QBManager

Straight Line

88747 If $p$ is the length of perpendicular from origin to the line whose intercepts on the axes are $a$ and $b$ , then $\frac{1}{a^{2}} + \frac{1}{b^{2}}$ is equal to

1

p^{2}

2

\frac{1}{p^{2}}

3

2 p^{2}

4

\frac{1}{2 p^{2}}

Explanation:

(B) : Equation of line in intercept form is
$\frac{x}{a} + \frac{y}{b} = 1 \Rightarrow \frac{x}{a} + \frac{y}{b} - 1 = 0$
Its distance from origin is $p$
$∴ | \frac{0 + 0 - 1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} | = p$
$\Rightarrow p = \frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}$
$\Rightarrow \frac{1}{p} = \sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}$
Now, squaring both sides, we get $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$

COMEDK-2019

Straight Line

88748 The point on the line $4 x + 3 y = 5$ , which is equidistant from $(1, 2)$ and $(3, 4)$ , is

1

(7, - 4)

2

(- 10, 15)

3

(\frac{1}{7}, \frac{8}{7})

4

(0, \frac{5}{4})

Explanation:

(B) : Let the point $(x, y)$ be on the line $4 x + 3 y = 5$ $∴ 4 x_{1} + 3 y_{1} = 5$
Distance of point $(x_{1}, y_{1})$ from $(1, 2)$
$∴ {(x_{1} - 1)}^{2} + {(y_{1} - 2)}^{2}$
And,
Distance of point $(x_{1}, y_{1})$ from $(3, 4)$
$∴ {(x_{1} - 3)}^{2} + {(y_{1} - 4)}^{2}$
Given that the point $(x_{1}, y_{1})$ is equidistant from $(1, 2)$ and $(3, 4)$
$∴ {(x_{1} - 1)}^{2} + {(y_{1} - 2)}^{2} = {(x_{1} - 3)}^{2} + {(y_{1} - 4)}^{2}$
$x_{1}^{2} + 1 - 2 x, + y^{2} + 4 - 4 y_{1} = x_{1}^{2} + 9 - 6 x_{1} + y_{1}^{2} + 16 -$
$8 y$
$4 x_{1} + 4 y_{1} = 20$
$x_{1} + y_{1} = 5$
Form equation (i) and (ii) we get,
$y_{1} = 15 and x_{1} = - 10$
Hence, the point will be $(- 10, 15)$

UPSEE-2015

Straight Line

88749 Let $P Q R$ be a right angled isosceles triangle, right angled at $P (2, 1)$ . If the equation of the line $Q R$ is $2 x + y = 3$ , Then the equation representing the pair of lines $P Q$ and $P R$ is

1

3 x^{2} - 3 y^{2} + 8 x y + 20 x + 10 y + 25 = 0

2

3 x^{2} - 3 y^{2} + 8 x y - 20 x - 10 y + 25 = 0

3

3 x^{2} - 3 y^{2} + 8 x y + 10 x + 15 y + 20 = 0

4

3 x^{2} - 3 y^{2} - 8 x y - 10 x - 15 y - 20 = 0

Explanation:

(B) : The equation of PQ and PR are given by,
$y - 1 = \frac{- 2 \pm \tan 45^{\circ}}{1 \pm (- 2 \tan 45^{\circ})} (x - 2)$
$y - 1 = (\frac{- 2 \mp 1}{1 \pm 2}) (x - 2)$
$y - 1 = - \frac{1}{3} (x - 2)$
and $y - 1 = 3 (x - 2)$
$x + 3 y = 5$ and $3 x - y = 5$
The combined equation of two line is,
$(x + 3 y - 5) (3 x - y - 5) = 0$
Hence, $3 x^{2} - 3 y^{2} + 8 x y - 20 x - 10 y + 25 = 0$

UPSEE-2010

Straight Line

88750 The distance of the point $A (a, b, c)$ from the $X$ axis is

1 a

2

\sqrt{b^{2} + c^{2}}

3

\sqrt{a^{2} + b^{2}}

4

a^{2} + b^{2}

Explanation:

(B) : Given the distance between the point
$P (a, b, c)$ and $x$ -axis
Distance $(d) = \sqrt{(a - a)^{2} + (0 - b)^{2} + (0 - c)^{2}}$
$d = \sqrt{0^{2} + (- b)^{2} + (- c)^{2}}$
$d = \sqrt{b^{2} + c^{2}}$

[JCECE-2017]

Straight Line

88747 If $p$ is the length of perpendicular from origin to the line whose intercepts on the axes are $a$ and $b$ , then $\frac{1}{a^{2}} + \frac{1}{b^{2}}$ is equal to

1

p^{2}

2

\frac{1}{p^{2}}

3

2 p^{2}

4

\frac{1}{2 p^{2}}

Explanation:

(B) : Equation of line in intercept form is
$\frac{x}{a} + \frac{y}{b} = 1 \Rightarrow \frac{x}{a} + \frac{y}{b} - 1 = 0$
Its distance from origin is $p$
$∴ | \frac{0 + 0 - 1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} | = p$
$\Rightarrow p = \frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}$
$\Rightarrow \frac{1}{p} = \sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}$
Now, squaring both sides, we get $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$

COMEDK-2019

Straight Line

88748 The point on the line $4 x + 3 y = 5$ , which is equidistant from $(1, 2)$ and $(3, 4)$ , is

1

(7, - 4)

2

(- 10, 15)

3

(\frac{1}{7}, \frac{8}{7})

4

(0, \frac{5}{4})

Explanation:

(B) : Let the point $(x, y)$ be on the line $4 x + 3 y = 5$ $∴ 4 x_{1} + 3 y_{1} = 5$
Distance of point $(x_{1}, y_{1})$ from $(1, 2)$
$∴ {(x_{1} - 1)}^{2} + {(y_{1} - 2)}^{2}$
And,
Distance of point $(x_{1}, y_{1})$ from $(3, 4)$
$∴ {(x_{1} - 3)}^{2} + {(y_{1} - 4)}^{2}$
Given that the point $(x_{1}, y_{1})$ is equidistant from $(1, 2)$ and $(3, 4)$
$∴ {(x_{1} - 1)}^{2} + {(y_{1} - 2)}^{2} = {(x_{1} - 3)}^{2} + {(y_{1} - 4)}^{2}$
$x_{1}^{2} + 1 - 2 x, + y^{2} + 4 - 4 y_{1} = x_{1}^{2} + 9 - 6 x_{1} + y_{1}^{2} + 16 -$
$8 y$
$4 x_{1} + 4 y_{1} = 20$
$x_{1} + y_{1} = 5$
Form equation (i) and (ii) we get,
$y_{1} = 15 and x_{1} = - 10$
Hence, the point will be $(- 10, 15)$

UPSEE-2015

Straight Line

88749 Let $P Q R$ be a right angled isosceles triangle, right angled at $P (2, 1)$ . If the equation of the line $Q R$ is $2 x + y = 3$ , Then the equation representing the pair of lines $P Q$ and $P R$ is

1

3 x^{2} - 3 y^{2} + 8 x y + 20 x + 10 y + 25 = 0

2

3 x^{2} - 3 y^{2} + 8 x y - 20 x - 10 y + 25 = 0

3

3 x^{2} - 3 y^{2} + 8 x y + 10 x + 15 y + 20 = 0

4

3 x^{2} - 3 y^{2} - 8 x y - 10 x - 15 y - 20 = 0

Explanation:

(B) : The equation of PQ and PR are given by,
$y - 1 = \frac{- 2 \pm \tan 45^{\circ}}{1 \pm (- 2 \tan 45^{\circ})} (x - 2)$
$y - 1 = (\frac{- 2 \mp 1}{1 \pm 2}) (x - 2)$
$y - 1 = - \frac{1}{3} (x - 2)$
and $y - 1 = 3 (x - 2)$
$x + 3 y = 5$ and $3 x - y = 5$
The combined equation of two line is,
$(x + 3 y - 5) (3 x - y - 5) = 0$
Hence, $3 x^{2} - 3 y^{2} + 8 x y - 20 x - 10 y + 25 = 0$

UPSEE-2010

Straight Line

88750 The distance of the point $A (a, b, c)$ from the $X$ axis is

1 a

2

\sqrt{b^{2} + c^{2}}

3

\sqrt{a^{2} + b^{2}}

4

a^{2} + b^{2}

Explanation:

(B) : Given the distance between the point
$P (a, b, c)$ and $x$ -axis
Distance $(d) = \sqrt{(a - a)^{2} + (0 - b)^{2} + (0 - c)^{2}}$
$d = \sqrt{0^{2} + (- b)^{2} + (- c)^{2}}$
$d = \sqrt{b^{2} + c^{2}}$

[JCECE-2017]

Straight Line

88747 If $p$ is the length of perpendicular from origin to the line whose intercepts on the axes are $a$ and $b$ , then $\frac{1}{a^{2}} + \frac{1}{b^{2}}$ is equal to

1

p^{2}

2

\frac{1}{p^{2}}

3

2 p^{2}

4

\frac{1}{2 p^{2}}

Explanation:

(B) : Equation of line in intercept form is
$\frac{x}{a} + \frac{y}{b} = 1 \Rightarrow \frac{x}{a} + \frac{y}{b} - 1 = 0$
Its distance from origin is $p$
$∴ | \frac{0 + 0 - 1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} | = p$
$\Rightarrow p = \frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}$
$\Rightarrow \frac{1}{p} = \sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}$
Now, squaring both sides, we get $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$

COMEDK-2019

Straight Line

88748 The point on the line $4 x + 3 y = 5$ , which is equidistant from $(1, 2)$ and $(3, 4)$ , is

1

(7, - 4)

2

(- 10, 15)

3

(\frac{1}{7}, \frac{8}{7})

4

(0, \frac{5}{4})

Explanation:

(B) : Let the point $(x, y)$ be on the line $4 x + 3 y = 5$ $∴ 4 x_{1} + 3 y_{1} = 5$
Distance of point $(x_{1}, y_{1})$ from $(1, 2)$
$∴ {(x_{1} - 1)}^{2} + {(y_{1} - 2)}^{2}$
And,
Distance of point $(x_{1}, y_{1})$ from $(3, 4)$
$∴ {(x_{1} - 3)}^{2} + {(y_{1} - 4)}^{2}$
Given that the point $(x_{1}, y_{1})$ is equidistant from $(1, 2)$ and $(3, 4)$
$∴ {(x_{1} - 1)}^{2} + {(y_{1} - 2)}^{2} = {(x_{1} - 3)}^{2} + {(y_{1} - 4)}^{2}$
$x_{1}^{2} + 1 - 2 x, + y^{2} + 4 - 4 y_{1} = x_{1}^{2} + 9 - 6 x_{1} + y_{1}^{2} + 16 -$
$8 y$
$4 x_{1} + 4 y_{1} = 20$
$x_{1} + y_{1} = 5$
Form equation (i) and (ii) we get,
$y_{1} = 15 and x_{1} = - 10$
Hence, the point will be $(- 10, 15)$

UPSEE-2015

Straight Line

88749 Let $P Q R$ be a right angled isosceles triangle, right angled at $P (2, 1)$ . If the equation of the line $Q R$ is $2 x + y = 3$ , Then the equation representing the pair of lines $P Q$ and $P R$ is

1

3 x^{2} - 3 y^{2} + 8 x y + 20 x + 10 y + 25 = 0

2

3 x^{2} - 3 y^{2} + 8 x y - 20 x - 10 y + 25 = 0

3

3 x^{2} - 3 y^{2} + 8 x y + 10 x + 15 y + 20 = 0

4

3 x^{2} - 3 y^{2} - 8 x y - 10 x - 15 y - 20 = 0

Explanation:

(B) : The equation of PQ and PR are given by,
$y - 1 = \frac{- 2 \pm \tan 45^{\circ}}{1 \pm (- 2 \tan 45^{\circ})} (x - 2)$
$y - 1 = (\frac{- 2 \mp 1}{1 \pm 2}) (x - 2)$
$y - 1 = - \frac{1}{3} (x - 2)$
and $y - 1 = 3 (x - 2)$
$x + 3 y = 5$ and $3 x - y = 5$
The combined equation of two line is,
$(x + 3 y - 5) (3 x - y - 5) = 0$
Hence, $3 x^{2} - 3 y^{2} + 8 x y - 20 x - 10 y + 25 = 0$

UPSEE-2010

Straight Line

88750 The distance of the point $A (a, b, c)$ from the $X$ axis is

1 a

2

\sqrt{b^{2} + c^{2}}

3

\sqrt{a^{2} + b^{2}}

4

a^{2} + b^{2}

Explanation:

(B) : Given the distance between the point
$P (a, b, c)$ and $x$ -axis
Distance $(d) = \sqrt{(a - a)^{2} + (0 - b)^{2} + (0 - c)^{2}}$
$d = \sqrt{0^{2} + (- b)^{2} + (- c)^{2}}$
$d = \sqrt{b^{2} + c^{2}}$

[JCECE-2017]

Straight Line

88747 If $p$ is the length of perpendicular from origin to the line whose intercepts on the axes are $a$ and $b$ , then $\frac{1}{a^{2}} + \frac{1}{b^{2}}$ is equal to

1

p^{2}

2

\frac{1}{p^{2}}

3

2 p^{2}

4

\frac{1}{2 p^{2}}

Explanation:

(B) : Equation of line in intercept form is
$\frac{x}{a} + \frac{y}{b} = 1 \Rightarrow \frac{x}{a} + \frac{y}{b} - 1 = 0$
Its distance from origin is $p$
$∴ | \frac{0 + 0 - 1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} | = p$
$\Rightarrow p = \frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}$
$\Rightarrow \frac{1}{p} = \sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}$
Now, squaring both sides, we get $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$

COMEDK-2019

Straight Line

88748 The point on the line $4 x + 3 y = 5$ , which is equidistant from $(1, 2)$ and $(3, 4)$ , is

1

(7, - 4)

2

(- 10, 15)

3

(\frac{1}{7}, \frac{8}{7})

4

(0, \frac{5}{4})

Explanation:

(B) : Let the point $(x, y)$ be on the line $4 x + 3 y = 5$ $∴ 4 x_{1} + 3 y_{1} = 5$
Distance of point $(x_{1}, y_{1})$ from $(1, 2)$
$∴ {(x_{1} - 1)}^{2} + {(y_{1} - 2)}^{2}$
And,
Distance of point $(x_{1}, y_{1})$ from $(3, 4)$
$∴ {(x_{1} - 3)}^{2} + {(y_{1} - 4)}^{2}$
Given that the point $(x_{1}, y_{1})$ is equidistant from $(1, 2)$ and $(3, 4)$
$∴ {(x_{1} - 1)}^{2} + {(y_{1} - 2)}^{2} = {(x_{1} - 3)}^{2} + {(y_{1} - 4)}^{2}$
$x_{1}^{2} + 1 - 2 x, + y^{2} + 4 - 4 y_{1} = x_{1}^{2} + 9 - 6 x_{1} + y_{1}^{2} + 16 -$
$8 y$
$4 x_{1} + 4 y_{1} = 20$
$x_{1} + y_{1} = 5$
Form equation (i) and (ii) we get,
$y_{1} = 15 and x_{1} = - 10$
Hence, the point will be $(- 10, 15)$

UPSEE-2015

Straight Line

88749 Let $P Q R$ be a right angled isosceles triangle, right angled at $P (2, 1)$ . If the equation of the line $Q R$ is $2 x + y = 3$ , Then the equation representing the pair of lines $P Q$ and $P R$ is

1

3 x^{2} - 3 y^{2} + 8 x y + 20 x + 10 y + 25 = 0

2

3 x^{2} - 3 y^{2} + 8 x y - 20 x - 10 y + 25 = 0

3

3 x^{2} - 3 y^{2} + 8 x y + 10 x + 15 y + 20 = 0

4

3 x^{2} - 3 y^{2} - 8 x y - 10 x - 15 y - 20 = 0

Explanation:

(B) : The equation of PQ and PR are given by,
$y - 1 = \frac{- 2 \pm \tan 45^{\circ}}{1 \pm (- 2 \tan 45^{\circ})} (x - 2)$
$y - 1 = (\frac{- 2 \mp 1}{1 \pm 2}) (x - 2)$
$y - 1 = - \frac{1}{3} (x - 2)$
and $y - 1 = 3 (x - 2)$
$x + 3 y = 5$ and $3 x - y = 5$
The combined equation of two line is,
$(x + 3 y - 5) (3 x - y - 5) = 0$
Hence, $3 x^{2} - 3 y^{2} + 8 x y - 20 x - 10 y + 25 = 0$

UPSEE-2010

Straight Line

88750 The distance of the point $A (a, b, c)$ from the $X$ axis is

1 a

2

\sqrt{b^{2} + c^{2}}

3

\sqrt{a^{2} + b^{2}}

4

a^{2} + b^{2}

Explanation:

(B) : Given the distance between the point
$P (a, b, c)$ and $x$ -axis
Distance $(d) = \sqrt{(a - a)^{2} + (0 - b)^{2} + (0 - c)^{2}}$
$d = \sqrt{0^{2} + (- b)^{2} + (- c)^{2}}$
$d = \sqrt{b^{2} + c^{2}}$

[JCECE-2017]

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: