QBManager

Straight Line

88766 Let $P \equiv (- 1, 0), Q \equiv (0, 0)$ and $R \equiv (3, 3 \sqrt{3})$ be three points. The equation of the bisector of the angle $PQR$ is

1

\sqrt{3} x - y = 0

2

x - \sqrt{3} y = 0

3

\sqrt{3} x + y = 0

4

x + \sqrt{3} y = 0

Explanation:

(C) : Given, $P (- 1, 0), Q (0, 0), R (3, 3 \sqrt{3})$
taking $Q (0, 0)$ and $P (- 1, 0)$ ,
$y - 0 = \frac{0 - 0}{0 + 1} (x + 1)$
$y = 0$
Now, taking $Q (0, 0)$ and $R (3, 3 \sqrt{3})$ ,
$y - 3 \sqrt{3} = \frac{3 \sqrt{3}}{3} (x - 3)$
$y - 3 \sqrt{3} = \sqrt{3} x - 3 \sqrt{3}$
$y = \sqrt{3} x \Rightarrow \sqrt{3} x - y = 0$

Karnataka CET-2010

Straight Line

88765 The quadratic equation whose roots are the $x$ and $y$ intercepts of the line passing through (1, 1) and making a triangle of area $A$ with the coordinate axes is

1

x^{2} + Ax + 2 A = 0

2

x^{2} - 2 Ax + 2 A = 0

3

x^{2} - Ax + 2 A = 0

4 None of these

Explanation:

(B): Equation of the line making intercepts a and
$b$ on the axes is $\frac{x}{a} + \frac{y}{b} = 1$
Since, it passes through $(1, 1)$
$\frac{1}{a} + \frac{1}{b} = 1$
$a + b = a b$
Also the area of the triangle formed by the line and the axes is A.
$∴ \frac{1}{2} ab = A \Rightarrow ab = 2 A$
From equation (i) and (ii), we get, $a + b = 2 A$
Hence, $a$ and $b$ are the roots of the equation $x^{2} - (a + b) x + a b = 0 \Rightarrow x^{2} - 2 A x + 2 A = 0$

BITSAT-2013

Straight Line

88767 If the lines represented by $x^{2} - 2 p x y - y^{2}$ are rotated about the origin through an angle $θ$ , one in clockwise direction and other in anticlockwise direction. Then, the equation of bisectors of the angles between the lines in the new position is

1

p x_{2}^{2} + 2 x y + p y_{2}^{2} = 0

2

p x^{2} - 2 x y + p y^{2} = 0

3

p x^{2} + 2 x y - p y^{2} = 0

4 None of the above

Explanation:

(C) : The given equation is,
$x - 2$ pxy $- y^{2} = 0$
Comparing equation -
we get, $a = 1$
$a x^{2} + 2 h x y + b y^{2} = 0$
$b = - 1$
$h = - p$
Equation of the bisector of angles is,
$\frac{x^{2} - y^{2}}{a - b} = \frac{x y}{h} \Rightarrow \frac{x^{2} - y^{2}}{1 + 1} = \frac{x y}{- p}$
$x^{2} - y^{2} = - \frac{2 x y}{p} \Rightarrow p x^{2} - p y^{2} = - 2 x y$
$p x^{2} + 2 x y - p y^{2} = 0$

JCECE-2012

Straight Line

88768 The equation of the bisector of the angle between the lines $2 x + y - 6 = 0$ and $2 x - 4 y + 7$
$= 0$ which contains the point $(1, 2)$ is

1

6 x - 2 y - 5 = 0

2

2 x + 6 y - 19 = 0

3

6 x + 2 y - 5 = 0

4

2 x + 6 y + 19 = 0

Explanation:

(A) : Given, the equation of lines are,
$2 x + y - 6 = 0$
$2 x - 4 y + 7 = 0$
The equation of the bisector of the angle between the lines $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$
$\frac{a_{1} x + b_{1} y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}} = \pm \frac{a_{2} x + b_{2} y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}$
Then the equation of the bisectors between the lines are,
$\frac{2 x + y - 6}{\sqrt{5}} = \pm \frac{2 x - 4 y + 7}{2 \sqrt{5}}$
$2 x + 6 y - 19 = 0 and 6 x - 2 y - 5 = 0$
Substitute the point $(1, 2)$ On equation of the two given lines are,
$2 + 12 - 19 = - 5 < 0$
$and 6 - 12 - 5 < 0$
$∴$ The equation which is obtained by taking -ve sign Hence, $6 x - 2 y - 5 = 0$

BCECE-2018

Straight Line

88769 The perpendicular bisector of line segment joining the points $P (1, 4)$ and $Q (k, 3)$ has $y$ intercept ' -4 '. Then a possible value of ' $k$ ' among the following is

1 2

2 -2

3 -4

4 -1

Explanation:

(C) : Two lines are perpendicular then $m_{1} m_{2} = - 1$

$m_{2} = \frac{4 - 3}{1 - k} = \frac{1}{1 - k}$
$AB ⊥ PQ$
$m_{1} m_{2} = - 1$
$\Rightarrow m_{1} = - 1 / m^{2}$
$\Rightarrow m_{1} = k - 1$
$C = (\frac{1 + k}{2}, \frac{4 + 3}{2}) = (\frac{1 + k}{2}, \frac{7}{2})$
Equation of line $AB$ ,
$y - \frac{7}{2} = (k - 1) [x - \frac{(1 + k)}{2}]$
at $x = 0, y = - 4$
$k = - 4$

APEAPCET-2021-23.08.2021

Straight Line

88766 Let $P \equiv (- 1, 0), Q \equiv (0, 0)$ and $R \equiv (3, 3 \sqrt{3})$ be three points. The equation of the bisector of the angle $PQR$ is

1

\sqrt{3} x - y = 0

2

x - \sqrt{3} y = 0

3

\sqrt{3} x + y = 0

4

x + \sqrt{3} y = 0

Explanation:

(C) : Given, $P (- 1, 0), Q (0, 0), R (3, 3 \sqrt{3})$
taking $Q (0, 0)$ and $P (- 1, 0)$ ,
$y - 0 = \frac{0 - 0}{0 + 1} (x + 1)$
$y = 0$
Now, taking $Q (0, 0)$ and $R (3, 3 \sqrt{3})$ ,
$y - 3 \sqrt{3} = \frac{3 \sqrt{3}}{3} (x - 3)$
$y - 3 \sqrt{3} = \sqrt{3} x - 3 \sqrt{3}$
$y = \sqrt{3} x \Rightarrow \sqrt{3} x - y = 0$

Karnataka CET-2010

Straight Line

88765 The quadratic equation whose roots are the $x$ and $y$ intercepts of the line passing through (1, 1) and making a triangle of area $A$ with the coordinate axes is

1

x^{2} + Ax + 2 A = 0

2

x^{2} - 2 Ax + 2 A = 0

3

x^{2} - Ax + 2 A = 0

4 None of these

Explanation:

(B): Equation of the line making intercepts a and
$b$ on the axes is $\frac{x}{a} + \frac{y}{b} = 1$
Since, it passes through $(1, 1)$
$\frac{1}{a} + \frac{1}{b} = 1$
$a + b = a b$
Also the area of the triangle formed by the line and the axes is A.
$∴ \frac{1}{2} ab = A \Rightarrow ab = 2 A$
From equation (i) and (ii), we get, $a + b = 2 A$
Hence, $a$ and $b$ are the roots of the equation $x^{2} - (a + b) x + a b = 0 \Rightarrow x^{2} - 2 A x + 2 A = 0$

BITSAT-2013

Straight Line

88767 If the lines represented by $x^{2} - 2 p x y - y^{2}$ are rotated about the origin through an angle $θ$ , one in clockwise direction and other in anticlockwise direction. Then, the equation of bisectors of the angles between the lines in the new position is

1

p x_{2}^{2} + 2 x y + p y_{2}^{2} = 0

2

p x^{2} - 2 x y + p y^{2} = 0

3

p x^{2} + 2 x y - p y^{2} = 0

4 None of the above

Explanation:

(C) : The given equation is,
$x - 2$ pxy $- y^{2} = 0$
Comparing equation -
we get, $a = 1$
$a x^{2} + 2 h x y + b y^{2} = 0$
$b = - 1$
$h = - p$
Equation of the bisector of angles is,
$\frac{x^{2} - y^{2}}{a - b} = \frac{x y}{h} \Rightarrow \frac{x^{2} - y^{2}}{1 + 1} = \frac{x y}{- p}$
$x^{2} - y^{2} = - \frac{2 x y}{p} \Rightarrow p x^{2} - p y^{2} = - 2 x y$
$p x^{2} + 2 x y - p y^{2} = 0$

JCECE-2012

Straight Line

88768 The equation of the bisector of the angle between the lines $2 x + y - 6 = 0$ and $2 x - 4 y + 7$
$= 0$ which contains the point $(1, 2)$ is

1

6 x - 2 y - 5 = 0

2

2 x + 6 y - 19 = 0

3

6 x + 2 y - 5 = 0

4

2 x + 6 y + 19 = 0

Explanation:

(A) : Given, the equation of lines are,
$2 x + y - 6 = 0$
$2 x - 4 y + 7 = 0$
The equation of the bisector of the angle between the lines $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$
$\frac{a_{1} x + b_{1} y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}} = \pm \frac{a_{2} x + b_{2} y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}$
Then the equation of the bisectors between the lines are,
$\frac{2 x + y - 6}{\sqrt{5}} = \pm \frac{2 x - 4 y + 7}{2 \sqrt{5}}$
$2 x + 6 y - 19 = 0 and 6 x - 2 y - 5 = 0$
Substitute the point $(1, 2)$ On equation of the two given lines are,
$2 + 12 - 19 = - 5 < 0$
$and 6 - 12 - 5 < 0$
$∴$ The equation which is obtained by taking -ve sign Hence, $6 x - 2 y - 5 = 0$

BCECE-2018

Straight Line

88769 The perpendicular bisector of line segment joining the points $P (1, 4)$ and $Q (k, 3)$ has $y$ intercept ' -4 '. Then a possible value of ' $k$ ' among the following is

1 2

2 -2

3 -4

4 -1

Explanation:

(C) : Two lines are perpendicular then $m_{1} m_{2} = - 1$

$m_{2} = \frac{4 - 3}{1 - k} = \frac{1}{1 - k}$
$AB ⊥ PQ$
$m_{1} m_{2} = - 1$
$\Rightarrow m_{1} = - 1 / m^{2}$
$\Rightarrow m_{1} = k - 1$
$C = (\frac{1 + k}{2}, \frac{4 + 3}{2}) = (\frac{1 + k}{2}, \frac{7}{2})$
Equation of line $AB$ ,
$y - \frac{7}{2} = (k - 1) [x - \frac{(1 + k)}{2}]$
at $x = 0, y = - 4$
$k = - 4$

APEAPCET-2021-23.08.2021

Straight Line

88766 Let $P \equiv (- 1, 0), Q \equiv (0, 0)$ and $R \equiv (3, 3 \sqrt{3})$ be three points. The equation of the bisector of the angle $PQR$ is

1

\sqrt{3} x - y = 0

2

x - \sqrt{3} y = 0

3

\sqrt{3} x + y = 0

4

x + \sqrt{3} y = 0

Explanation:

(C) : Given, $P (- 1, 0), Q (0, 0), R (3, 3 \sqrt{3})$
taking $Q (0, 0)$ and $P (- 1, 0)$ ,
$y - 0 = \frac{0 - 0}{0 + 1} (x + 1)$
$y = 0$
Now, taking $Q (0, 0)$ and $R (3, 3 \sqrt{3})$ ,
$y - 3 \sqrt{3} = \frac{3 \sqrt{3}}{3} (x - 3)$
$y - 3 \sqrt{3} = \sqrt{3} x - 3 \sqrt{3}$
$y = \sqrt{3} x \Rightarrow \sqrt{3} x - y = 0$

Karnataka CET-2010

Straight Line

88765 The quadratic equation whose roots are the $x$ and $y$ intercepts of the line passing through (1, 1) and making a triangle of area $A$ with the coordinate axes is

1

x^{2} + Ax + 2 A = 0

2

x^{2} - 2 Ax + 2 A = 0

3

x^{2} - Ax + 2 A = 0

4 None of these

Explanation:

(B): Equation of the line making intercepts a and
$b$ on the axes is $\frac{x}{a} + \frac{y}{b} = 1$
Since, it passes through $(1, 1)$
$\frac{1}{a} + \frac{1}{b} = 1$
$a + b = a b$
Also the area of the triangle formed by the line and the axes is A.
$∴ \frac{1}{2} ab = A \Rightarrow ab = 2 A$
From equation (i) and (ii), we get, $a + b = 2 A$
Hence, $a$ and $b$ are the roots of the equation $x^{2} - (a + b) x + a b = 0 \Rightarrow x^{2} - 2 A x + 2 A = 0$

BITSAT-2013

Straight Line

88767 If the lines represented by $x^{2} - 2 p x y - y^{2}$ are rotated about the origin through an angle $θ$ , one in clockwise direction and other in anticlockwise direction. Then, the equation of bisectors of the angles between the lines in the new position is

1

p x_{2}^{2} + 2 x y + p y_{2}^{2} = 0

2

p x^{2} - 2 x y + p y^{2} = 0

3

p x^{2} + 2 x y - p y^{2} = 0

4 None of the above

Explanation:

(C) : The given equation is,
$x - 2$ pxy $- y^{2} = 0$
Comparing equation -
we get, $a = 1$
$a x^{2} + 2 h x y + b y^{2} = 0$
$b = - 1$
$h = - p$
Equation of the bisector of angles is,
$\frac{x^{2} - y^{2}}{a - b} = \frac{x y}{h} \Rightarrow \frac{x^{2} - y^{2}}{1 + 1} = \frac{x y}{- p}$
$x^{2} - y^{2} = - \frac{2 x y}{p} \Rightarrow p x^{2} - p y^{2} = - 2 x y$
$p x^{2} + 2 x y - p y^{2} = 0$

JCECE-2012

Straight Line

88768 The equation of the bisector of the angle between the lines $2 x + y - 6 = 0$ and $2 x - 4 y + 7$
$= 0$ which contains the point $(1, 2)$ is

1

6 x - 2 y - 5 = 0

2

2 x + 6 y - 19 = 0

3

6 x + 2 y - 5 = 0

4

2 x + 6 y + 19 = 0

Explanation:

(A) : Given, the equation of lines are,
$2 x + y - 6 = 0$
$2 x - 4 y + 7 = 0$
The equation of the bisector of the angle between the lines $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$
$\frac{a_{1} x + b_{1} y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}} = \pm \frac{a_{2} x + b_{2} y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}$
Then the equation of the bisectors between the lines are,
$\frac{2 x + y - 6}{\sqrt{5}} = \pm \frac{2 x - 4 y + 7}{2 \sqrt{5}}$
$2 x + 6 y - 19 = 0 and 6 x - 2 y - 5 = 0$
Substitute the point $(1, 2)$ On equation of the two given lines are,
$2 + 12 - 19 = - 5 < 0$
$and 6 - 12 - 5 < 0$
$∴$ The equation which is obtained by taking -ve sign Hence, $6 x - 2 y - 5 = 0$

BCECE-2018

Straight Line

88769 The perpendicular bisector of line segment joining the points $P (1, 4)$ and $Q (k, 3)$ has $y$ intercept ' -4 '. Then a possible value of ' $k$ ' among the following is

1 2

2 -2

3 -4

4 -1

Explanation:

(C) : Two lines are perpendicular then $m_{1} m_{2} = - 1$

$m_{2} = \frac{4 - 3}{1 - k} = \frac{1}{1 - k}$
$AB ⊥ PQ$
$m_{1} m_{2} = - 1$
$\Rightarrow m_{1} = - 1 / m^{2}$
$\Rightarrow m_{1} = k - 1$
$C = (\frac{1 + k}{2}, \frac{4 + 3}{2}) = (\frac{1 + k}{2}, \frac{7}{2})$
Equation of line $AB$ ,
$y - \frac{7}{2} = (k - 1) [x - \frac{(1 + k)}{2}]$
at $x = 0, y = - 4$
$k = - 4$

APEAPCET-2021-23.08.2021

Straight Line

88766 Let $P \equiv (- 1, 0), Q \equiv (0, 0)$ and $R \equiv (3, 3 \sqrt{3})$ be three points. The equation of the bisector of the angle $PQR$ is

1

\sqrt{3} x - y = 0

2

x - \sqrt{3} y = 0

3

\sqrt{3} x + y = 0

4

x + \sqrt{3} y = 0

Explanation:

(C) : Given, $P (- 1, 0), Q (0, 0), R (3, 3 \sqrt{3})$
taking $Q (0, 0)$ and $P (- 1, 0)$ ,
$y - 0 = \frac{0 - 0}{0 + 1} (x + 1)$
$y = 0$
Now, taking $Q (0, 0)$ and $R (3, 3 \sqrt{3})$ ,
$y - 3 \sqrt{3} = \frac{3 \sqrt{3}}{3} (x - 3)$
$y - 3 \sqrt{3} = \sqrt{3} x - 3 \sqrt{3}$
$y = \sqrt{3} x \Rightarrow \sqrt{3} x - y = 0$

Karnataka CET-2010

Straight Line

88765 The quadratic equation whose roots are the $x$ and $y$ intercepts of the line passing through (1, 1) and making a triangle of area $A$ with the coordinate axes is

1

x^{2} + Ax + 2 A = 0

2

x^{2} - 2 Ax + 2 A = 0

3

x^{2} - Ax + 2 A = 0

4 None of these

Explanation:

(B): Equation of the line making intercepts a and
$b$ on the axes is $\frac{x}{a} + \frac{y}{b} = 1$
Since, it passes through $(1, 1)$
$\frac{1}{a} + \frac{1}{b} = 1$
$a + b = a b$
Also the area of the triangle formed by the line and the axes is A.
$∴ \frac{1}{2} ab = A \Rightarrow ab = 2 A$
From equation (i) and (ii), we get, $a + b = 2 A$
Hence, $a$ and $b$ are the roots of the equation $x^{2} - (a + b) x + a b = 0 \Rightarrow x^{2} - 2 A x + 2 A = 0$

BITSAT-2013

Straight Line

88767 If the lines represented by $x^{2} - 2 p x y - y^{2}$ are rotated about the origin through an angle $θ$ , one in clockwise direction and other in anticlockwise direction. Then, the equation of bisectors of the angles between the lines in the new position is

1

p x_{2}^{2} + 2 x y + p y_{2}^{2} = 0

2

p x^{2} - 2 x y + p y^{2} = 0

3

p x^{2} + 2 x y - p y^{2} = 0

4 None of the above

Explanation:

(C) : The given equation is,
$x - 2$ pxy $- y^{2} = 0$
Comparing equation -
we get, $a = 1$
$a x^{2} + 2 h x y + b y^{2} = 0$
$b = - 1$
$h = - p$
Equation of the bisector of angles is,
$\frac{x^{2} - y^{2}}{a - b} = \frac{x y}{h} \Rightarrow \frac{x^{2} - y^{2}}{1 + 1} = \frac{x y}{- p}$
$x^{2} - y^{2} = - \frac{2 x y}{p} \Rightarrow p x^{2} - p y^{2} = - 2 x y$
$p x^{2} + 2 x y - p y^{2} = 0$

JCECE-2012

Straight Line

88768 The equation of the bisector of the angle between the lines $2 x + y - 6 = 0$ and $2 x - 4 y + 7$
$= 0$ which contains the point $(1, 2)$ is

1

6 x - 2 y - 5 = 0

2

2 x + 6 y - 19 = 0

3

6 x + 2 y - 5 = 0

4

2 x + 6 y + 19 = 0

Explanation:

(A) : Given, the equation of lines are,
$2 x + y - 6 = 0$
$2 x - 4 y + 7 = 0$
The equation of the bisector of the angle between the lines $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$
$\frac{a_{1} x + b_{1} y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}} = \pm \frac{a_{2} x + b_{2} y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}$
Then the equation of the bisectors between the lines are,
$\frac{2 x + y - 6}{\sqrt{5}} = \pm \frac{2 x - 4 y + 7}{2 \sqrt{5}}$
$2 x + 6 y - 19 = 0 and 6 x - 2 y - 5 = 0$
Substitute the point $(1, 2)$ On equation of the two given lines are,
$2 + 12 - 19 = - 5 < 0$
$and 6 - 12 - 5 < 0$
$∴$ The equation which is obtained by taking -ve sign Hence, $6 x - 2 y - 5 = 0$

BCECE-2018

Straight Line

88769 The perpendicular bisector of line segment joining the points $P (1, 4)$ and $Q (k, 3)$ has $y$ intercept ' -4 '. Then a possible value of ' $k$ ' among the following is

1 2

2 -2

3 -4

4 -1

Explanation:

(C) : Two lines are perpendicular then $m_{1} m_{2} = - 1$

$m_{2} = \frac{4 - 3}{1 - k} = \frac{1}{1 - k}$
$AB ⊥ PQ$
$m_{1} m_{2} = - 1$
$\Rightarrow m_{1} = - 1 / m^{2}$
$\Rightarrow m_{1} = k - 1$
$C = (\frac{1 + k}{2}, \frac{4 + 3}{2}) = (\frac{1 + k}{2}, \frac{7}{2})$
Equation of line $AB$ ,
$y - \frac{7}{2} = (k - 1) [x - \frac{(1 + k)}{2}]$
at $x = 0, y = - 4$
$k = - 4$

APEAPCET-2021-23.08.2021

Straight Line

88766 Let $P \equiv (- 1, 0), Q \equiv (0, 0)$ and $R \equiv (3, 3 \sqrt{3})$ be three points. The equation of the bisector of the angle $PQR$ is

1

\sqrt{3} x - y = 0

2

x - \sqrt{3} y = 0

3

\sqrt{3} x + y = 0

4

x + \sqrt{3} y = 0

Explanation:

(C) : Given, $P (- 1, 0), Q (0, 0), R (3, 3 \sqrt{3})$
taking $Q (0, 0)$ and $P (- 1, 0)$ ,
$y - 0 = \frac{0 - 0}{0 + 1} (x + 1)$
$y = 0$
Now, taking $Q (0, 0)$ and $R (3, 3 \sqrt{3})$ ,
$y - 3 \sqrt{3} = \frac{3 \sqrt{3}}{3} (x - 3)$
$y - 3 \sqrt{3} = \sqrt{3} x - 3 \sqrt{3}$
$y = \sqrt{3} x \Rightarrow \sqrt{3} x - y = 0$

Karnataka CET-2010

Straight Line

88765 The quadratic equation whose roots are the $x$ and $y$ intercepts of the line passing through (1, 1) and making a triangle of area $A$ with the coordinate axes is

1

x^{2} + Ax + 2 A = 0

2

x^{2} - 2 Ax + 2 A = 0

3

x^{2} - Ax + 2 A = 0

4 None of these

Explanation:

(B): Equation of the line making intercepts a and
$b$ on the axes is $\frac{x}{a} + \frac{y}{b} = 1$
Since, it passes through $(1, 1)$
$\frac{1}{a} + \frac{1}{b} = 1$
$a + b = a b$
Also the area of the triangle formed by the line and the axes is A.
$∴ \frac{1}{2} ab = A \Rightarrow ab = 2 A$
From equation (i) and (ii), we get, $a + b = 2 A$
Hence, $a$ and $b$ are the roots of the equation $x^{2} - (a + b) x + a b = 0 \Rightarrow x^{2} - 2 A x + 2 A = 0$

BITSAT-2013

Straight Line

88767 If the lines represented by $x^{2} - 2 p x y - y^{2}$ are rotated about the origin through an angle $θ$ , one in clockwise direction and other in anticlockwise direction. Then, the equation of bisectors of the angles between the lines in the new position is

1

p x_{2}^{2} + 2 x y + p y_{2}^{2} = 0

2

p x^{2} - 2 x y + p y^{2} = 0

3

p x^{2} + 2 x y - p y^{2} = 0

4 None of the above

Explanation:

(C) : The given equation is,
$x - 2$ pxy $- y^{2} = 0$
Comparing equation -
we get, $a = 1$
$a x^{2} + 2 h x y + b y^{2} = 0$
$b = - 1$
$h = - p$
Equation of the bisector of angles is,
$\frac{x^{2} - y^{2}}{a - b} = \frac{x y}{h} \Rightarrow \frac{x^{2} - y^{2}}{1 + 1} = \frac{x y}{- p}$
$x^{2} - y^{2} = - \frac{2 x y}{p} \Rightarrow p x^{2} - p y^{2} = - 2 x y$
$p x^{2} + 2 x y - p y^{2} = 0$

JCECE-2012

Straight Line

88768 The equation of the bisector of the angle between the lines $2 x + y - 6 = 0$ and $2 x - 4 y + 7$
$= 0$ which contains the point $(1, 2)$ is

1

6 x - 2 y - 5 = 0

2

2 x + 6 y - 19 = 0

3

6 x + 2 y - 5 = 0

4

2 x + 6 y + 19 = 0

Explanation:

(A) : Given, the equation of lines are,
$2 x + y - 6 = 0$
$2 x - 4 y + 7 = 0$
The equation of the bisector of the angle between the lines $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$
$\frac{a_{1} x + b_{1} y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}} = \pm \frac{a_{2} x + b_{2} y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}$
Then the equation of the bisectors between the lines are,
$\frac{2 x + y - 6}{\sqrt{5}} = \pm \frac{2 x - 4 y + 7}{2 \sqrt{5}}$
$2 x + 6 y - 19 = 0 and 6 x - 2 y - 5 = 0$
Substitute the point $(1, 2)$ On equation of the two given lines are,
$2 + 12 - 19 = - 5 < 0$
$and 6 - 12 - 5 < 0$
$∴$ The equation which is obtained by taking -ve sign Hence, $6 x - 2 y - 5 = 0$

BCECE-2018

Straight Line

88769 The perpendicular bisector of line segment joining the points $P (1, 4)$ and $Q (k, 3)$ has $y$ intercept ' -4 '. Then a possible value of ' $k$ ' among the following is

1 2

2 -2

3 -4

4 -1

Explanation:

(C) : Two lines are perpendicular then $m_{1} m_{2} = - 1$

$m_{2} = \frac{4 - 3}{1 - k} = \frac{1}{1 - k}$
$AB ⊥ PQ$
$m_{1} m_{2} = - 1$
$\Rightarrow m_{1} = - 1 / m^{2}$
$\Rightarrow m_{1} = k - 1$
$C = (\frac{1 + k}{2}, \frac{4 + 3}{2}) = (\frac{1 + k}{2}, \frac{7}{2})$
Equation of line $AB$ ,
$y - \frac{7}{2} = (k - 1) [x - \frac{(1 + k)}{2}]$
at $x = 0, y = - 4$
$k = - 4$

APEAPCET-2021-23.08.2021