Thus, the required co-ordinates of the points on x-axis are (26, 0) and (-4, 0).

Question 5

Find the coordinates of the points on the y-axis, which are at a distance of 10 units from the point (-8, 4).

Sol:

Let the coordinates of the point on y-axis be (0, y).
From the given information, we have:
${\displaystyle \sqrt{{(0+8)}^2+{(y-4)}^2}}$ = 10
(0 + 8)² + (y - 4)² = 100
64 + y² + 16 - 8y = 100
y² - 8y - 20 = 0
y² - 10y + 2y - 20 = 0
y(y - 10) + 2(y - 10) = 0
(y - 10)(y + 2) = 0
y = 10, - 2
Thus, the required co-ordinates of the points on y-axis are (0, 10) and (0, -2).

Question 6

A point A is at a distance of ${\displaystyle \sqrt{10}}$ unit from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice its abscissa.

Sol:

It is given that the coordinates of point A are such that its ordinate is twice its abscissa.
So, let the coordinates of point A be (x, 2x).
We have:
${\displaystyle \sqrt{{(x-4)}^2+{(2x-3)}^2}=\sqrt{10}}$
(x - 4)² + (2x - 3)² = 10
x² + 16 - 8x + 4x² + 9 - 12x = 10
5x² - 20x + 15 = 10
x² - 4x + 3 = 0
x² - x - 3x + 3 = 0
x(x - 1) -3(x - 1) = 0
(x - 1)(x - 3) = 0
x = 1, 3
Thus, the co-ordinates of the point A are (1, 2) and (3, 6).

Question 7

A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.

Sol:

We know that the distance between the two points (x₁​, y₁​) and (x₂​, y₂​) is

d = ${\displaystyle \sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}}$

Let the given points be A = (a, 7) and B = (−3, a) and the third point given is P(2, −1).

We first find the distance between P(2, −1) and A =(a, 7) as follows:

PA = ${\displaystyle \sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}}$

​= ${\displaystyle \sqrt{{(a-2)}^2+{(7-(-1))}^2}}$

​= ${\displaystyle \sqrt{{(a-2)}^2+{(7+1)}^2}}$

​= ${\displaystyle \sqrt{{(a-2)}^2+8^2}}$

= ${\displaystyle \sqrt{{(a-2)}^2+64}}$​ 

Similarly, the distance between P(2,−1) and B = (−3, a) is:

PB = ${\displaystyle \sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}}$

= ${\displaystyle \sqrt{{(-3-2)}^2+{(a-(-1))}^2}}$

​= ${\displaystyle \sqrt{{(-5)}^2+{(a+1)}^2}}$

​= ${\displaystyle \sqrt{25+{(a+1)}^2}}$ 

Since the point P(2,−1) is equidistant from the points A(a, 7) and B = (−3, a), therefore, PA = PB that is:

${\displaystyle \sqrt{{(a-2)}^2+64}=\sqrt{25+{(a+1)}^2}}$

⇒ ${\displaystyle \left(\sqrt{{(a-2)}^2+64{}^2}\right)=(\sqrt{25+{({(a+1)}^2)}^2}}$

⇒ (a − 2)² + 64 = 25 + (a + 1)²

⇒ (a − 2)² − (a + 1)² = 25 − 64

⇒ (a² + 4 − 4a) − (a² + 1 + 2a) = −39   ...( ∵ (a − b)² = a² + b² − 2ab, (a + b)² = a² + b² + 2ab)

⇒ a² + 4 − 4a − a² − 1 − 2a = −39

⇒ −6a + 3 = −39

⇒− 6a =−39 −3

⇒ −6a = −42

⇒ a =${\displaystyle \frac{-42}{-6}}$ ​=7

Hence, a = 7.

Question 8

What point on the x-axis is equidistant from the points (7, 6) and (-3, 4)?

Sol:

Let the co-ordinates of the required point on x-axis be P (x, 0).
The given points are A (7, 6) and B (-3, 4).
Given, PA = PB
PA² = PB²
(x - 7)² + (0 - 6)² = (x + 3)² + (0 - 4)²
x² + 49 - 14x + 36 = x² + 9 + 6x + 16
60 = 20x
x = 3
Thus, the required point is (3, 0).

Question 9

Find a point on the y-axis which is equidistant from the points (5, 2) and (-4, 3).

Sol:

Let the co-ordinates of the required point on y-axis be P (0, y).
The given points are A (5, 2) and B (-4, 3).
Given, PA = PB
PA² = PB²
(0 -5)² + (y -2)² = (0 + 4)² + (y - 3)²
25 + y² + 4 - 4y = 16 + y² + 9 - 6y
2y = -4
y = -2
Thus, the required point is (0, -2).

Question 10.1

A point P lies on the x-axis and another point Q lies on the y-axis.
Write the ordinate of point P.

Sol:

Since, the point P lies on the x-axis, its ordinate is 0.

Question 10.2

A point P lies on the x-axis and another point Q lies on the y-axis.
Write the abscissa of point Q.

Sol:

Since, the point Q lies on the y-axis, its abscissa is 0.

Question 10.3

A point P lies on the x-axis and another point Q lies on the y-axis.
If the abscissa of point P is -12 and the ordinate of point Q is -16; calculate the length of line segment PQ.

Sol:

The co-ordinates of P and Q are (-12, 0) and (0, -16) respectively.
PQ = ${\displaystyle \sqrt{{(-12-0)}^2+{(0+16)}^2}}$
= ${\displaystyle \sqrt{144+256}}$
= ${\displaystyle \sqrt{400}}$
= 20

Question 11

Show that the points P (0, 5), Q (5, 10) and R (6, 3) are the vertices of an isosceles triangle.

Sol:

PQ = ${\displaystyle \sqrt{{(5-0)}^2+{(10-5)}^2}}$
= ${\displaystyle \sqrt{25+25}}$
= ${\displaystyle \sqrt{50}}$
= 5${\displaystyle \sqrt{2}}$

QR = ${\displaystyle \sqrt{{(6-5)}^2+{(3-10)}^2}}$
= ${\displaystyle \sqrt{1+49}}$
= ${\displaystyle \sqrt{50}}$
= 5${\displaystyle \sqrt{2}}$

RP = ${\displaystyle \sqrt{{(0-6)}^2+{(5-3)}^2}}$
= ${\displaystyle \sqrt{36+4}}$
= ${\displaystyle \sqrt{40}}$
= 2${\displaystyle \sqrt{10}}$

Since, PQ = QR, ΔPQR is an isosceles triangle.

Question 12

PQ = ${\displaystyle \sqrt{{(5-0)}^2+{(10-5)}^2}}$
= ${\displaystyle \sqrt{25+25}}$
= ${\displaystyle \sqrt{50}}$
= 5${\displaystyle \sqrt{2}}$

QR = ${\displaystyle \sqrt{{(6-5)}^2+{(3-10)}^2}}$
= ${\displaystyle \sqrt{1+49}}$
= ${\displaystyle \sqrt{50}}$
= 5${\displaystyle \sqrt{2}}$

RP = ${\displaystyle \sqrt{{(0-6)}^2+{(5-3)}^2}}$
= ${\displaystyle \sqrt{36+4}}$
= ${\displaystyle \sqrt{40}}$
= 2${\displaystyle \sqrt{10}}$

Since, PQ = QR, ΔPQR is an isosceles triangle.

Sol:

PQ = ${\displaystyle \sqrt{{(6-0)}^2+{(2+4)}^2}=6\sqrt{2}\text{units}}$

QR = ${\displaystyle \sqrt{{(6-3)}^2+{(2-5)}^2}=3\sqrt{2}\text{units}}$

RS = ${\displaystyle \sqrt{{(3+3)}^2+{(5+1)}^2}=6\sqrt{2}\text{units}}$

PS = ${\displaystyle \sqrt{{(-3-0)}^2+{(-1+4)}^2}=3\sqrt{2}\text{units}}$

PR = ${\displaystyle \sqrt{{(3-0)}^2+{(5+4)}^2}=3\sqrt{10}\text{units}}$

QS = ${\displaystyle \sqrt{{(6+3)}^2+{(2+1)}^2}=3\sqrt{10}\text{units}}$

∵ PQ = RS and QR = PS,
Also PR = QS
∴ PQRS is a rectangle.

Question 13

Prove that the points A (1, -3), B (-3, 0) and C (4, 1) are the vertices of an isosceles right-angled triangle. Find the area of the triangle.

Sol:

AB =${\displaystyle \sqrt{{(-3-1)}^2+{(0+3)}^2}=\sqrt{16+9}=\sqrt{25}}$ = 5

BC =${\displaystyle \sqrt{{(4+3)}^2+{(1+0)}^2}=\sqrt{49+1}=\sqrt{50}=5\sqrt{2}}$

CA =${\displaystyle \sqrt{{(1-4)}^2+{(-3-1)}^2}=\sqrt{9+16}=\sqrt{25}}$ = 5

∵ AB = CA
A, B, C are the vertices of an isosceless triangle.
AB² + CA²  = 25 + 25 = 50

BC²  = ${\displaystyle {\left(5\sqrt{2}\right)}^2}$ = 50

∴ AB² + CA² = BC²

Hence, A, B., C are the vertices of a right - angled triangle.

Hence, ΔABC is an isosceles right-angled triangle.

Area of ΔABC = ${\displaystyle \frac{1}{2}\times \text{AB}\times \text{CA}}$

= ${\displaystyle \frac{1}{2}\times 5\times 5}$

= 12.5 sq.units

Question 14

Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.

Sol:

AB = ${\displaystyle \sqrt{{(1-5)}^2+{(5-6)}^2}=\sqrt{16+1}=\sqrt{17}}$

BC = ${\displaystyle \sqrt{{(2-1)}^2+{(1-5)}^2}=\sqrt{1+16}=\sqrt{17}}$

CD = = ${\displaystyle \sqrt{{(6-2)}^2+{(2-1)}^2}=\sqrt{16+1}=\sqrt{17}}$

DA = = ${\displaystyle \sqrt{{(5-6)}^2+{(6-2)}^2}=\sqrt{1+16}=\sqrt{17}}$

AC = = ${\displaystyle \sqrt{{(2-5)}^2+{(1-6)}^2}=\sqrt{9+25}=\sqrt{34}}$

BD = = ${\displaystyle \sqrt{{(6-1)}^2+{(2-5)}^2}=\sqrt{25+9} =\sqrt{34}}$

Since, AB = BC = CD = DA and AC = BD,

A, B, C and D are the vertices of a square.

Question 15

Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.

Sol:

Let the given points be A (-3, 2), B (-5, -5), C (2, -3) and D (4, 4).

AB = ${\displaystyle \sqrt{{(-5+3)}^2+{(-5-2)}^2}=\sqrt{4+49}=\sqrt{53}}$

BC = = ${\displaystyle \sqrt{{(2+5)}^2+{(-3+5)}^2}=\sqrt{49+4}=\sqrt{53}}$

CD= = ${\displaystyle \sqrt{{(4-2)}^2+{(4+3)}^2}=\sqrt{4+49}=\sqrt{53}}$

DA = ${\displaystyle \sqrt{{(-3-4)}^2+{(2-4)}^2}=\sqrt{49+4}=\sqrt{53}}$

AC =${\displaystyle \sqrt{{(2+3)}^2+{(-3-2)}^2}=\sqrt{25+25}=5\sqrt{2}}$

BD =${\displaystyle \sqrt{{(4+5)}^2+{(4+5)}^2}=\sqrt{81+81}=9\sqrt{2}}$

Since, AB = BC = CD = DA and AC ≠ BD

The given vertices are the vertices of a rhombus.

Question 16

Points A (-3, -2), B (-6, a), C (-3, -4) and D (0, -1) are the vertices of quadrilateral ABCD; find a if 'a' is negative and AB = CD.

Sol:

AB = CD
AB² = CD²
(- 6 + 3)² + (a + 2)² = (0 + 3)² + (- 1 +  4)²
9 + a² + 4 + 4a = 9 + 9
a² + 4a - 5 = 0
a² - a + 5a - 5 = 0
a(a - 1) + 5 (a - 1) = 0
(a - 1) (a + 5) = 0
a = 1 or - 5
It is given that a is negative, thus the value of a is - 5.

Question 17

The vertices of a triangle are (5, 1), (11, 1) and (11, 9). Find the co-ordinates of the circumcentre of the triangle.

Sol:

Let the circumcentre be P (x, y).
Then, PA = PB
PA² = PB²
(x - 5)² + (y - 1)² = (x - 11)² + (y - 1)²
x² + 25 - 10x = x² + 121 - 22x
12x = 96
x = 8

Also, PA = PC
PA² = PC²
(x - 5)² + (y - 1)² = (x - 11)² + (y - 9)²
x² + 25 - 10x + y² + 1 - 2y = x² + 121 - 22x + y² + 81 - 18y
12x + 16y = 176
3x + 4y = 44
24 + 4y = 44
4y = 20
y = 5
Thus, the coordinates of the circumcentre of the triangle are (8, 5).

Question 18

Given A = (3, 1) and B = (0, y - 1). Find y if AB = 5.

Sol:

AB = 5
AB² = 25
(0 - 3)² + (y - 1 - 1)² = 25
9 + y² + 4 - 4y = 25
y² - 4y - 12 = 0
y² - 6y + 2y - 12 = 0
y(y - 6) + 2(y - 6) = 0
(y - 6) (y + 2) = 0
y = 6, -2

Question 19

Given A = (x + 2, -2) and B (11, 6). Find x if AB = 17.

Sol:

AB = 17
AB² = 289
(11 - x - 2)² + (6 + 2)² = 289
x² + 81 - 18x + 64 = 289
x² - 18x - 144 = 0
x² - 24x + 6x - 144 = 0
x(x - 24) + 6(x - 24) = 0
(x - 24) (x + 6) = 0
x = 24, -6

Question 20

The centre of a circle is (2x - 1, 3x + 1). Find x if the circle passes through (-3, -1) and the length of its diameter is 20 unit.

Sol:

Distance between the points A (2x - 1, 3x + 1) and B (- 3, - 1) = Radius of circle
AB = 10 (Since, diameter = 20 units, given)
AB² = 100
(-3 - 2x + 1)² + (-1 - 3x - 1)² = 100

(-2 - 2x)² + (-2 - 3x)² = 100

4 + 4x² + 8x + 4 + 9x² + 12x = 100

13x² + 20x - 92 = 0

x = ${\displaystyle \frac{-20\pm \sqrt{400+4784}}{26}}$

x = ${\displaystyle \frac{-20\pm 72}{26}}$

x = 2, - ${\displaystyle \frac{46}{13}}$.

QUestion 21

The length of line PQ is 10 units and the co-ordinates of P are (2, -3); calculate the co-ordinates of point Q, if its abscissa is 10.

Sol:

Let the co-ordinates of point Q be (10, y).
PQ = 10
PQ² = 100
(10 - 2)² + (y + 3)² = 100
64 + y² + 9 + 6y = 100
y² + 6y - 27 = 0
y² + 9y - 3y - 27 = 0
y(y + 9) - 3(y + 9) = 0
(y + 9) (y - 3) = 0
y = -9, 3
Thus, the required co-ordinates of point Q are (10, -9) and (10, 3).

QUestion 22.1

Point P (2, -7) is the center of a circle with radius 13 unit, PT is perpendicular to chord AB and T = (-2, -4); calculate the length of: AT 

Sol:

Given, radius = 13 units
PA = PB = 13 units
Using distance formula,
PT = ${\displaystyle \sqrt{{(-2-2)}^2 +{(-4+7)}^2}}$
= ${\displaystyle \sqrt{16+9}}$
= ${\displaystyle \sqrt{25}}$
= 5
Using Pythagoras theorem in Δ PAT,
AT² = PA² - PT² 
AT² = 169 - 25
AT² = 144
AT = 12 units.

Question 22.2

Point P (2, -7) is the centre of a circle with radius 13 unit, PT is perpendicular to chord AB and T = (-2, -4); calculate the length of AB.

Sol:

We know that the perpendicular from the center of a circle to a chord bisects the chord.
∴ AB = 2AT
= 2 x 12 units
= 24 units.

Question 23

Calculate the distance between the points P (2, 2) and Q (5, 4) correct to three significant figures.

Sol:

PQ = ${\displaystyle \sqrt{{(5-2)}^2+{(4-2)}^2}}$

= ${\displaystyle \sqrt{9+4}}$

= ${\displaystyle \sqrt{13}}$

= 3.6055

= 3.61 units

Question 24

Calculate the distance between A (7, 3) and B on the x-axis whose abscissa is 11.

Sol:

We know that any point on x-axis has coordinates of the form (x, 0).
Abscissa of point B = 11
Since, B lies of x-axis, so its co-ordinates are (11, 0).
AB = ${\displaystyle \sqrt{{(11-7)}^2+{(0-3)}^2}}$
= ${\displaystyle \sqrt{16+9}}$
= ${\displaystyle \sqrt{25}}$
= 5 units

Question 25

Calculate the distance between A (5, -3) and B on the y-axis whose ordinate is 9.

' Sol:

We know that any point on y-axis has coordinates of the form (0, y).
Ordinate of point B = 9
Since, B lies of y-axis, so its co-ordinates are (0, 9).
AB = ${\displaystyle \sqrt{{(0-5)}^2+{(9+3)}^2}}$
= ${\displaystyle \sqrt{25+144}}$
= ${\displaystyle \sqrt{169}}$
= 13 units

Question 26

Find the point on y-axis whose distances from the points A (6, 7) and B (4, -3) are in the ratio 1: 2.

Sol:

Let the required point on y-axis be P (0, y).
PA = ${\displaystyle \sqrt{{(0-6)}^2+{(y-7)}^2}}$
= ${\displaystyle \sqrt{36+y^2+49-14y}}$
= ${\displaystyle \sqrt{y^2-14y+85}}$
PB = ${\displaystyle \sqrt{{(0-4)}^2+{(y+3)}^2}}$
= ${\displaystyle \sqrt{16+y^2+9+6y}}$
= ${\displaystyle \sqrt{y^2+6y+25}}$
From the given information, we have:
${\displaystyle \frac{\text{PA}}{\text{PB}}=\frac{1}{2}}$
${\displaystyle \frac{{\text{PA}}^2}{{\text{PB}}^2}=\frac{1}{4}}$
${\displaystyle \frac{y^2-14y+85}{y^2+6y+25}=\frac{1}{4}}$
4y² - 56y + 340
= y² + 6y + 25
3y² - 62y + 315
= 0
y = ${\displaystyle \frac{62\pm \sqrt{3844-3780}}{6}}$
y = ${\displaystyle \frac{62\pm 8}{6}}$
y = ${\displaystyle 9,\frac{35}{3}}$
Thus, the required points on y-axis are (0, 9) and ${\displaystyle (0,\frac{35}{3})}$.

Question 27

The distances of point P (x, y) from the points A (1, - 3) and B (- 2, 2) are in the ratio 2: 3.
Show that: 5x² + 5y² - 34x + 70y + 58 = 0.

Sol:

It is given that PA: PB = 2: 3

${\displaystyle \frac{\text{PA}}{\text{PB}}=\frac{2}{3}}$

${\displaystyle \frac{{\text{PA}}^2}{{\text{PB}}^2}=\frac{4}{9}}$

${\displaystyle \frac{{(x-1)}^2+{(y+3)}^2}{{(x+2)}^2+{(y-2)}^2}=\frac{4}{9}}$

${\displaystyle \frac{x^2+1-2x+y^2+9+6y}{x^2+4+4x+y^2+4-4y}=\frac{4}{9}}$

9(x² - 2x + y² + 10 + 6y) = 4(x² + 4x + y² + 8 - 4y)

9x² - 18x + 9y² + 90 + 54y = 4x² + 16x + 4y² + 32 - 16y

5x² + 5y² - 34x + 70y + 58 = 0

Hence, proved.

Question 28

The points A (3, 0), B (a, -2) and C (4, -1) are the vertices of triangle ABC right angled at vertex A. Find the value of a.

Sol:

AB = ${\displaystyle \sqrt{{(\text{a}-3)}^2+{(-2-0)}^2}}$
= ${\displaystyle \sqrt{{\text{a}}^2+9-6\text{a}+4}}$
= ${\displaystyle \sqrt{{\text{a}}^2-6\text{a}+13}}$

BC = ${\displaystyle \sqrt{{(4-\text{a})}^2+{(-1+2)}^2}}$
= ${\displaystyle \sqrt{{\text{a}}^2+16-8\text{a}+1}}$
= ${\displaystyle \sqrt{{\text{a}}^2-8\text{a}+17}}$

CA = ${\displaystyle \sqrt{{(3-4)}^2+{(0+1)}^2}}$
= ${\displaystyle \sqrt{1+1}}$
= ${\displaystyle \sqrt{2}}$

Since, triangle ABC is a right-angled at A, we have:

AB² + AC² = BC²
⇒ a² - 6a + 13 + 2 = a² - 8a + 17
⇒ 2a = 2
⇒ a = 1