QBManager

Straight Line

88633 The joint equation of the lines passing through origin and having slopes $(1 + \sqrt{2})$ and $\frac{- 1}{1 + \sqrt{2}}$ is

1

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

2

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

3

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

4

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

Explanation:

(C) :
Let $m_{1} = 1 + \sqrt{2}$ and $m_{2} = \frac{- 1}{1 + \sqrt{2}}$
The equations of lines passing the origin and having slopes $m_{1}$ and $m_{2}$
$y = (1 + \sqrt{2}) x$ and $y = \frac{- 1}{1 + \sqrt{2}} x$
$(1 + \sqrt{2}) x - y = 0$ and $(1 + \sqrt{2}) y + x = 0$
Their joint equation is
$[(1 + \sqrt{2}) x - y] [(1 + \sqrt{2}) y + x] = 0$
$(1 + \sqrt{2})^{2} x y + (1 + \sqrt{2}) x^{2} - (1 + \sqrt{2}) y^{2} - x y = 0$
$(1 + \sqrt{2}) x^{2} + [1 + 2 \sqrt{2} + 2 - 1] xy - (1 + \sqrt{2}) y^{2} = 0$
$(1 + \sqrt{2}) x^{2} + 2 (1 + \sqrt{2}) x y - (1 + \sqrt{2}) y^{2} = 0$
(Dividing both sides by $(1 + \sqrt{2})$ )
$x^{2} + 2 x y - y^{2} = 0$

MHT CET-2019

Straight Line

88634 The reflection of the point $(4, - 13)$ in the line $5 x + y + 6 = 0$ , is

1

(- 1, - 14)

2

(3, 4)

3

(1, 2)

4

- 4, 13)

Explanation:

(A) :

Let $Q (a, b)$ be the reflection of $P (4, - 13)$ in the line $5 x + y + 6 = 0$
Then the mid-point $R (\frac{a + 4}{2}, \frac{b - 13}{2})$ lies
on $5 x + y + 6 = 0$
$∴ 5 (\frac{a + 4}{2}) + \frac{b - 13}{2} + 6 = 0$
$5 a + 20 + b - 13 + 12 = 0$
$\Rightarrow 5 a + b + 19 = 0$
Also PQ is perpendicular to $5 x + y + 6 = 0$
Therefore $\frac{b + 13}{a - 4} \times (- \frac{5}{1}) = - 1$
$\Rightarrow a - 5 b - 69 = 0$
Solving (i) and (ii), we get $a = - 1, b = - 14$

BITSAT-2010

Straight Line

88635 Two equal sides of an isosceles triangle are $7 x - y + 3 = 0$ and $x + y - 3 = 0$ and its third side passes through the point $(1, - 10)$ . Find the equation of the third side

1

x - 3 y = - 31

2

x - 3 y = 31

3

x + 3 y = 31

4

x + 3 y = - 31

Explanation:

(B):
Let $A B = A C$
Slope of equation $AB = 7$
Slope of equation $AC = - 1$
Let $m$ be the slope of $B C$ .
Then,
$\frac{m - 7}{1 + 7 m} = \pm \frac{m + 1}{m - 1}$
Taking +ve sign,

$m^{2} - 8 m + 7 = 7 m^{2} + 8 m + 1$
$- 6 m^{2} - 16 m + 6 = 0$
$3 m^{2} + 8 m - 3 = 0$
$3 m^{2} + 9 m - m - 3 = 0$
$(m + 3) (3 m - 1) = 0$
$m = - 3, \frac{1}{3}$
Now, the equation of line passing through $(1, - 10)$ is, $y + 10 = m (x - 1)$
Hence, the equation of third side is,
$y + 10 = - 3 (x - 1) and y + 10 = \frac{1}{3} (x - 1)$
$3 x + y + 7 = 0 and x - 3 y - 31 = 0$
$∴$ equation of third side is $x - 3 y = 31$

BITSAT-2017

Straight Line

88671 If the lines $3 x + 4 y - 5 = 0, 2 x + 3 y - 4 = 0$ and $p x + 4 y - 6 = 0$ all meet at the same point, then $p$ is equal to

1 -2

2 0

3 1

4 2

Explanation:

(D) : We have the equation of line
$3 x + 4 y - 5 = 0$
$2 x + 3 y - 4 = 0$
$px + 4 y - 6 = 0$
.....(iii) From Eqs. (i) and (ii) we get
$x = - 1, y = 2$
Put this value in equation (iii),
So, $- p + 8 - 6 = 0$
$p = 2$

AP EAMCET-2020-17.09.2020

Straight Line

88633 The joint equation of the lines passing through origin and having slopes $(1 + \sqrt{2})$ and $\frac{- 1}{1 + \sqrt{2}}$ is

1

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

2

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

3

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

4

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

Explanation:

(C) :
Let $m_{1} = 1 + \sqrt{2}$ and $m_{2} = \frac{- 1}{1 + \sqrt{2}}$
The equations of lines passing the origin and having slopes $m_{1}$ and $m_{2}$
$y = (1 + \sqrt{2}) x$ and $y = \frac{- 1}{1 + \sqrt{2}} x$
$(1 + \sqrt{2}) x - y = 0$ and $(1 + \sqrt{2}) y + x = 0$
Their joint equation is
$[(1 + \sqrt{2}) x - y] [(1 + \sqrt{2}) y + x] = 0$
$(1 + \sqrt{2})^{2} x y + (1 + \sqrt{2}) x^{2} - (1 + \sqrt{2}) y^{2} - x y = 0$
$(1 + \sqrt{2}) x^{2} + [1 + 2 \sqrt{2} + 2 - 1] xy - (1 + \sqrt{2}) y^{2} = 0$
$(1 + \sqrt{2}) x^{2} + 2 (1 + \sqrt{2}) x y - (1 + \sqrt{2}) y^{2} = 0$
(Dividing both sides by $(1 + \sqrt{2})$ )
$x^{2} + 2 x y - y^{2} = 0$

MHT CET-2019

Straight Line

88634 The reflection of the point $(4, - 13)$ in the line $5 x + y + 6 = 0$ , is

1

(- 1, - 14)

2

(3, 4)

3

(1, 2)

4

- 4, 13)

Explanation:

(A) :

Let $Q (a, b)$ be the reflection of $P (4, - 13)$ in the line $5 x + y + 6 = 0$
Then the mid-point $R (\frac{a + 4}{2}, \frac{b - 13}{2})$ lies
on $5 x + y + 6 = 0$
$∴ 5 (\frac{a + 4}{2}) + \frac{b - 13}{2} + 6 = 0$
$5 a + 20 + b - 13 + 12 = 0$
$\Rightarrow 5 a + b + 19 = 0$
Also PQ is perpendicular to $5 x + y + 6 = 0$
Therefore $\frac{b + 13}{a - 4} \times (- \frac{5}{1}) = - 1$
$\Rightarrow a - 5 b - 69 = 0$
Solving (i) and (ii), we get $a = - 1, b = - 14$

BITSAT-2010

Straight Line

88635 Two equal sides of an isosceles triangle are $7 x - y + 3 = 0$ and $x + y - 3 = 0$ and its third side passes through the point $(1, - 10)$ . Find the equation of the third side

1

x - 3 y = - 31

2

x - 3 y = 31

3

x + 3 y = 31

4

x + 3 y = - 31

Explanation:

(B):
Let $A B = A C$
Slope of equation $AB = 7$
Slope of equation $AC = - 1$
Let $m$ be the slope of $B C$ .
Then,
$\frac{m - 7}{1 + 7 m} = \pm \frac{m + 1}{m - 1}$
Taking +ve sign,

$m^{2} - 8 m + 7 = 7 m^{2} + 8 m + 1$
$- 6 m^{2} - 16 m + 6 = 0$
$3 m^{2} + 8 m - 3 = 0$
$3 m^{2} + 9 m - m - 3 = 0$
$(m + 3) (3 m - 1) = 0$
$m = - 3, \frac{1}{3}$
Now, the equation of line passing through $(1, - 10)$ is, $y + 10 = m (x - 1)$
Hence, the equation of third side is,
$y + 10 = - 3 (x - 1) and y + 10 = \frac{1}{3} (x - 1)$
$3 x + y + 7 = 0 and x - 3 y - 31 = 0$
$∴$ equation of third side is $x - 3 y = 31$

BITSAT-2017

Straight Line

88671 If the lines $3 x + 4 y - 5 = 0, 2 x + 3 y - 4 = 0$ and $p x + 4 y - 6 = 0$ all meet at the same point, then $p$ is equal to

1 -2

2 0

3 1

4 2

Explanation:

(D) : We have the equation of line
$3 x + 4 y - 5 = 0$
$2 x + 3 y - 4 = 0$
$px + 4 y - 6 = 0$
.....(iii) From Eqs. (i) and (ii) we get
$x = - 1, y = 2$
Put this value in equation (iii),
So, $- p + 8 - 6 = 0$
$p = 2$

AP EAMCET-2020-17.09.2020

Straight Line

88633 The joint equation of the lines passing through origin and having slopes $(1 + \sqrt{2})$ and $\frac{- 1}{1 + \sqrt{2}}$ is

1

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

2

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

3

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

4

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

Explanation:

(C) :
Let $m_{1} = 1 + \sqrt{2}$ and $m_{2} = \frac{- 1}{1 + \sqrt{2}}$
The equations of lines passing the origin and having slopes $m_{1}$ and $m_{2}$
$y = (1 + \sqrt{2}) x$ and $y = \frac{- 1}{1 + \sqrt{2}} x$
$(1 + \sqrt{2}) x - y = 0$ and $(1 + \sqrt{2}) y + x = 0$
Their joint equation is
$[(1 + \sqrt{2}) x - y] [(1 + \sqrt{2}) y + x] = 0$
$(1 + \sqrt{2})^{2} x y + (1 + \sqrt{2}) x^{2} - (1 + \sqrt{2}) y^{2} - x y = 0$
$(1 + \sqrt{2}) x^{2} + [1 + 2 \sqrt{2} + 2 - 1] xy - (1 + \sqrt{2}) y^{2} = 0$
$(1 + \sqrt{2}) x^{2} + 2 (1 + \sqrt{2}) x y - (1 + \sqrt{2}) y^{2} = 0$
(Dividing both sides by $(1 + \sqrt{2})$ )
$x^{2} + 2 x y - y^{2} = 0$

MHT CET-2019

Straight Line

88634 The reflection of the point $(4, - 13)$ in the line $5 x + y + 6 = 0$ , is

1

(- 1, - 14)

2

(3, 4)

3

(1, 2)

4

- 4, 13)

Explanation:

(A) :

Let $Q (a, b)$ be the reflection of $P (4, - 13)$ in the line $5 x + y + 6 = 0$
Then the mid-point $R (\frac{a + 4}{2}, \frac{b - 13}{2})$ lies
on $5 x + y + 6 = 0$
$∴ 5 (\frac{a + 4}{2}) + \frac{b - 13}{2} + 6 = 0$
$5 a + 20 + b - 13 + 12 = 0$
$\Rightarrow 5 a + b + 19 = 0$
Also PQ is perpendicular to $5 x + y + 6 = 0$
Therefore $\frac{b + 13}{a - 4} \times (- \frac{5}{1}) = - 1$
$\Rightarrow a - 5 b - 69 = 0$
Solving (i) and (ii), we get $a = - 1, b = - 14$

BITSAT-2010

Straight Line

88635 Two equal sides of an isosceles triangle are $7 x - y + 3 = 0$ and $x + y - 3 = 0$ and its third side passes through the point $(1, - 10)$ . Find the equation of the third side

1

x - 3 y = - 31

2

x - 3 y = 31

3

x + 3 y = 31

4

x + 3 y = - 31

Explanation:

(B):
Let $A B = A C$
Slope of equation $AB = 7$
Slope of equation $AC = - 1$
Let $m$ be the slope of $B C$ .
Then,
$\frac{m - 7}{1 + 7 m} = \pm \frac{m + 1}{m - 1}$
Taking +ve sign,

$m^{2} - 8 m + 7 = 7 m^{2} + 8 m + 1$
$- 6 m^{2} - 16 m + 6 = 0$
$3 m^{2} + 8 m - 3 = 0$
$3 m^{2} + 9 m - m - 3 = 0$
$(m + 3) (3 m - 1) = 0$
$m = - 3, \frac{1}{3}$
Now, the equation of line passing through $(1, - 10)$ is, $y + 10 = m (x - 1)$
Hence, the equation of third side is,
$y + 10 = - 3 (x - 1) and y + 10 = \frac{1}{3} (x - 1)$
$3 x + y + 7 = 0 and x - 3 y - 31 = 0$
$∴$ equation of third side is $x - 3 y = 31$

BITSAT-2017

Straight Line

88671 If the lines $3 x + 4 y - 5 = 0, 2 x + 3 y - 4 = 0$ and $p x + 4 y - 6 = 0$ all meet at the same point, then $p$ is equal to

1 -2

2 0

3 1

4 2

Explanation:

(D) : We have the equation of line
$3 x + 4 y - 5 = 0$
$2 x + 3 y - 4 = 0$
$px + 4 y - 6 = 0$
.....(iii) From Eqs. (i) and (ii) we get
$x = - 1, y = 2$
Put this value in equation (iii),
So, $- p + 8 - 6 = 0$
$p = 2$

AP EAMCET-2020-17.09.2020

Straight Line

88633 The joint equation of the lines passing through origin and having slopes $(1 + \sqrt{2})$ and $\frac{- 1}{1 + \sqrt{2}}$ is

1

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

2

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

3

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

4

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

Explanation:

(C) :
Let $m_{1} = 1 + \sqrt{2}$ and $m_{2} = \frac{- 1}{1 + \sqrt{2}}$
The equations of lines passing the origin and having slopes $m_{1}$ and $m_{2}$
$y = (1 + \sqrt{2}) x$ and $y = \frac{- 1}{1 + \sqrt{2}} x$
$(1 + \sqrt{2}) x - y = 0$ and $(1 + \sqrt{2}) y + x = 0$
Their joint equation is
$[(1 + \sqrt{2}) x - y] [(1 + \sqrt{2}) y + x] = 0$
$(1 + \sqrt{2})^{2} x y + (1 + \sqrt{2}) x^{2} - (1 + \sqrt{2}) y^{2} - x y = 0$
$(1 + \sqrt{2}) x^{2} + [1 + 2 \sqrt{2} + 2 - 1] xy - (1 + \sqrt{2}) y^{2} = 0$
$(1 + \sqrt{2}) x^{2} + 2 (1 + \sqrt{2}) x y - (1 + \sqrt{2}) y^{2} = 0$
(Dividing both sides by $(1 + \sqrt{2})$ )
$x^{2} + 2 x y - y^{2} = 0$

MHT CET-2019

Straight Line

88634 The reflection of the point $(4, - 13)$ in the line $5 x + y + 6 = 0$ , is

1

(- 1, - 14)

2

(3, 4)

3

(1, 2)

4

- 4, 13)

Explanation:

(A) :

Let $Q (a, b)$ be the reflection of $P (4, - 13)$ in the line $5 x + y + 6 = 0$
Then the mid-point $R (\frac{a + 4}{2}, \frac{b - 13}{2})$ lies
on $5 x + y + 6 = 0$
$∴ 5 (\frac{a + 4}{2}) + \frac{b - 13}{2} + 6 = 0$
$5 a + 20 + b - 13 + 12 = 0$
$\Rightarrow 5 a + b + 19 = 0$
Also PQ is perpendicular to $5 x + y + 6 = 0$
Therefore $\frac{b + 13}{a - 4} \times (- \frac{5}{1}) = - 1$
$\Rightarrow a - 5 b - 69 = 0$
Solving (i) and (ii), we get $a = - 1, b = - 14$

BITSAT-2010

Straight Line

88635 Two equal sides of an isosceles triangle are $7 x - y + 3 = 0$ and $x + y - 3 = 0$ and its third side passes through the point $(1, - 10)$ . Find the equation of the third side

1

x - 3 y = - 31

2

x - 3 y = 31

3

x + 3 y = 31

4

x + 3 y = - 31

Explanation:

(B):
Let $A B = A C$
Slope of equation $AB = 7$
Slope of equation $AC = - 1$
Let $m$ be the slope of $B C$ .
Then,
$\frac{m - 7}{1 + 7 m} = \pm \frac{m + 1}{m - 1}$
Taking +ve sign,

$m^{2} - 8 m + 7 = 7 m^{2} + 8 m + 1$
$- 6 m^{2} - 16 m + 6 = 0$
$3 m^{2} + 8 m - 3 = 0$
$3 m^{2} + 9 m - m - 3 = 0$
$(m + 3) (3 m - 1) = 0$
$m = - 3, \frac{1}{3}$
Now, the equation of line passing through $(1, - 10)$ is, $y + 10 = m (x - 1)$
Hence, the equation of third side is,
$y + 10 = - 3 (x - 1) and y + 10 = \frac{1}{3} (x - 1)$
$3 x + y + 7 = 0 and x - 3 y - 31 = 0$
$∴$ equation of third side is $x - 3 y = 31$

BITSAT-2017

Straight Line

88671 If the lines $3 x + 4 y - 5 = 0, 2 x + 3 y - 4 = 0$ and $p x + 4 y - 6 = 0$ all meet at the same point, then $p$ is equal to

1 -2

2 0

3 1

4 2

Explanation:

(D) : We have the equation of line
$3 x + 4 y - 5 = 0$
$2 x + 3 y - 4 = 0$
$px + 4 y - 6 = 0$
.....(iii) From Eqs. (i) and (ii) we get
$x = - 1, y = 2$
Put this value in equation (iii),
So, $- p + 8 - 6 = 0$
$p = 2$

AP EAMCET-2020-17.09.2020

Question Bank Series

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