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Co-Ordinate system

88426 Lines represented by $4 x^{2} - y^{2} - 8 x + 4 y = 0$ intersect each other at

1

(0, 0)

2

(2, 1)

3

(2, 4)

4

(2, 2)

Explanation:

(D) : Given the equation,
$4 x^{2} - y^{2} - 8 x + 4 y = 0$
Comparing the equation, $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y$ $+ c = 0$
Then, $a = 2, b = - 1$
$g = - 4, f = 2, h = 0, c = 0$
Point of intersection $=$
$(\frac{hf - bg}{ab - h^{2}}, \frac{gh - af}{ab - h^{2}}) = (\frac{0 - 4}{- 2}, \frac{0 - 4}{- 2}) \equiv (2, 2)$

MHT CET-2019

Co-Ordinate system

88427 If lines represented by equation $p^{2} - q y^{2} = 0$ are distinct, then

1

pq > 0

2

pq < 0

3

pq = 0

4

p + q = 0

Explanation:

(A) : Given,
${px}^{2} - {qy}^{2} = 0$ and comparing with ${ax}^{2} + 2 hxy + {by}^{2} = 0$ ,
we get $a = p, b = - q, h = 0$
We know that line are real and distinct if $h^{2} - a b > 0$
$\Rightarrow 0 + pq > 0 \Rightarrow pq > 0$

MHT CET-2017

Co-Ordinate system

88416 If $(- 4, 0)$ and $(1, - 1)$ are two vertices of a triangle of area 4 square units, then its third vertex lies on

1

y = x

2

5 x + y + 12 = 0

3

x + 5 y + 12 = 0

4 None of these

Explanation:

(C) : Let the third vertex of triangle be (h, k).
Area of triangle
$= \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$
$4 = \frac{1}{2} | - 4 (- 1 - k) + 1 (k - 0) + h (0 + 1) |$
$4 = \frac{1}{2} | (5 k + h + 4) |$
$= 5 k + h + 4 = \pm 8$
$h + 5 k + 12 = 0$ or $h + 5 k - 4 = 0$
Hence, third vertex of triangle lies on either
$x + 5 y + 12 = 0$ or $x + 5 y - 4 = 0$

SRM JEE-2011

Co-Ordinate system

88417 The joint equation of two lines through the origin each making an angle of $30^{\circ}$ with the $Y$ axis is

1

2 x^{2} - 3 y^{2} = 0

2

x^{2} + 3 y^{2} = 0

3

3 x^{2} - y^{2} = 0

4

x^{2} - 3 y^{2} = 0

Explanation:

(C) : From diagram

$∴$ Slope of lines are $= m_{1} = \tan 60^{\circ} = \sqrt{3}$ and
$m_{2} = \tan 120^{\circ} = \tan (90^{\circ} + 30^{\circ}) = - \sqrt{3}$
$y = \sqrt{3} x$ and $y = - \sqrt{3} x$
Their equations are $y - \sqrt{3} x = 0$ and $y + \sqrt{3} x = 0$
Joint equation is $(\sqrt{3} x - y) (\sqrt{3} x + y) = 0 \Rightarrow 3 x^{2} - y^{2} = 0$

MHT CET-2020

Co-Ordinate system

88418 The straight lines represented by the equation $9 x^{2} - 12 x y + 4 y^{2} = 0$ are

1 Intersect at

60^{\circ}

2 Perpendicular

3 Coincident

4 Parallel

Explanation:

(C) : Given,
$9 x^{2} - 12 x y + 4 y^{2} = 0$
The given pair of lines are of the form
${ax}^{2} + 2 hxy + {by}^{2} = 0$
$a = 9, 2 h = - 12$
$h = - 6, b = 4$
$h^{2} - ab = (- 6)^{2} - 36 = 0$
$∴$ The lines are coincident.

MHT CET-2020

Co-Ordinate system

88426 Lines represented by $4 x^{2} - y^{2} - 8 x + 4 y = 0$ intersect each other at

1

(0, 0)

2

(2, 1)

3

(2, 4)

4

(2, 2)

Explanation:

(D) : Given the equation,
$4 x^{2} - y^{2} - 8 x + 4 y = 0$
Comparing the equation, $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y$ $+ c = 0$
Then, $a = 2, b = - 1$
$g = - 4, f = 2, h = 0, c = 0$
Point of intersection $=$
$(\frac{hf - bg}{ab - h^{2}}, \frac{gh - af}{ab - h^{2}}) = (\frac{0 - 4}{- 2}, \frac{0 - 4}{- 2}) \equiv (2, 2)$

MHT CET-2019

Co-Ordinate system

88427 If lines represented by equation $p^{2} - q y^{2} = 0$ are distinct, then

1

pq > 0

2

pq < 0

3

pq = 0

4

p + q = 0

Explanation:

(A) : Given,
${px}^{2} - {qy}^{2} = 0$ and comparing with ${ax}^{2} + 2 hxy + {by}^{2} = 0$ ,
we get $a = p, b = - q, h = 0$
We know that line are real and distinct if $h^{2} - a b > 0$
$\Rightarrow 0 + pq > 0 \Rightarrow pq > 0$

MHT CET-2017

Co-Ordinate system

88416 If $(- 4, 0)$ and $(1, - 1)$ are two vertices of a triangle of area 4 square units, then its third vertex lies on

1

y = x

2

5 x + y + 12 = 0

3

x + 5 y + 12 = 0

4 None of these

Explanation:

(C) : Let the third vertex of triangle be (h, k).
Area of triangle
$= \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$
$4 = \frac{1}{2} | - 4 (- 1 - k) + 1 (k - 0) + h (0 + 1) |$
$4 = \frac{1}{2} | (5 k + h + 4) |$
$= 5 k + h + 4 = \pm 8$
$h + 5 k + 12 = 0$ or $h + 5 k - 4 = 0$
Hence, third vertex of triangle lies on either
$x + 5 y + 12 = 0$ or $x + 5 y - 4 = 0$

SRM JEE-2011

Co-Ordinate system

88417 The joint equation of two lines through the origin each making an angle of $30^{\circ}$ with the $Y$ axis is

1

2 x^{2} - 3 y^{2} = 0

2

x^{2} + 3 y^{2} = 0

3

3 x^{2} - y^{2} = 0

4

x^{2} - 3 y^{2} = 0

Explanation:

(C) : From diagram

$∴$ Slope of lines are $= m_{1} = \tan 60^{\circ} = \sqrt{3}$ and
$m_{2} = \tan 120^{\circ} = \tan (90^{\circ} + 30^{\circ}) = - \sqrt{3}$
$y = \sqrt{3} x$ and $y = - \sqrt{3} x$
Their equations are $y - \sqrt{3} x = 0$ and $y + \sqrt{3} x = 0$
Joint equation is $(\sqrt{3} x - y) (\sqrt{3} x + y) = 0 \Rightarrow 3 x^{2} - y^{2} = 0$

MHT CET-2020

Co-Ordinate system

88418 The straight lines represented by the equation $9 x^{2} - 12 x y + 4 y^{2} = 0$ are

1 Intersect at

60^{\circ}

2 Perpendicular

3 Coincident

4 Parallel

Explanation:

(C) : Given,
$9 x^{2} - 12 x y + 4 y^{2} = 0$
The given pair of lines are of the form
${ax}^{2} + 2 hxy + {by}^{2} = 0$
$a = 9, 2 h = - 12$
$h = - 6, b = 4$
$h^{2} - ab = (- 6)^{2} - 36 = 0$
$∴$ The lines are coincident.

MHT CET-2020

Co-Ordinate system

88426 Lines represented by $4 x^{2} - y^{2} - 8 x + 4 y = 0$ intersect each other at

1

(0, 0)

2

(2, 1)

3

(2, 4)

4

(2, 2)

Explanation:

(D) : Given the equation,
$4 x^{2} - y^{2} - 8 x + 4 y = 0$
Comparing the equation, $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y$ $+ c = 0$
Then, $a = 2, b = - 1$
$g = - 4, f = 2, h = 0, c = 0$
Point of intersection $=$
$(\frac{hf - bg}{ab - h^{2}}, \frac{gh - af}{ab - h^{2}}) = (\frac{0 - 4}{- 2}, \frac{0 - 4}{- 2}) \equiv (2, 2)$

MHT CET-2019

Co-Ordinate system

88427 If lines represented by equation $p^{2} - q y^{2} = 0$ are distinct, then

1

pq > 0

2

pq < 0

3

pq = 0

4

p + q = 0

Explanation:

(A) : Given,
${px}^{2} - {qy}^{2} = 0$ and comparing with ${ax}^{2} + 2 hxy + {by}^{2} = 0$ ,
we get $a = p, b = - q, h = 0$
We know that line are real and distinct if $h^{2} - a b > 0$
$\Rightarrow 0 + pq > 0 \Rightarrow pq > 0$

MHT CET-2017

Co-Ordinate system

88416 If $(- 4, 0)$ and $(1, - 1)$ are two vertices of a triangle of area 4 square units, then its third vertex lies on

1

y = x

2

5 x + y + 12 = 0

3

x + 5 y + 12 = 0

4 None of these

Explanation:

(C) : Let the third vertex of triangle be (h, k).
Area of triangle
$= \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$
$4 = \frac{1}{2} | - 4 (- 1 - k) + 1 (k - 0) + h (0 + 1) |$
$4 = \frac{1}{2} | (5 k + h + 4) |$
$= 5 k + h + 4 = \pm 8$
$h + 5 k + 12 = 0$ or $h + 5 k - 4 = 0$
Hence, third vertex of triangle lies on either
$x + 5 y + 12 = 0$ or $x + 5 y - 4 = 0$

SRM JEE-2011

Co-Ordinate system

88417 The joint equation of two lines through the origin each making an angle of $30^{\circ}$ with the $Y$ axis is

1

2 x^{2} - 3 y^{2} = 0

2

x^{2} + 3 y^{2} = 0

3

3 x^{2} - y^{2} = 0

4

x^{2} - 3 y^{2} = 0

Explanation:

(C) : From diagram

$∴$ Slope of lines are $= m_{1} = \tan 60^{\circ} = \sqrt{3}$ and
$m_{2} = \tan 120^{\circ} = \tan (90^{\circ} + 30^{\circ}) = - \sqrt{3}$
$y = \sqrt{3} x$ and $y = - \sqrt{3} x$
Their equations are $y - \sqrt{3} x = 0$ and $y + \sqrt{3} x = 0$
Joint equation is $(\sqrt{3} x - y) (\sqrt{3} x + y) = 0 \Rightarrow 3 x^{2} - y^{2} = 0$

MHT CET-2020

Co-Ordinate system

88418 The straight lines represented by the equation $9 x^{2} - 12 x y + 4 y^{2} = 0$ are

1 Intersect at

60^{\circ}

2 Perpendicular

3 Coincident

4 Parallel

Explanation:

(C) : Given,
$9 x^{2} - 12 x y + 4 y^{2} = 0$
The given pair of lines are of the form
${ax}^{2} + 2 hxy + {by}^{2} = 0$
$a = 9, 2 h = - 12$
$h = - 6, b = 4$
$h^{2} - ab = (- 6)^{2} - 36 = 0$
$∴$ The lines are coincident.

MHT CET-2020

Co-Ordinate system

88426 Lines represented by $4 x^{2} - y^{2} - 8 x + 4 y = 0$ intersect each other at

1

(0, 0)

2

(2, 1)

3

(2, 4)

4

(2, 2)

Explanation:

(D) : Given the equation,
$4 x^{2} - y^{2} - 8 x + 4 y = 0$
Comparing the equation, $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y$ $+ c = 0$
Then, $a = 2, b = - 1$
$g = - 4, f = 2, h = 0, c = 0$
Point of intersection $=$
$(\frac{hf - bg}{ab - h^{2}}, \frac{gh - af}{ab - h^{2}}) = (\frac{0 - 4}{- 2}, \frac{0 - 4}{- 2}) \equiv (2, 2)$

MHT CET-2019

Co-Ordinate system

88427 If lines represented by equation $p^{2} - q y^{2} = 0$ are distinct, then

1

pq > 0

2

pq < 0

3

pq = 0

4

p + q = 0

Explanation:

(A) : Given,
${px}^{2} - {qy}^{2} = 0$ and comparing with ${ax}^{2} + 2 hxy + {by}^{2} = 0$ ,
we get $a = p, b = - q, h = 0$
We know that line are real and distinct if $h^{2} - a b > 0$
$\Rightarrow 0 + pq > 0 \Rightarrow pq > 0$

MHT CET-2017

Co-Ordinate system

88416 If $(- 4, 0)$ and $(1, - 1)$ are two vertices of a triangle of area 4 square units, then its third vertex lies on

1

y = x

2

5 x + y + 12 = 0

3

x + 5 y + 12 = 0

4 None of these

Explanation:

(C) : Let the third vertex of triangle be (h, k).
Area of triangle
$= \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$
$4 = \frac{1}{2} | - 4 (- 1 - k) + 1 (k - 0) + h (0 + 1) |$
$4 = \frac{1}{2} | (5 k + h + 4) |$
$= 5 k + h + 4 = \pm 8$
$h + 5 k + 12 = 0$ or $h + 5 k - 4 = 0$
Hence, third vertex of triangle lies on either
$x + 5 y + 12 = 0$ or $x + 5 y - 4 = 0$

SRM JEE-2011

Co-Ordinate system

88417 The joint equation of two lines through the origin each making an angle of $30^{\circ}$ with the $Y$ axis is

1

2 x^{2} - 3 y^{2} = 0

2

x^{2} + 3 y^{2} = 0

3

3 x^{2} - y^{2} = 0

4

x^{2} - 3 y^{2} = 0

Explanation:

(C) : From diagram

$∴$ Slope of lines are $= m_{1} = \tan 60^{\circ} = \sqrt{3}$ and
$m_{2} = \tan 120^{\circ} = \tan (90^{\circ} + 30^{\circ}) = - \sqrt{3}$
$y = \sqrt{3} x$ and $y = - \sqrt{3} x$
Their equations are $y - \sqrt{3} x = 0$ and $y + \sqrt{3} x = 0$
Joint equation is $(\sqrt{3} x - y) (\sqrt{3} x + y) = 0 \Rightarrow 3 x^{2} - y^{2} = 0$

MHT CET-2020

Co-Ordinate system

88418 The straight lines represented by the equation $9 x^{2} - 12 x y + 4 y^{2} = 0$ are

1 Intersect at

60^{\circ}

2 Perpendicular

3 Coincident

4 Parallel

Explanation:

(C) : Given,
$9 x^{2} - 12 x y + 4 y^{2} = 0$
The given pair of lines are of the form
${ax}^{2} + 2 hxy + {by}^{2} = 0$
$a = 9, 2 h = - 12$
$h = - 6, b = 4$
$h^{2} - ab = (- 6)^{2} - 36 = 0$
$∴$ The lines are coincident.

MHT CET-2020

Co-Ordinate system

88426 Lines represented by $4 x^{2} - y^{2} - 8 x + 4 y = 0$ intersect each other at

1

(0, 0)

2

(2, 1)

3

(2, 4)

4

(2, 2)

Explanation:

(D) : Given the equation,
$4 x^{2} - y^{2} - 8 x + 4 y = 0$
Comparing the equation, $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y$ $+ c = 0$
Then, $a = 2, b = - 1$
$g = - 4, f = 2, h = 0, c = 0$
Point of intersection $=$
$(\frac{hf - bg}{ab - h^{2}}, \frac{gh - af}{ab - h^{2}}) = (\frac{0 - 4}{- 2}, \frac{0 - 4}{- 2}) \equiv (2, 2)$

MHT CET-2019

Co-Ordinate system

88427 If lines represented by equation $p^{2} - q y^{2} = 0$ are distinct, then

1

pq > 0

2

pq < 0

3

pq = 0

4

p + q = 0

Explanation:

(A) : Given,
${px}^{2} - {qy}^{2} = 0$ and comparing with ${ax}^{2} + 2 hxy + {by}^{2} = 0$ ,
we get $a = p, b = - q, h = 0$
We know that line are real and distinct if $h^{2} - a b > 0$
$\Rightarrow 0 + pq > 0 \Rightarrow pq > 0$

MHT CET-2017

Co-Ordinate system

88416 If $(- 4, 0)$ and $(1, - 1)$ are two vertices of a triangle of area 4 square units, then its third vertex lies on

1

y = x

2

5 x + y + 12 = 0

3

x + 5 y + 12 = 0

4 None of these

Explanation:

(C) : Let the third vertex of triangle be (h, k).
Area of triangle
$= \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$
$4 = \frac{1}{2} | - 4 (- 1 - k) + 1 (k - 0) + h (0 + 1) |$
$4 = \frac{1}{2} | (5 k + h + 4) |$
$= 5 k + h + 4 = \pm 8$
$h + 5 k + 12 = 0$ or $h + 5 k - 4 = 0$
Hence, third vertex of triangle lies on either
$x + 5 y + 12 = 0$ or $x + 5 y - 4 = 0$

SRM JEE-2011

Co-Ordinate system

88417 The joint equation of two lines through the origin each making an angle of $30^{\circ}$ with the $Y$ axis is

1

2 x^{2} - 3 y^{2} = 0

2

x^{2} + 3 y^{2} = 0

3

3 x^{2} - y^{2} = 0

4

x^{2} - 3 y^{2} = 0

Explanation:

(C) : From diagram

$∴$ Slope of lines are $= m_{1} = \tan 60^{\circ} = \sqrt{3}$ and
$m_{2} = \tan 120^{\circ} = \tan (90^{\circ} + 30^{\circ}) = - \sqrt{3}$
$y = \sqrt{3} x$ and $y = - \sqrt{3} x$
Their equations are $y - \sqrt{3} x = 0$ and $y + \sqrt{3} x = 0$
Joint equation is $(\sqrt{3} x - y) (\sqrt{3} x + y) = 0 \Rightarrow 3 x^{2} - y^{2} = 0$

MHT CET-2020

Co-Ordinate system

88418 The straight lines represented by the equation $9 x^{2} - 12 x y + 4 y^{2} = 0$ are

1 Intersect at

60^{\circ}

2 Perpendicular

3 Coincident

4 Parallel

Explanation:

(C) : Given,
$9 x^{2} - 12 x y + 4 y^{2} = 0$
The given pair of lines are of the form
${ax}^{2} + 2 hxy + {by}^{2} = 0$
$a = 9, 2 h = - 12$
$h = - 6, b = 4$
$h^{2} - ab = (- 6)^{2} - 36 = 0$
$∴$ The lines are coincident.

MHT CET-2020