QBManager

Straight Line

88805 The point of intersection of lines represented by $x^{2} - y^{2} + x + 3 y - 2 = 0$ is

1

(1, 0)

2

(0, 2)

3

(- \frac{1}{2}, \frac{3}{2})

4

(\frac{1}{2}, \frac{1}{2})

Explanation:

(C) :
We have $x^{2} - y^{2} + x + 3 y - 2 = 0$ , where
$a = 1, h = 0, b = - 1, g = \frac{1}{2}, f = \frac{3}{2}, c = - 2$
Point of intersection is given by $(\frac{hf - bg}{ab - h^{2}}, \frac{gh - af}{ab - h^{2}})$
$\equiv (\frac{0 + \frac{1}{2}}{- 1}, \frac{0 - \frac{3}{2}}{- 1}) \equiv (- \frac{1}{2}, \frac{3}{2})$

MHT CET-2018

Straight Line

88806 Which of the following equation does not represent a pair of lines?

1

x^{2} - x = 0

2

x y - x = 0

3

y^{2} - x + 1 = 0

4

x y + x + y + 1 = 0

Explanation:

(C) :
Check the options,
Option A :
$x^{2} - x = 0 \Rightarrow x (x - 1) = 0 \Rightarrow x = 0$ and $x = 1$ are to lines
Option B :
$xy - x = 0 \Rightarrow x (y - 1) = 0 \Rightarrow x = 0$ and $y - 1$ are two lines
Option D :
$x y + x + y + 1 \Rightarrow x (y + 1) + 1 (y + 1) = 0 \Rightarrow x + 1 = 0$
and $y + 1 = 0$ are two lines
Option c :
$y^{2} - x + 1 = 0 \Rightarrow y^{2} = x - 1 \Rightarrow$ This is equation of parabola
Hence, option (c) does not represent a pair of lines.

MHT CET-2016

Straight Line

88807 The joint equation of lines passing through the origin and trisecting the first quadrant is

1

x^{2} + \sqrt{3} x y - y^{2} = 0

2

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

3

\sqrt{3} x^{2} - 4 x y + \sqrt{3} y^{2} = 0

4

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

Explanation:

(C) :
As given, both lines passing through origin $(0, 0)$ and $θ_{1} = \frac{π}{6}, θ_{2} = \frac{π}{3}$
$∴$ Equation of first line is :
$y - 0 = \tan (\frac{π}{6}) (x - 0)$
$y = \frac{1}{\sqrt{3}} x$

$\sqrt{3} y = x$
$∴ x - \sqrt{3} y = 0$
Equation of second line is :
$y - 0 = \tan (\frac{π}{3}) (x - 0) \Rightarrow y = \sqrt{3} x$
$∴ \sqrt{3} x - y = 0$
Hence, joint equation of these lines is
$(x - \sqrt{3} y) (\sqrt{3} x - y) = 0$
$∴ \sqrt{3} x^{2} - x y - 3 x y + \sqrt{3} y^{2} = 0 \Rightarrow \sqrt{3} x^{2} + \sqrt{3} y^{2} - 4 x y = 0$

MHT CET-2016

Straight Line

88808 If the equation $k x y + 10 x + 4 = 0$ represents a pair of lines then value of $k$ is

1 0

2 15

3 -15

4

\frac{1}{15}

Explanation:

(B) : Given equation is, $kxy + 10 x + 4 = 0$
Comparing given equation with
${ax}^{2} + 2 hxy + {by}^{2} + 2 gx + 2 fy + c = 0$ , we get
$a = 0, 2 h = k, b = 0, 2 g = 10, 2 f = 6, c = 4$
The given equation represents pair of lines, if $a b c + 2 f g h - a f^{2} - b^{2} - c h^{2} = 0$
$(0) (0) (4) + 2 (3) (5) (\frac{k}{2}) - 0 (3)^{2} - 0 (5)^{2} - 4 {(\frac{k}{2})}^{2} = 0$
$0 + 15 k - 0 - 0 - k^{2} = 0 \Rightarrow k^{2} - 15 k = 0 \Rightarrow k = 0 or$
$k = 15$
If $k = 0$ , then the given equation becomes $10 x + 6 y + 4 = 0$ which does not represent a pair of lines.

MHT CET-2013

Straight Line

88809 The joint equation of pair of lines through origin each of which makes an angle of $60^{\circ}$ with the $Y$ -axis is

1

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

2

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

3

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

4

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

Explanation:

(A) :
Let $L_{1}$ and $L_{2}$ be the lines which makes an angle of $60^{\circ}$ with the $Y$ -axis.
$∴$ The inclination of lines $L_{1}$ and $L_{2}$ are $30^{\circ}$ and $150^{\circ}$ respectively.
Their slopes are $m_{1} = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$ and
$m_{2} = \tan 150^{\circ} = \tan (90^{\circ} + 60^{\circ}) = - \cot 60^{\circ} = - \frac{1}{\sqrt{3}}$
$∴$ Their joint equation is $(x - \sqrt{3 y}) (x + \sqrt{3} y) = 0 \Rightarrow x^{2} - 3 y^{2} = 0$

MHT CET-2013

Straight Line

88805 The point of intersection of lines represented by $x^{2} - y^{2} + x + 3 y - 2 = 0$ is

1

(1, 0)

2

(0, 2)

3

(- \frac{1}{2}, \frac{3}{2})

4

(\frac{1}{2}, \frac{1}{2})

Explanation:

(C) :
We have $x^{2} - y^{2} + x + 3 y - 2 = 0$ , where
$a = 1, h = 0, b = - 1, g = \frac{1}{2}, f = \frac{3}{2}, c = - 2$
Point of intersection is given by $(\frac{hf - bg}{ab - h^{2}}, \frac{gh - af}{ab - h^{2}})$
$\equiv (\frac{0 + \frac{1}{2}}{- 1}, \frac{0 - \frac{3}{2}}{- 1}) \equiv (- \frac{1}{2}, \frac{3}{2})$

MHT CET-2018

Straight Line

88806 Which of the following equation does not represent a pair of lines?

1

x^{2} - x = 0

2

x y - x = 0

3

y^{2} - x + 1 = 0

4

x y + x + y + 1 = 0

Explanation:

(C) :
Check the options,
Option A :
$x^{2} - x = 0 \Rightarrow x (x - 1) = 0 \Rightarrow x = 0$ and $x = 1$ are to lines
Option B :
$xy - x = 0 \Rightarrow x (y - 1) = 0 \Rightarrow x = 0$ and $y - 1$ are two lines
Option D :
$x y + x + y + 1 \Rightarrow x (y + 1) + 1 (y + 1) = 0 \Rightarrow x + 1 = 0$
and $y + 1 = 0$ are two lines
Option c :
$y^{2} - x + 1 = 0 \Rightarrow y^{2} = x - 1 \Rightarrow$ This is equation of parabola
Hence, option (c) does not represent a pair of lines.

MHT CET-2016

Straight Line

88807 The joint equation of lines passing through the origin and trisecting the first quadrant is

1

x^{2} + \sqrt{3} x y - y^{2} = 0

2

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

3

\sqrt{3} x^{2} - 4 x y + \sqrt{3} y^{2} = 0

4

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

Explanation:

(C) :
As given, both lines passing through origin $(0, 0)$ and $θ_{1} = \frac{π}{6}, θ_{2} = \frac{π}{3}$
$∴$ Equation of first line is :
$y - 0 = \tan (\frac{π}{6}) (x - 0)$
$y = \frac{1}{\sqrt{3}} x$

$\sqrt{3} y = x$
$∴ x - \sqrt{3} y = 0$
Equation of second line is :
$y - 0 = \tan (\frac{π}{3}) (x - 0) \Rightarrow y = \sqrt{3} x$
$∴ \sqrt{3} x - y = 0$
Hence, joint equation of these lines is
$(x - \sqrt{3} y) (\sqrt{3} x - y) = 0$
$∴ \sqrt{3} x^{2} - x y - 3 x y + \sqrt{3} y^{2} = 0 \Rightarrow \sqrt{3} x^{2} + \sqrt{3} y^{2} - 4 x y = 0$

MHT CET-2016

Straight Line

88808 If the equation $k x y + 10 x + 4 = 0$ represents a pair of lines then value of $k$ is

1 0

2 15

3 -15

4

\frac{1}{15}

Explanation:

(B) : Given equation is, $kxy + 10 x + 4 = 0$
Comparing given equation with
${ax}^{2} + 2 hxy + {by}^{2} + 2 gx + 2 fy + c = 0$ , we get
$a = 0, 2 h = k, b = 0, 2 g = 10, 2 f = 6, c = 4$
The given equation represents pair of lines, if $a b c + 2 f g h - a f^{2} - b^{2} - c h^{2} = 0$
$(0) (0) (4) + 2 (3) (5) (\frac{k}{2}) - 0 (3)^{2} - 0 (5)^{2} - 4 {(\frac{k}{2})}^{2} = 0$
$0 + 15 k - 0 - 0 - k^{2} = 0 \Rightarrow k^{2} - 15 k = 0 \Rightarrow k = 0 or$
$k = 15$
If $k = 0$ , then the given equation becomes $10 x + 6 y + 4 = 0$ which does not represent a pair of lines.

MHT CET-2013

Straight Line

88809 The joint equation of pair of lines through origin each of which makes an angle of $60^{\circ}$ with the $Y$ -axis is

1

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

2

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

3

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

4

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

Explanation:

(A) :
Let $L_{1}$ and $L_{2}$ be the lines which makes an angle of $60^{\circ}$ with the $Y$ -axis.
$∴$ The inclination of lines $L_{1}$ and $L_{2}$ are $30^{\circ}$ and $150^{\circ}$ respectively.
Their slopes are $m_{1} = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$ and
$m_{2} = \tan 150^{\circ} = \tan (90^{\circ} + 60^{\circ}) = - \cot 60^{\circ} = - \frac{1}{\sqrt{3}}$
$∴$ Their joint equation is $(x - \sqrt{3 y}) (x + \sqrt{3} y) = 0 \Rightarrow x^{2} - 3 y^{2} = 0$

MHT CET-2013

Straight Line

88805 The point of intersection of lines represented by $x^{2} - y^{2} + x + 3 y - 2 = 0$ is

1

(1, 0)

2

(0, 2)

3

(- \frac{1}{2}, \frac{3}{2})

4

(\frac{1}{2}, \frac{1}{2})

Explanation:

(C) :
We have $x^{2} - y^{2} + x + 3 y - 2 = 0$ , where
$a = 1, h = 0, b = - 1, g = \frac{1}{2}, f = \frac{3}{2}, c = - 2$
Point of intersection is given by $(\frac{hf - bg}{ab - h^{2}}, \frac{gh - af}{ab - h^{2}})$
$\equiv (\frac{0 + \frac{1}{2}}{- 1}, \frac{0 - \frac{3}{2}}{- 1}) \equiv (- \frac{1}{2}, \frac{3}{2})$

MHT CET-2018

Straight Line

88806 Which of the following equation does not represent a pair of lines?

1

x^{2} - x = 0

2

x y - x = 0

3

y^{2} - x + 1 = 0

4

x y + x + y + 1 = 0

Explanation:

(C) :
Check the options,
Option A :
$x^{2} - x = 0 \Rightarrow x (x - 1) = 0 \Rightarrow x = 0$ and $x = 1$ are to lines
Option B :
$xy - x = 0 \Rightarrow x (y - 1) = 0 \Rightarrow x = 0$ and $y - 1$ are two lines
Option D :
$x y + x + y + 1 \Rightarrow x (y + 1) + 1 (y + 1) = 0 \Rightarrow x + 1 = 0$
and $y + 1 = 0$ are two lines
Option c :
$y^{2} - x + 1 = 0 \Rightarrow y^{2} = x - 1 \Rightarrow$ This is equation of parabola
Hence, option (c) does not represent a pair of lines.

MHT CET-2016

Straight Line

88807 The joint equation of lines passing through the origin and trisecting the first quadrant is

1

x^{2} + \sqrt{3} x y - y^{2} = 0

2

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

3

\sqrt{3} x^{2} - 4 x y + \sqrt{3} y^{2} = 0

4

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

Explanation:

(C) :
As given, both lines passing through origin $(0, 0)$ and $θ_{1} = \frac{π}{6}, θ_{2} = \frac{π}{3}$
$∴$ Equation of first line is :
$y - 0 = \tan (\frac{π}{6}) (x - 0)$
$y = \frac{1}{\sqrt{3}} x$

$\sqrt{3} y = x$
$∴ x - \sqrt{3} y = 0$
Equation of second line is :
$y - 0 = \tan (\frac{π}{3}) (x - 0) \Rightarrow y = \sqrt{3} x$
$∴ \sqrt{3} x - y = 0$
Hence, joint equation of these lines is
$(x - \sqrt{3} y) (\sqrt{3} x - y) = 0$
$∴ \sqrt{3} x^{2} - x y - 3 x y + \sqrt{3} y^{2} = 0 \Rightarrow \sqrt{3} x^{2} + \sqrt{3} y^{2} - 4 x y = 0$

MHT CET-2016

Straight Line

88808 If the equation $k x y + 10 x + 4 = 0$ represents a pair of lines then value of $k$ is

1 0

2 15

3 -15

4

\frac{1}{15}

Explanation:

(B) : Given equation is, $kxy + 10 x + 4 = 0$
Comparing given equation with
${ax}^{2} + 2 hxy + {by}^{2} + 2 gx + 2 fy + c = 0$ , we get
$a = 0, 2 h = k, b = 0, 2 g = 10, 2 f = 6, c = 4$
The given equation represents pair of lines, if $a b c + 2 f g h - a f^{2} - b^{2} - c h^{2} = 0$
$(0) (0) (4) + 2 (3) (5) (\frac{k}{2}) - 0 (3)^{2} - 0 (5)^{2} - 4 {(\frac{k}{2})}^{2} = 0$
$0 + 15 k - 0 - 0 - k^{2} = 0 \Rightarrow k^{2} - 15 k = 0 \Rightarrow k = 0 or$
$k = 15$
If $k = 0$ , then the given equation becomes $10 x + 6 y + 4 = 0$ which does not represent a pair of lines.

MHT CET-2013

Straight Line

88809 The joint equation of pair of lines through origin each of which makes an angle of $60^{\circ}$ with the $Y$ -axis is

1

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

2

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

3

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

4

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

Explanation:

(A) :
Let $L_{1}$ and $L_{2}$ be the lines which makes an angle of $60^{\circ}$ with the $Y$ -axis.
$∴$ The inclination of lines $L_{1}$ and $L_{2}$ are $30^{\circ}$ and $150^{\circ}$ respectively.
Their slopes are $m_{1} = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$ and
$m_{2} = \tan 150^{\circ} = \tan (90^{\circ} + 60^{\circ}) = - \cot 60^{\circ} = - \frac{1}{\sqrt{3}}$
$∴$ Their joint equation is $(x - \sqrt{3 y}) (x + \sqrt{3} y) = 0 \Rightarrow x^{2} - 3 y^{2} = 0$

MHT CET-2013

Straight Line

88805 The point of intersection of lines represented by $x^{2} - y^{2} + x + 3 y - 2 = 0$ is

1

(1, 0)

2

(0, 2)

3

(- \frac{1}{2}, \frac{3}{2})

4

(\frac{1}{2}, \frac{1}{2})

Explanation:

(C) :
We have $x^{2} - y^{2} + x + 3 y - 2 = 0$ , where
$a = 1, h = 0, b = - 1, g = \frac{1}{2}, f = \frac{3}{2}, c = - 2$
Point of intersection is given by $(\frac{hf - bg}{ab - h^{2}}, \frac{gh - af}{ab - h^{2}})$
$\equiv (\frac{0 + \frac{1}{2}}{- 1}, \frac{0 - \frac{3}{2}}{- 1}) \equiv (- \frac{1}{2}, \frac{3}{2})$

MHT CET-2018

Straight Line

88806 Which of the following equation does not represent a pair of lines?

1

x^{2} - x = 0

2

x y - x = 0

3

y^{2} - x + 1 = 0

4

x y + x + y + 1 = 0

Explanation:

(C) :
Check the options,
Option A :
$x^{2} - x = 0 \Rightarrow x (x - 1) = 0 \Rightarrow x = 0$ and $x = 1$ are to lines
Option B :
$xy - x = 0 \Rightarrow x (y - 1) = 0 \Rightarrow x = 0$ and $y - 1$ are two lines
Option D :
$x y + x + y + 1 \Rightarrow x (y + 1) + 1 (y + 1) = 0 \Rightarrow x + 1 = 0$
and $y + 1 = 0$ are two lines
Option c :
$y^{2} - x + 1 = 0 \Rightarrow y^{2} = x - 1 \Rightarrow$ This is equation of parabola
Hence, option (c) does not represent a pair of lines.

MHT CET-2016

Straight Line

88807 The joint equation of lines passing through the origin and trisecting the first quadrant is

1

x^{2} + \sqrt{3} x y - y^{2} = 0

2

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

3

\sqrt{3} x^{2} - 4 x y + \sqrt{3} y^{2} = 0

4

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

Explanation:

(C) :
As given, both lines passing through origin $(0, 0)$ and $θ_{1} = \frac{π}{6}, θ_{2} = \frac{π}{3}$
$∴$ Equation of first line is :
$y - 0 = \tan (\frac{π}{6}) (x - 0)$
$y = \frac{1}{\sqrt{3}} x$

$\sqrt{3} y = x$
$∴ x - \sqrt{3} y = 0$
Equation of second line is :
$y - 0 = \tan (\frac{π}{3}) (x - 0) \Rightarrow y = \sqrt{3} x$
$∴ \sqrt{3} x - y = 0$
Hence, joint equation of these lines is
$(x - \sqrt{3} y) (\sqrt{3} x - y) = 0$
$∴ \sqrt{3} x^{2} - x y - 3 x y + \sqrt{3} y^{2} = 0 \Rightarrow \sqrt{3} x^{2} + \sqrt{3} y^{2} - 4 x y = 0$

MHT CET-2016

Straight Line

88808 If the equation $k x y + 10 x + 4 = 0$ represents a pair of lines then value of $k$ is

1 0

2 15

3 -15

4

\frac{1}{15}

Explanation:

(B) : Given equation is, $kxy + 10 x + 4 = 0$
Comparing given equation with
${ax}^{2} + 2 hxy + {by}^{2} + 2 gx + 2 fy + c = 0$ , we get
$a = 0, 2 h = k, b = 0, 2 g = 10, 2 f = 6, c = 4$
The given equation represents pair of lines, if $a b c + 2 f g h - a f^{2} - b^{2} - c h^{2} = 0$
$(0) (0) (4) + 2 (3) (5) (\frac{k}{2}) - 0 (3)^{2} - 0 (5)^{2} - 4 {(\frac{k}{2})}^{2} = 0$
$0 + 15 k - 0 - 0 - k^{2} = 0 \Rightarrow k^{2} - 15 k = 0 \Rightarrow k = 0 or$
$k = 15$
If $k = 0$ , then the given equation becomes $10 x + 6 y + 4 = 0$ which does not represent a pair of lines.

MHT CET-2013

Straight Line

88809 The joint equation of pair of lines through origin each of which makes an angle of $60^{\circ}$ with the $Y$ -axis is

1

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

2

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

3

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

4

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

Explanation:

(A) :
Let $L_{1}$ and $L_{2}$ be the lines which makes an angle of $60^{\circ}$ with the $Y$ -axis.
$∴$ The inclination of lines $L_{1}$ and $L_{2}$ are $30^{\circ}$ and $150^{\circ}$ respectively.
Their slopes are $m_{1} = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$ and
$m_{2} = \tan 150^{\circ} = \tan (90^{\circ} + 60^{\circ}) = - \cot 60^{\circ} = - \frac{1}{\sqrt{3}}$
$∴$ Their joint equation is $(x - \sqrt{3 y}) (x + \sqrt{3} y) = 0 \Rightarrow x^{2} - 3 y^{2} = 0$

MHT CET-2013

Straight Line

88805 The point of intersection of lines represented by $x^{2} - y^{2} + x + 3 y - 2 = 0$ is

1

(1, 0)

2

(0, 2)

3

(- \frac{1}{2}, \frac{3}{2})

4

(\frac{1}{2}, \frac{1}{2})

Explanation:

(C) :
We have $x^{2} - y^{2} + x + 3 y - 2 = 0$ , where
$a = 1, h = 0, b = - 1, g = \frac{1}{2}, f = \frac{3}{2}, c = - 2$
Point of intersection is given by $(\frac{hf - bg}{ab - h^{2}}, \frac{gh - af}{ab - h^{2}})$
$\equiv (\frac{0 + \frac{1}{2}}{- 1}, \frac{0 - \frac{3}{2}}{- 1}) \equiv (- \frac{1}{2}, \frac{3}{2})$

MHT CET-2018

Straight Line

88806 Which of the following equation does not represent a pair of lines?

1

x^{2} - x = 0

2

x y - x = 0

3

y^{2} - x + 1 = 0

4

x y + x + y + 1 = 0

Explanation:

(C) :
Check the options,
Option A :
$x^{2} - x = 0 \Rightarrow x (x - 1) = 0 \Rightarrow x = 0$ and $x = 1$ are to lines
Option B :
$xy - x = 0 \Rightarrow x (y - 1) = 0 \Rightarrow x = 0$ and $y - 1$ are two lines
Option D :
$x y + x + y + 1 \Rightarrow x (y + 1) + 1 (y + 1) = 0 \Rightarrow x + 1 = 0$
and $y + 1 = 0$ are two lines
Option c :
$y^{2} - x + 1 = 0 \Rightarrow y^{2} = x - 1 \Rightarrow$ This is equation of parabola
Hence, option (c) does not represent a pair of lines.

MHT CET-2016

Straight Line

88807 The joint equation of lines passing through the origin and trisecting the first quadrant is

1

x^{2} + \sqrt{3} x y - y^{2} = 0

2

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

3

\sqrt{3} x^{2} - 4 x y + \sqrt{3} y^{2} = 0

4

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

Explanation:

(C) :
As given, both lines passing through origin $(0, 0)$ and $θ_{1} = \frac{π}{6}, θ_{2} = \frac{π}{3}$
$∴$ Equation of first line is :
$y - 0 = \tan (\frac{π}{6}) (x - 0)$
$y = \frac{1}{\sqrt{3}} x$

$\sqrt{3} y = x$
$∴ x - \sqrt{3} y = 0$
Equation of second line is :
$y - 0 = \tan (\frac{π}{3}) (x - 0) \Rightarrow y = \sqrt{3} x$
$∴ \sqrt{3} x - y = 0$
Hence, joint equation of these lines is
$(x - \sqrt{3} y) (\sqrt{3} x - y) = 0$
$∴ \sqrt{3} x^{2} - x y - 3 x y + \sqrt{3} y^{2} = 0 \Rightarrow \sqrt{3} x^{2} + \sqrt{3} y^{2} - 4 x y = 0$

MHT CET-2016

Straight Line

88808 If the equation $k x y + 10 x + 4 = 0$ represents a pair of lines then value of $k$ is

1 0

2 15

3 -15

4

\frac{1}{15}

Explanation:

(B) : Given equation is, $kxy + 10 x + 4 = 0$
Comparing given equation with
${ax}^{2} + 2 hxy + {by}^{2} + 2 gx + 2 fy + c = 0$ , we get
$a = 0, 2 h = k, b = 0, 2 g = 10, 2 f = 6, c = 4$
The given equation represents pair of lines, if $a b c + 2 f g h - a f^{2} - b^{2} - c h^{2} = 0$
$(0) (0) (4) + 2 (3) (5) (\frac{k}{2}) - 0 (3)^{2} - 0 (5)^{2} - 4 {(\frac{k}{2})}^{2} = 0$
$0 + 15 k - 0 - 0 - k^{2} = 0 \Rightarrow k^{2} - 15 k = 0 \Rightarrow k = 0 or$
$k = 15$
If $k = 0$ , then the given equation becomes $10 x + 6 y + 4 = 0$ which does not represent a pair of lines.

MHT CET-2013

Straight Line

88809 The joint equation of pair of lines through origin each of which makes an angle of $60^{\circ}$ with the $Y$ -axis is

1

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

2

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

3

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

4

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

Explanation:

(A) :
Let $L_{1}$ and $L_{2}$ be the lines which makes an angle of $60^{\circ}$ with the $Y$ -axis.
$∴$ The inclination of lines $L_{1}$ and $L_{2}$ are $30^{\circ}$ and $150^{\circ}$ respectively.
Their slopes are $m_{1} = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$ and
$m_{2} = \tan 150^{\circ} = \tan (90^{\circ} + 60^{\circ}) = - \cot 60^{\circ} = - \frac{1}{\sqrt{3}}$
$∴$ Their joint equation is $(x - \sqrt{3 y}) (x + \sqrt{3} y) = 0 \Rightarrow x^{2} - 3 y^{2} = 0$

MHT CET-2013