QBManager

Straight Line

88733 The equations of the lines on which the perpendiculars from the origin make $30^{\circ}$ angle with $x$ -axis and which form a triangle of area $\frac{50}{\sqrt{3}}$ with axes, are

1

x + \sqrt{3} y \pm 10 = 0

2

\sqrt{3} x + y \pm 10 = 0

3

x \pm \sqrt{3} y - 10 = 0

4 None of above

Explanation:

(B) : Let $AB$ be the line which perpendicular to $P$ drawn from the origin on the line
Given $α = 30^{\circ}$
the equation of the normal form is, $x \cos 30^{\circ} + y \sin 30^{\circ} = P$
$\sqrt{3} x + y = 2 P$
$\sqrt{3} x + y - 2 P = 0$

Area $= 50 \sqrt{3}$
(Given)
This meet $x$ - axis in point $[\frac{2 P}{\sqrt{3}}, 0]$ ie point $A$ .
This meet $y$ -axis in point $(0, 2 P)$ i.e point $B$
Area of triangle $= \frac{1}{2} \times (\frac{2 P}{\sqrt{3}}) \times 2 P$
$\frac{50}{\sqrt{3}} = \frac{2 P^{2}}{\sqrt{3}}$
$P = \pm 5$
$∴$ The equation of line is $\sqrt{3} x + y \pm 10 = 0$

UPSEE-2010

Straight Line

88734 If the line $p x - q y = r$ intersects the coordinate axes at $(a, 0)$ and $(0, b)$ then the value of $(a + b)$ is equal to

1

\frac{r (q + p)}{p q}

2

\frac{r (q - p)}{p q}

3

\frac{r (p - q)}{pq}

4

\frac{r (p - q)}{p + q}

Explanation:

(B): The equation of line,
$p x - q y = r$
According to question
$p \cdot a - q \cdot 0 = r$
$a p = r$
$p \cdot 0 - q \cdot b = r$
$- b q = r$
Value of $a + b = r / p + (- r / q)$
$= r (1 / p - 1 / q) = r (\frac{q - p}{p q})$

Shift-II]

Straight Line

88735 if $2 a + 3 b + 6 c = 0$ , then the equation $a x^{2} + b x +$ $c = 0$ has at least one real root in

1

(0, 1)

2

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

3

(\frac{1}{4}, \frac{1}{2})

4 None of these

Explanation:

(A) : Given, $2 a + 3 b + 6 c = 0$
Let, $f^{'} (x) = a x^{2} + b x + c$
$f (x) = \frac{a x^{3}}{3} + \frac{b x^{2}}{2} + c x + d$
$f (x) = \frac{2 a x^{3} + 3 b x^{2} + 6 c x + 6 d}{6}$
Now,
$f (1) = \frac{2 a + 3 b + 6 c + 6 d}{6}$
$= \frac{6 d}{6} = d [\begin{matrix} ∵ 2 a + 3 b + 6 c = 0 \\ given \end{matrix}]$
and,
$f (0) = \frac{6 d}{6} = d$
$∴ f (0) = f (1)$
$f^{'} (x)$ will vanish at least once between 0 and 1 (by Relle's theorem)
Therefore, one of the roots of the equation ${ax}^{2} + bx + c$ $= 0$ lies between 0 and 1 .

Manipal UGET-2020

Straight Line

88736 The equations of the lines passing through the point
$(1, 0)$ and at a distance $\frac{\sqrt{3}}{2}$ from the origin are

1

\sqrt{3} x + y - \sqrt{3} = 0, \sqrt{3} x - y - \sqrt{3} = 0

2

\sqrt{3} x + y + \sqrt{3} = 0, \sqrt{3} x - y + \sqrt{3} = 0

3

x + \sqrt{3} y - \sqrt{3} = 0, x - \sqrt{3} y - \sqrt{3} = 0

4 None of the above

Explanation:

(A) : Let slope of a line be $m$ .
Now, the equation of a line passing through $(1, 0)$ is
$y - 0 = m (x - 1)$
$\Rightarrow mx - y - m = 0$
Distance from origin $= \frac{\sqrt{3}}{2}$
$\Rightarrow \frac{| - m |}{\sqrt{1 + m^{2}}} = \frac{\sqrt{3}}{2}$
$\Rightarrow 4 m^{2} = 3 (1 + m^{2})$
$\Rightarrow m^{2} = 3$
$\Rightarrow m = \pm \sqrt{3}$
$∴$ Equation of lines are
$\sqrt{3} x - y - \sqrt{3} = 0$
and $- \sqrt{3} x - y + \sqrt{3} = 0$
$\Rightarrow \sqrt{3} x - y - \sqrt{3} = 0$
and $\sqrt{3} x + y - \sqrt{3} = 0$
hence, option (a) is correct.

Manipal-2014

Straight Line

88733 The equations of the lines on which the perpendiculars from the origin make $30^{\circ}$ angle with $x$ -axis and which form a triangle of area $\frac{50}{\sqrt{3}}$ with axes, are

1

x + \sqrt{3} y \pm 10 = 0

2

\sqrt{3} x + y \pm 10 = 0

3

x \pm \sqrt{3} y - 10 = 0

4 None of above

Explanation:

(B) : Let $AB$ be the line which perpendicular to $P$ drawn from the origin on the line
Given $α = 30^{\circ}$
the equation of the normal form is, $x \cos 30^{\circ} + y \sin 30^{\circ} = P$
$\sqrt{3} x + y = 2 P$
$\sqrt{3} x + y - 2 P = 0$

Area $= 50 \sqrt{3}$
(Given)
This meet $x$ - axis in point $[\frac{2 P}{\sqrt{3}}, 0]$ ie point $A$ .
This meet $y$ -axis in point $(0, 2 P)$ i.e point $B$
Area of triangle $= \frac{1}{2} \times (\frac{2 P}{\sqrt{3}}) \times 2 P$
$\frac{50}{\sqrt{3}} = \frac{2 P^{2}}{\sqrt{3}}$
$P = \pm 5$
$∴$ The equation of line is $\sqrt{3} x + y \pm 10 = 0$

UPSEE-2010

Straight Line

88734 If the line $p x - q y = r$ intersects the coordinate axes at $(a, 0)$ and $(0, b)$ then the value of $(a + b)$ is equal to

1

\frac{r (q + p)}{p q}

2

\frac{r (q - p)}{p q}

3

\frac{r (p - q)}{pq}

4

\frac{r (p - q)}{p + q}

Explanation:

(B): The equation of line,
$p x - q y = r$
According to question
$p \cdot a - q \cdot 0 = r$
$a p = r$
$p \cdot 0 - q \cdot b = r$
$- b q = r$
Value of $a + b = r / p + (- r / q)$
$= r (1 / p - 1 / q) = r (\frac{q - p}{p q})$

Shift-II]

Straight Line

88735 if $2 a + 3 b + 6 c = 0$ , then the equation $a x^{2} + b x +$ $c = 0$ has at least one real root in

1

(0, 1)

2

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

3

(\frac{1}{4}, \frac{1}{2})

4 None of these

Explanation:

(A) : Given, $2 a + 3 b + 6 c = 0$
Let, $f^{'} (x) = a x^{2} + b x + c$
$f (x) = \frac{a x^{3}}{3} + \frac{b x^{2}}{2} + c x + d$
$f (x) = \frac{2 a x^{3} + 3 b x^{2} + 6 c x + 6 d}{6}$
Now,
$f (1) = \frac{2 a + 3 b + 6 c + 6 d}{6}$
$= \frac{6 d}{6} = d [\begin{matrix} ∵ 2 a + 3 b + 6 c = 0 \\ given \end{matrix}]$
and,
$f (0) = \frac{6 d}{6} = d$
$∴ f (0) = f (1)$
$f^{'} (x)$ will vanish at least once between 0 and 1 (by Relle's theorem)
Therefore, one of the roots of the equation ${ax}^{2} + bx + c$ $= 0$ lies between 0 and 1 .

Manipal UGET-2020

Straight Line

88736 The equations of the lines passing through the point
$(1, 0)$ and at a distance $\frac{\sqrt{3}}{2}$ from the origin are

1

\sqrt{3} x + y - \sqrt{3} = 0, \sqrt{3} x - y - \sqrt{3} = 0

2

\sqrt{3} x + y + \sqrt{3} = 0, \sqrt{3} x - y + \sqrt{3} = 0

3

x + \sqrt{3} y - \sqrt{3} = 0, x - \sqrt{3} y - \sqrt{3} = 0

4 None of the above

Explanation:

(A) : Let slope of a line be $m$ .
Now, the equation of a line passing through $(1, 0)$ is
$y - 0 = m (x - 1)$
$\Rightarrow mx - y - m = 0$
Distance from origin $= \frac{\sqrt{3}}{2}$
$\Rightarrow \frac{| - m |}{\sqrt{1 + m^{2}}} = \frac{\sqrt{3}}{2}$
$\Rightarrow 4 m^{2} = 3 (1 + m^{2})$
$\Rightarrow m^{2} = 3$
$\Rightarrow m = \pm \sqrt{3}$
$∴$ Equation of lines are
$\sqrt{3} x - y - \sqrt{3} = 0$
and $- \sqrt{3} x - y + \sqrt{3} = 0$
$\Rightarrow \sqrt{3} x - y - \sqrt{3} = 0$
and $\sqrt{3} x + y - \sqrt{3} = 0$
hence, option (a) is correct.

Manipal-2014

Straight Line

88733 The equations of the lines on which the perpendiculars from the origin make $30^{\circ}$ angle with $x$ -axis and which form a triangle of area $\frac{50}{\sqrt{3}}$ with axes, are

1

x + \sqrt{3} y \pm 10 = 0

2

\sqrt{3} x + y \pm 10 = 0

3

x \pm \sqrt{3} y - 10 = 0

4 None of above

Explanation:

(B) : Let $AB$ be the line which perpendicular to $P$ drawn from the origin on the line
Given $α = 30^{\circ}$
the equation of the normal form is, $x \cos 30^{\circ} + y \sin 30^{\circ} = P$
$\sqrt{3} x + y = 2 P$
$\sqrt{3} x + y - 2 P = 0$

Area $= 50 \sqrt{3}$
(Given)
This meet $x$ - axis in point $[\frac{2 P}{\sqrt{3}}, 0]$ ie point $A$ .
This meet $y$ -axis in point $(0, 2 P)$ i.e point $B$
Area of triangle $= \frac{1}{2} \times (\frac{2 P}{\sqrt{3}}) \times 2 P$
$\frac{50}{\sqrt{3}} = \frac{2 P^{2}}{\sqrt{3}}$
$P = \pm 5$
$∴$ The equation of line is $\sqrt{3} x + y \pm 10 = 0$

UPSEE-2010

Straight Line

88734 If the line $p x - q y = r$ intersects the coordinate axes at $(a, 0)$ and $(0, b)$ then the value of $(a + b)$ is equal to

1

\frac{r (q + p)}{p q}

2

\frac{r (q - p)}{p q}

3

\frac{r (p - q)}{pq}

4

\frac{r (p - q)}{p + q}

Explanation:

(B): The equation of line,
$p x - q y = r$
According to question
$p \cdot a - q \cdot 0 = r$
$a p = r$
$p \cdot 0 - q \cdot b = r$
$- b q = r$
Value of $a + b = r / p + (- r / q)$
$= r (1 / p - 1 / q) = r (\frac{q - p}{p q})$

Shift-II]

Straight Line

88735 if $2 a + 3 b + 6 c = 0$ , then the equation $a x^{2} + b x +$ $c = 0$ has at least one real root in

1

(0, 1)

2

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

3

(\frac{1}{4}, \frac{1}{2})

4 None of these

Explanation:

(A) : Given, $2 a + 3 b + 6 c = 0$
Let, $f^{'} (x) = a x^{2} + b x + c$
$f (x) = \frac{a x^{3}}{3} + \frac{b x^{2}}{2} + c x + d$
$f (x) = \frac{2 a x^{3} + 3 b x^{2} + 6 c x + 6 d}{6}$
Now,
$f (1) = \frac{2 a + 3 b + 6 c + 6 d}{6}$
$= \frac{6 d}{6} = d [\begin{matrix} ∵ 2 a + 3 b + 6 c = 0 \\ given \end{matrix}]$
and,
$f (0) = \frac{6 d}{6} = d$
$∴ f (0) = f (1)$
$f^{'} (x)$ will vanish at least once between 0 and 1 (by Relle's theorem)
Therefore, one of the roots of the equation ${ax}^{2} + bx + c$ $= 0$ lies between 0 and 1 .

Manipal UGET-2020

Straight Line

88736 The equations of the lines passing through the point
$(1, 0)$ and at a distance $\frac{\sqrt{3}}{2}$ from the origin are

1

\sqrt{3} x + y - \sqrt{3} = 0, \sqrt{3} x - y - \sqrt{3} = 0

2

\sqrt{3} x + y + \sqrt{3} = 0, \sqrt{3} x - y + \sqrt{3} = 0

3

x + \sqrt{3} y - \sqrt{3} = 0, x - \sqrt{3} y - \sqrt{3} = 0

4 None of the above

Explanation:

(A) : Let slope of a line be $m$ .
Now, the equation of a line passing through $(1, 0)$ is
$y - 0 = m (x - 1)$
$\Rightarrow mx - y - m = 0$
Distance from origin $= \frac{\sqrt{3}}{2}$
$\Rightarrow \frac{| - m |}{\sqrt{1 + m^{2}}} = \frac{\sqrt{3}}{2}$
$\Rightarrow 4 m^{2} = 3 (1 + m^{2})$
$\Rightarrow m^{2} = 3$
$\Rightarrow m = \pm \sqrt{3}$
$∴$ Equation of lines are
$\sqrt{3} x - y - \sqrt{3} = 0$
and $- \sqrt{3} x - y + \sqrt{3} = 0$
$\Rightarrow \sqrt{3} x - y - \sqrt{3} = 0$
and $\sqrt{3} x + y - \sqrt{3} = 0$
hence, option (a) is correct.

Manipal-2014

Straight Line

88733 The equations of the lines on which the perpendiculars from the origin make $30^{\circ}$ angle with $x$ -axis and which form a triangle of area $\frac{50}{\sqrt{3}}$ with axes, are

1

x + \sqrt{3} y \pm 10 = 0

2

\sqrt{3} x + y \pm 10 = 0

3

x \pm \sqrt{3} y - 10 = 0

4 None of above

Explanation:

(B) : Let $AB$ be the line which perpendicular to $P$ drawn from the origin on the line
Given $α = 30^{\circ}$
the equation of the normal form is, $x \cos 30^{\circ} + y \sin 30^{\circ} = P$
$\sqrt{3} x + y = 2 P$
$\sqrt{3} x + y - 2 P = 0$

Area $= 50 \sqrt{3}$
(Given)
This meet $x$ - axis in point $[\frac{2 P}{\sqrt{3}}, 0]$ ie point $A$ .
This meet $y$ -axis in point $(0, 2 P)$ i.e point $B$
Area of triangle $= \frac{1}{2} \times (\frac{2 P}{\sqrt{3}}) \times 2 P$
$\frac{50}{\sqrt{3}} = \frac{2 P^{2}}{\sqrt{3}}$
$P = \pm 5$
$∴$ The equation of line is $\sqrt{3} x + y \pm 10 = 0$

UPSEE-2010

Straight Line

88734 If the line $p x - q y = r$ intersects the coordinate axes at $(a, 0)$ and $(0, b)$ then the value of $(a + b)$ is equal to

1

\frac{r (q + p)}{p q}

2

\frac{r (q - p)}{p q}

3

\frac{r (p - q)}{pq}

4

\frac{r (p - q)}{p + q}

Explanation:

(B): The equation of line,
$p x - q y = r$
According to question
$p \cdot a - q \cdot 0 = r$
$a p = r$
$p \cdot 0 - q \cdot b = r$
$- b q = r$
Value of $a + b = r / p + (- r / q)$
$= r (1 / p - 1 / q) = r (\frac{q - p}{p q})$

Shift-II]

Straight Line

88735 if $2 a + 3 b + 6 c = 0$ , then the equation $a x^{2} + b x +$ $c = 0$ has at least one real root in

1

(0, 1)

2

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

3

(\frac{1}{4}, \frac{1}{2})

4 None of these

Explanation:

(A) : Given, $2 a + 3 b + 6 c = 0$
Let, $f^{'} (x) = a x^{2} + b x + c$
$f (x) = \frac{a x^{3}}{3} + \frac{b x^{2}}{2} + c x + d$
$f (x) = \frac{2 a x^{3} + 3 b x^{2} + 6 c x + 6 d}{6}$
Now,
$f (1) = \frac{2 a + 3 b + 6 c + 6 d}{6}$
$= \frac{6 d}{6} = d [\begin{matrix} ∵ 2 a + 3 b + 6 c = 0 \\ given \end{matrix}]$
and,
$f (0) = \frac{6 d}{6} = d$
$∴ f (0) = f (1)$
$f^{'} (x)$ will vanish at least once between 0 and 1 (by Relle's theorem)
Therefore, one of the roots of the equation ${ax}^{2} + bx + c$ $= 0$ lies between 0 and 1 .

Manipal UGET-2020

Straight Line

88736 The equations of the lines passing through the point
$(1, 0)$ and at a distance $\frac{\sqrt{3}}{2}$ from the origin are

1

\sqrt{3} x + y - \sqrt{3} = 0, \sqrt{3} x - y - \sqrt{3} = 0

2

\sqrt{3} x + y + \sqrt{3} = 0, \sqrt{3} x - y + \sqrt{3} = 0

3

x + \sqrt{3} y - \sqrt{3} = 0, x - \sqrt{3} y - \sqrt{3} = 0

4 None of the above

Explanation:

(A) : Let slope of a line be $m$ .
Now, the equation of a line passing through $(1, 0)$ is
$y - 0 = m (x - 1)$
$\Rightarrow mx - y - m = 0$
Distance from origin $= \frac{\sqrt{3}}{2}$
$\Rightarrow \frac{| - m |}{\sqrt{1 + m^{2}}} = \frac{\sqrt{3}}{2}$
$\Rightarrow 4 m^{2} = 3 (1 + m^{2})$
$\Rightarrow m^{2} = 3$
$\Rightarrow m = \pm \sqrt{3}$
$∴$ Equation of lines are
$\sqrt{3} x - y - \sqrt{3} = 0$
and $- \sqrt{3} x - y + \sqrt{3} = 0$
$\Rightarrow \sqrt{3} x - y - \sqrt{3} = 0$
and $\sqrt{3} x + y - \sqrt{3} = 0$
hence, option (a) is correct.

Manipal-2014

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Question Bank Series

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