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Slope of the line through the origin and (-2, 9)

= \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{9-0}{-2-0} = -\frac{9}{2}\)

Then the slope of the required line is \(\frac{2}{9}\).

Hence the equation is y – y₁ = m(x – x₁)

⇒ y – 9 = \(\frac{2}{9}(x - (-2))\)

⇒ 9y – 81 = 2x + 4 ⇒ 2x – 9x + 85 = 0.

Let equation of the line is \(\frac{x}{a} + \frac{y}{b}\) = 1 ______(1)

Given a + b = 9

Since (1) passes through (2, 2) we have;

\(\frac{2}{b} + \frac{2}{b}\) = 1 

⇒ 2a + 2b – ab 

⇒ 2 (a + b) = ab

⇒ 18 = ab. 

Then the numbers are 3 and 6.

Hence the equation of the line is

\(\frac{x}{3} + \frac{y}{6}\) = 1 

⇒ 6x + 3y = 18 

⇒ 2x + y = 6

OR

\(\frac{x}{3} + \frac{y}{6}\) = 1 

⇒ 3x + 6y = 18 

⇒ x + 2y =6.

The slope of the lines joining the points (x, -1) and (2, 1); (2, 1) and (4, 5) are same.

\(\frac{1-(-1)}{2-x}= \frac{5-1}{4-2}\)

\(\Rightarrow\) \(\frac{2}{2-x} = \frac{4}{2} \)

\(\Rightarrow\) 2 - x = 1 \(\Rightarrow\) x = 1

1. Equation of the parallel line 4x – 3y + k = 0

Passing through (1, 2)

4(1) – 3(2) + k = 0 ⇒ k = 2

⇒ 4x – 3y + 2 = 0.

2. Distance \(|\frac{12-2}{\sqrt{25}}| = \frac{10}{5} = 2.\)

3. 3x + 4y + 9 = 0

The equation of the perpendicular line will be 7x + y + k = 0.

Since x-intercept is 3, the line passes through the point (3, 0). So we have;

7(3) + 0 + k = 0 

⇒ 21 + 0 + k = 0 

⇒ k = -21

Therefore the equation is 7x + y – 21 = 0.

Slope of the line through the points (2, 5) and (-3, 6)

\(\frac{y_2 - y_1}{x_2 -x_1} = \frac{6-(5)}{-3-2} = -\frac{1}{5}\)

Then the slope of the required line is 5. Hence the equation is y – y₁ = m(x – x₁)

⇒ y – 5 = 5(x – (-3)

⇒ y – 5 = 5x + 15 

⇒ 5x – y + 20 = 0.

Since the P(a, b) is the mid-point of the line segment, then the x-intercept and the y-intercept will be 2a and 2b. Hence the equation of the line is

\(\frac{x}{2a} + \frac{y}{2b} = 1\)

\(\Rightarrow\) \(\frac{x}{a} + \frac{y}{b} = 2\)

Let (x, 0) be the points on the x-axis. 

Then the distance will be same;

(x – 3)² + 16 = (x – 7)² + 36

⇒ x² – 6x + 9 + 16 = x² – 14x + 49 + 36

⇒ 14x – 6x = 49 + 36 – 9 – 16

⇒ 8x = 60 ⇒ x = \(\frac{15}{2}\)

Hence the point is (\(\frac{15}{2}\), 0).

The equation of the line parallel to the line 3x – 4y + 2 = 0 is of the form 3x – 4y + k = 0.

Since it passes through (-2, 3), we have;

3(-2) – 4(3) + k = 0 ⇒ -6 – 12 + k = 0

⇒ -18 + k = 0 ⇒ k = 18

Hence the equation is 3x – 4y + 18 = 0.

1. Figure b

2. 3x + 4y = 12

\(\Rightarrow \frac{3x + 4y}{12} = 1\)

\(\Rightarrow\) \(\frac{x}{4} + \frac{y}{3} = 1\)

x – intercept = 4; y – intercept = 3.

1. Slope

\(\frac{y_2 - y_1}{x_2 - x_1} = \frac{4-(-2)}{-1-3}\) = \(-\frac{3}{2}\)

2. Slope

\(\frac{y_2-y_1}{x_2 - x_1}\) = \(\frac{1-(-5)}{2-4}\) = \(-\frac{6}{2}\) = -3

3. Slope

\(\frac{y_2-y_1}{x_2 - x_1}\) = \(\frac{3-(-2)}{4-0} = \frac{5}{4}\)

Let the equation of the circle be

(x – h)² + (y – k)² = r²

center lies on x axis so let the center be (h, 0),

then (x – h)² + y² = 25

Since circle pass through (2, 3) we have;

(2 – h)² + 3² = 25

⇒ (2 – h)² = 16 ⇒ 2 – h = ±4

⇒ h = 6, -2

When h = 6 ; equation of circle is

(x – 6)² + (y – 0)² = 25

⇒ x² + 36 – 12x + y² = 25

⇒ x² + y² – 12x + 11 = 0

When h = -2; equation of circle is

(x + 2)² + (y – 0)² = 25

⇒ x² + 4 + 4x + y² = 25

⇒ x² + y² + 4x – 21 = 0

Major axis lie on the y-axis so the standard equation of the ellipse is of the form

\(\frac{x^2}{a^2}+ \frac{y^2}{b^2} = 1\)

Since the ellipse passes through (3, 2)

\(\frac{9}{a^2} + \frac{4}{b^2} = 1\) .........(1)

Since the ellipse passes through (1, 6)

\(\frac{1}{a^2} + \frac{36}{b^2} = 1\) ......... (2)

Solving (1) and (2), we have Since the ellipse passes through (3, 2)

a² = 40; b² = 10

Thus the equation of the ellipse is

\(\frac{x^2}{40}+ \frac{y^2}{10}\)= 1

1. Mid-point of AB is (2, 1).

2. Slope of line through AB

= \(\frac{2-0}{4-0} = \frac{1}{2}\)

Slope of perpendicular line is – 2

Equation of the perpendicular line to AB is

y – 1 = -2(x – 2) ⇒ 2x + y = 5.

3. The meeting point of perpendicular bisector of AB and AC will be the centre of the circum circle.

The line perpendicular to AC is x = 4

Solving and x = 4

We get y = 5 – 8 = -3 and x = 4

Hence center is (4, -3) and radius is

\(\sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = 5\)

Equation of the circle is

(x – 4)² + (y + 3)² = 5.

Let the equation of the circle be

x² + y² + 2gx + 2fy + c = 0 ______(1)

Since (1) passes through (4, 1)

16 + 1 + 8g + 2f + c = 0

⇒ 8g + 2f + c = -17 ______(2)

Since (1) passes through (6, 5)

36 + 25 + 12g + 10f + c = 0

⇒ 12 g + 10f + c = -61 _______(3)

(1) – (2) ⇒ -4g – 8f = 16

⇒ -g – 2f = 4 ______(4)

Since center is on the line 4x + y = 16, we have;

⇒ -4g – f = 16 ______(5)

Solving (4) and (5) We get; g = -3; f = -4

(2) ⇒ 8(-3) + 2(-4) + c = -17

⇒ -24 – 8 + c = -17 ⇒ c = 15

Then the equation of the circle is

x² + y² – 6x – 8y + 15 = 0.

1. Since 25 > 4 the standard equation of the ellipse is \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\)

⇒ a² = 25; b² = 4

c² = a² – b² = 25 – 4 = 21 

⇒ c = √21

Coordinate of foci are (0, ±√21)

Coordinate of vertex are (0, ±5)

Length of major axis = 2a = 2 × 5 = 10

Length of minor axis = 2b = 2 × 2 = 4

Eccentricity = e = \(\frac{c}{a}\) =\(\frac{\sqrt{21}}{5}\)

Length of latus rectum = \(\frac{2b^2}{a} = \frac{2 \times 4}{5} = \frac{8}{5}.\)

2. Since 16 > 9 the standard equation of the ellipse is \(\frac{x^2}{a} + \frac{y^2}{b^2} = 1\)

⇒ a² = 16; b² = 9

c² = a² – b² = 16 – 9 = 7 

⇒ c = √7

Coordinate of foci are (±√7, 0)

Coordinate of vertex are (±4, 0)

Length of major axis = 2a = 2 × 4 = 8

Length of minor axis = 2b = 2 × 3 = 6

Eccentricity = e = \(\frac{c}{a}\)=\(\frac{\sqrt{7}}{4}\)

Length of latus rectum = \(\frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{9}{2}.\)

1. Foci (±4, 0) lie on the x-axis. So the equation of the ellipse is of the form \(\frac{x^2}{a} + \frac{y^2}{b^2} = 1\)

Given; Vertex (±5, 0) ⇒ a = 5

Given; Foci(±4, 0) Foci ⇒ c = 4 = \(\sqrt{a^2 - b^2}\)

⇒ 4 = \(\sqrt{25-b^2}\) ⇒ 16 = 25 – b² ⇒ b² = 9

Therefore the equation of the ellipse is

\(\frac{x^2}{25} + \frac{y^2}{9} = 1\).

2. The ends of major axis lie on the x-axis. So the equation of the ellipse is of the form 

 \(\frac{x^2}{a^2}+ \frac{y^2}{b^2} = 1\)

Given; Ends of the major

axis (±3, 0) ⇒ a = 3, ends of minor axis

(0, ±2) ⇒ b = 2

Therefore the equation of the ellipse is 

\(\frac{x^2}{9} + \frac{y^2}{4} = 1.\)

3. Since foci (±5, 0) lie on x-axis, the standard form of ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) 

Given; 2a = 26 ⇒ a = 13

Given; c = 5 = \(\sqrt{a^2-b^2}\)

⇒ 25 = 169 – b² ⇒ b² = 144

Therefore the equation of the ellipse is

\(\frac{x^2}{169} + \frac{y^2}{144} = 1.\)

4. The standard form of ellipse is

\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

Given; c = 4 = \(\sqrt{a^2-b^2}\)

⇒ 16 = a² – 9 

⇒ a² =25

Therefore the equation of the ellipse is

\(\frac{x^2}{25} + \frac{y^2}{9} = 1\).

1. Since the focus (6, 0) lies on the x-axis, therefore x-axis is the axis of parabola.

Also the directrix is x = – 6, ie; x = – a And focus (6, 0), ie; (a, 0)

Therefore the equation of the parabola is

y² = 4ax ⇒ y² = 24x.

2. The vertex of the parabola is at (0, 0) and focus is at (3, 0). 

Then axis of parabola is along x-axis. So the parabola is of the form y² = 4ax . 

The equation of the parabola is y² = 12x.

3. The vertex of the parabola is at (0, 0) and the axis is along x-axis. 

So the equation of parabola is of the torn y² = 4ax .

Since the parabola passes through point (2, 3)

Therefore, 3² = 4a × 2 ⇒ a = \(\frac{9}{8}\)

The required equation of the parabola is

y² = 4 × \(\frac{9}{8}\) x ⇒ y² = \(\frac{9}{2}\)x.

Given; y² = 8x, we have 4a = 8 ⇒ a = 2

Focus = (2, 0); Vertex = (0, 0)

Latus rectum = 4a = 8

1. Comparing the equation with the general form we get; 

4a = 20 ⇒ a = 5

Coordinate of focus are (5, 0)

Axis of the parabola is y = 0

Equation of the directrix is x = -5

Length of latus rectum = 4 × 5 = 20.

2. Comparing the equation with the general form we get; 

4a = 8 ⇒ a = 2

Coordinate of focus are (0, 2)

Axis of the parabola is x = 0

Equation of the directrix is y = – 2

Length of latus rectum = 4 × 2 = 8.

3. Convert the equation into general form, we get x² = -5y. 

Comparing the equation with the general form we get;

4a = 5 ⇒ a = \(\frac{5}{4}\)

Coordinate of focus are (0, −\(\frac{5}{4}\))

Axis of the parabola is x = 0

Equation of the directrix is y = \(\frac{5}{4}\)

Length of latus rectum = \(\frac{4 \times 5}{4}\) = 5.

The parabola is y² - kx + 8 = 0

y² = kx - 8

= k(x - 8/k)

= 4AX

where 4A = K

direction of the parabola is X + A = 0

Comparing it with x - 1 = 0 we get

1 = 8/k - K/4

or k² + 4k - 32 = 0

(k + 8)(k - 4) = 0

k + 8 = 0   or  k - 4 = 0

k = -8  or k = 4

Hence, the value of k is 4.

Let the numbers are a and b.

Given that difference between both numbers is 5.

∴ a − b= 5. ... (1)

Given that difference between their squares is 65.

∴ a² − b² = 65.

⇒ (a – b)( a + b) = 65 (∵(a – b)( a + b) = a² − b² )

⇒ 5(a + b) = 65

⇒ a + b = \(\frac{65}{5}=13.\) … (2)

Now, adding equations (1) & (2), we get

(a – b) + ( a + b) = 5 + 13

⇒ 2a = 18

⇒ a = \(\frac{18}{2}=9.\)

Now, putting a = 9 in equation (1), 

we get 9 – b = 5 ⇒ b = 9 – 5 = 4.

Hence, the numbers are 4 and 9.

Correct option is: (d) 3 : 5

Given:

X and Y can finish a work in 9 and 8 days respectively.

X works for 4 days and then left.

Y works for another 3 days and left.

Then Z joined and finished the remaining work in 1 day.

Formula Used:

If A and B can together finish a work in t days and alone can finish the work in a and b days respectively,

Then, 1/a + 1/b = 1/t

Work = Time × Efficiency

Calculation:

LCM of 9, 8 = 72

Efficiency of X = 8 units

Efficiency of Y = 9 units

Work done by X in 4 days = 8 × 4 = 32

Work done by Y in 3 days = 3 × 9 = 27

Remaining work = 72 – 32 – 27 = 13

Z finished the remaining work in 1 day.

Efficiency of Z = 13 units

Time X and Z would take = total work/efficiency of X + Z = 72/(13 + 8) = 72/21 = 24/7 days

∴ Required time is 24/7 days.

1. (c) 9 m

2. (b) quadratic polynomial

3. (c) 1 sec.

4. (b) 2

5. (b) -2, 4

Correct option is: (b) b² - a²

Let the number of bees = x

Then the number of flowes = (x - 1)

If the number of bees sitting on each flower is 2, number of flowers is (x/2 + 1).

⇒ (x/2 + 1) = (x - 1)

⇒  x - x/2 = 2

⇒ x/2 = 2

⇒ x = 4

Therefore, number of bees is 4 and the number of flowers is (4 - 1) i.e. 3.

Correct option is: (d) 7 : 22

Correct option is D) are you going to take these

Correct option is (d) 25/36

\(y = x + \sqrt {x + \sqrt {x + \sqrt {x + \ldots \infty } } } \)

\(\Rightarrow y - x = \sqrt {x + \sqrt {x + \ldots } }\)

\(\Rightarrow y - x = \sqrt y\)

⇒ y = (y –x)²

⇒ y² + x² – 2xy – y = 0         

at x = 2, we get,

y² – 5y + 4 = 0 ⇒ (y - 4) (y - 1) = 0

⇒ y = 4, y = 1

But  At x=2, from the  given equation

 \(y = x + \sqrt {x + \sqrt {x + \sqrt {x + \ldots \infty } } }\),

the value of y will be greater than 2.

So at x=2, y=1 is not possible.

Given system of equation is

2x + 3y = 7 

And (k – 1)x + (k + 2)y = 3k. 

By comparing given system of equation with standard form of system of equations, 

we get a₁ = 2, b₁ = 3, c₁ = 7 

And a₂ = k – 1, b₂ = k + 2 and c₂ = 3k. 

Given that system of equations have infinitely many solutions.

∴ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.\)

Now, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\)

⇒ \(\frac{2}{k-1}=\frac{3}{k+2}\)

⇒ 2k + 4 = 3k – 3 

⇒ 3k – 2k = 4 + 3 

⇒ k = 7.

And \(\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

⇒ \(\frac{3}{k+2}=\frac{7}{3k}\)

⇒ 9k = 7k + 14 

⇒ 2k = 14 

⇒ k = 7.

Hence for k = 7, given system of equations have infinitely many solutions.

Correct option is D) Yes, but Peter can’t

Correct option is C) them

The edge of cubical ice-cream brick is a =22 cm. 

The radius and height of the ice-cream cone are 2cm and 7cm, respectively. 

(a) Surface area of ice-cream cube = 6a² = 6 × 22² = 6 × 484 = 2904 cm² . 

(∵ The surface area of the cube is 6a² ) 

Hence, option (ii) is correct.

(b) The volume of ice-cream cube = a³ = 22³ cm³ . (∵ The volume of the cube is a³) 

Hence, option (iii) is correct. 

(c) The volume of each cone = \(\frac{1}3πr^2h\) = \(\frac{1}3\) x \(\frac{22}7\) x 2 x 2 x 7 = \(\frac{22\times4}3\) = \(\frac{88}3\) cm³

∵ r = 2 cm, h = 7 cm & π = \(\frac{22}7\)

Hence, option (i) is correct. 

(d) Let n number of children will get the ice-cream cones which are filled from cubical ice-brick. 

∴ n × Volume of one cone = Volume of ice-cream cube

⇒ n × \(\frac{88}3\) = 22³

(∵ the volume of ice– cream cube = 22³ & the volume of one ice cone = \(\frac{88}3\) )

⇒ n = \(\frac{22\times22\times22\times3}{22\times4}\) = 11 × 11 × 3 = 363.

∴ Total ice-cones are 363. 

Hence, option (ii) is correct. 

(e) The slant height of the cone is l = \(\sqrt{r^2+h^2}\) = \(\sqrt{2^2+7^2}\) = \(\sqrt{4+49}\) = \(\sqrt{53}\) cm.

Hence, option (i) is correct.

Correct option is B) performance

Correct option is D) for

Correct option is C) one hundred and forty.

Correct option is D) All children

Correct option is C) Yes, it’s a quarter to twelve.

Correct option is A) have to