QBManager

Straight Line

88789 If the lines $x = a + m, y = - 2$ and $y = m x$ are concurrent, then least value of $| a |$ is

1

2 \sqrt{2}

2 0

3

\sqrt{2}

4 None of the above

Explanation:

(B) : The given lines are,
$x = a + m$
$y = - 2$
$y = m x$
These lines are concurrent
$∴ mx = ma + m^{2} \Rightarrow m^{2} + ma + 2 = 0$
$m = \frac{- a \pm \sqrt{a^{2} - 8}}{2} \Rightarrow a^{2} \geq 8$

[JCECE-2016]

Straight Line

88790 Three lines $p x + q y + r = 0, q x + r y + p = 0$ and $r x + p y + q = 0$ are concurrent, if

1

p_{2} + q + r = 0

2

p_{3}^{2} + q_{3}^{2} + r^{2} = pq + qr + rp

3

p^{3} + q^{3} + r^{3} = 3 pqr

4 None of the above

Explanation:

(B) : Gvien lines are,
$p x + q y + r = 0$
$q x + r y + p = 0$
$r x + p y + q = 0$
These lines are concurrent
then,
$| \begin{array}{ccc} p & q & r \\ q & r & p \\ r & p & q \end{array} | = 0$
$- (p^{3} + q^{3} + r^{3} - 3 p q r) = 0$
$p^{3} + q^{3} + r^{3} - 3 p q r = 0$
$(p + q + r) (p^{2} + q^{2} + r^{2} - pq - qr - rp) = 0$
$∴ (p + q + r) = 0$
and $p_{2}^{2} + q_{2}^{2} + r_{2}^{2} - pq - qr - rp = 0$
Hence $p^{2} + q^{2} + r^{2} = pq + qr + rp$

JCECE-2011

Straight Line

88791 If the line $y = m x$ meets the lines $x + 2 y - 1 = 0$ and $2 x - y + 3 = 0$ at the same point, then $m$ is equal to

1 1

2 2

3 -2

4 -1

Explanation:

(D) :

$x + 2 y = 1$
$2 x - y = - 3$
From equation (i) and (ii),
$x = - 1, y = 1$
$y = mx$
So, $1 = m (- 1)$
$m = - 1$

SCRA-2014

Straight Line

88792 The distance between the points of concurrency of the two families of straight lines given by $x + (5 λ + 1) y + 1 - 3 λ = 0$ and
$(5 μ + 2) x - 3 y + 3 + 6 μ = 0$ is

1 4

2

\frac{2 \sqrt{2}}{5}

3

\frac{\sqrt{2}}{5}

4 6

Explanation:

(B) : Family of straight lines given as-
$x + (5 λ + 1) y + 1 - 3 λ = 0 \dots .(i)$
$(5 μ + 2) x - 3 y + 3 + 6 μ = 0$
From equation (i)
$(x + y + 1) + λ (5 y - 3) = 0$
$\begin{array}{cl} Which gives x + y + 1 = 0 \\ and 5 y - 3 = 0 \end{array}$
Solving equations (iii) and (iv), we get-
Point $P (\frac{- 8}{5}, \frac{3}{5})$
Similarly from equation (ii)
Which gives
$(2 x - 3 y + 3) + μ (5 x + 6) = 0$
$2 x - 3 y + 3 = 0$
$x = \frac{- 6}{5}$
Solving equation (v) and (vi), we get the point $Q (\frac{- 6}{5}, \frac{1}{5})$
Hence, point of concurrency are $P (\frac{- 8}{5}, \frac{3}{5})$ and
$Q (\frac{- 6}{5}, \frac{1}{5})$
$∴ PQ = \sqrt{{(\frac{- 8}{5} + \frac{6}{5})}^{2} + {(\frac{3}{5} - \frac{1}{5})}^{2}}$
$= \frac{\sqrt{8}}{5} = \frac{2 \sqrt{2}}{5}$
$\Rightarrow p \cos α - q \sin α = \sqrt{3}$ and $p \sin α + q \cos α = - 1$
Adding and squaring, we get
$p^{2} + q^{2} = 3 + 1 \Rightarrow p^{2} + q^{2} = 4$

TS EAMCET-2022-19.07.2022

Straight Line

88793 The straight lines $x + 3 y - 9 = 0, 4 x + 5 y - 1 = 0$ , $p x + q y + 10 = 0$ are concurrent, if the line $5 x + 6 y + 10 = 0$ passes through the point

1

(q, - p)

2

(q, p)

3

(p, - q)

4

(p, q)

Explanation:

(A) : It is given that lines $x + 3 y - 9 = 0$ ,
$4 x + 5 y - 1 = 0, px + qy + 10 = 0$ are concurrent
$∴ | \begin{array}{lll} 1 & 3 & - 9 \\ 4 & 5 & - 1 \\ p & q & 10 \end{array} | = 0$
$\Rightarrow 50 + q - 3 (40 + p) - 9 (4 q - 5 p) = 0$
$\Rightarrow 50 + q - 120 - 3 p - 36 q + 45 p = 0$
$\Rightarrow 42 p - 35 q - 70 = 0$
$\Rightarrow - 6 p + 5 q + 10 = 0$
So, $5 x + 6 y + 10 = 0$ passes through $(q, - p)$ .

TS EAMCET-2020-10.09.2020

Straight Line

88789 If the lines $x = a + m, y = - 2$ and $y = m x$ are concurrent, then least value of $| a |$ is

1

2 \sqrt{2}

2 0

3

\sqrt{2}

4 None of the above

Explanation:

(B) : The given lines are,
$x = a + m$
$y = - 2$
$y = m x$
These lines are concurrent
$∴ mx = ma + m^{2} \Rightarrow m^{2} + ma + 2 = 0$
$m = \frac{- a \pm \sqrt{a^{2} - 8}}{2} \Rightarrow a^{2} \geq 8$

[JCECE-2016]

Straight Line

88790 Three lines $p x + q y + r = 0, q x + r y + p = 0$ and $r x + p y + q = 0$ are concurrent, if

1

p_{2} + q + r = 0

2

p_{3}^{2} + q_{3}^{2} + r^{2} = pq + qr + rp

3

p^{3} + q^{3} + r^{3} = 3 pqr

4 None of the above

Explanation:

(B) : Gvien lines are,
$p x + q y + r = 0$
$q x + r y + p = 0$
$r x + p y + q = 0$
These lines are concurrent
then,
$| \begin{array}{ccc} p & q & r \\ q & r & p \\ r & p & q \end{array} | = 0$
$- (p^{3} + q^{3} + r^{3} - 3 p q r) = 0$
$p^{3} + q^{3} + r^{3} - 3 p q r = 0$
$(p + q + r) (p^{2} + q^{2} + r^{2} - pq - qr - rp) = 0$
$∴ (p + q + r) = 0$
and $p_{2}^{2} + q_{2}^{2} + r_{2}^{2} - pq - qr - rp = 0$
Hence $p^{2} + q^{2} + r^{2} = pq + qr + rp$

JCECE-2011

Straight Line

88791 If the line $y = m x$ meets the lines $x + 2 y - 1 = 0$ and $2 x - y + 3 = 0$ at the same point, then $m$ is equal to

1 1

2 2

3 -2

4 -1

Explanation:

(D) :

$x + 2 y = 1$
$2 x - y = - 3$
From equation (i) and (ii),
$x = - 1, y = 1$
$y = mx$
So, $1 = m (- 1)$
$m = - 1$

SCRA-2014

Straight Line

88792 The distance between the points of concurrency of the two families of straight lines given by $x + (5 λ + 1) y + 1 - 3 λ = 0$ and
$(5 μ + 2) x - 3 y + 3 + 6 μ = 0$ is

1 4

2

\frac{2 \sqrt{2}}{5}

3

\frac{\sqrt{2}}{5}

4 6

Explanation:

(B) : Family of straight lines given as-
$x + (5 λ + 1) y + 1 - 3 λ = 0 \dots .(i)$
$(5 μ + 2) x - 3 y + 3 + 6 μ = 0$
From equation (i)
$(x + y + 1) + λ (5 y - 3) = 0$
$\begin{array}{cl} Which gives x + y + 1 = 0 \\ and 5 y - 3 = 0 \end{array}$
Solving equations (iii) and (iv), we get-
Point $P (\frac{- 8}{5}, \frac{3}{5})$
Similarly from equation (ii)
Which gives
$(2 x - 3 y + 3) + μ (5 x + 6) = 0$
$2 x - 3 y + 3 = 0$
$x = \frac{- 6}{5}$
Solving equation (v) and (vi), we get the point $Q (\frac{- 6}{5}, \frac{1}{5})$
Hence, point of concurrency are $P (\frac{- 8}{5}, \frac{3}{5})$ and
$Q (\frac{- 6}{5}, \frac{1}{5})$
$∴ PQ = \sqrt{{(\frac{- 8}{5} + \frac{6}{5})}^{2} + {(\frac{3}{5} - \frac{1}{5})}^{2}}$
$= \frac{\sqrt{8}}{5} = \frac{2 \sqrt{2}}{5}$
$\Rightarrow p \cos α - q \sin α = \sqrt{3}$ and $p \sin α + q \cos α = - 1$
Adding and squaring, we get
$p^{2} + q^{2} = 3 + 1 \Rightarrow p^{2} + q^{2} = 4$

TS EAMCET-2022-19.07.2022

Straight Line

88793 The straight lines $x + 3 y - 9 = 0, 4 x + 5 y - 1 = 0$ , $p x + q y + 10 = 0$ are concurrent, if the line $5 x + 6 y + 10 = 0$ passes through the point

1

(q, - p)

2

(q, p)

3

(p, - q)

4

(p, q)

Explanation:

(A) : It is given that lines $x + 3 y - 9 = 0$ ,
$4 x + 5 y - 1 = 0, px + qy + 10 = 0$ are concurrent
$∴ | \begin{array}{lll} 1 & 3 & - 9 \\ 4 & 5 & - 1 \\ p & q & 10 \end{array} | = 0$
$\Rightarrow 50 + q - 3 (40 + p) - 9 (4 q - 5 p) = 0$
$\Rightarrow 50 + q - 120 - 3 p - 36 q + 45 p = 0$
$\Rightarrow 42 p - 35 q - 70 = 0$
$\Rightarrow - 6 p + 5 q + 10 = 0$
So, $5 x + 6 y + 10 = 0$ passes through $(q, - p)$ .

TS EAMCET-2020-10.09.2020

Straight Line

88789 If the lines $x = a + m, y = - 2$ and $y = m x$ are concurrent, then least value of $| a |$ is

1

2 \sqrt{2}

2 0

3

\sqrt{2}

4 None of the above

Explanation:

(B) : The given lines are,
$x = a + m$
$y = - 2$
$y = m x$
These lines are concurrent
$∴ mx = ma + m^{2} \Rightarrow m^{2} + ma + 2 = 0$
$m = \frac{- a \pm \sqrt{a^{2} - 8}}{2} \Rightarrow a^{2} \geq 8$

[JCECE-2016]

Straight Line

88790 Three lines $p x + q y + r = 0, q x + r y + p = 0$ and $r x + p y + q = 0$ are concurrent, if

1

p_{2} + q + r = 0

2

p_{3}^{2} + q_{3}^{2} + r^{2} = pq + qr + rp

3

p^{3} + q^{3} + r^{3} = 3 pqr

4 None of the above

Explanation:

(B) : Gvien lines are,
$p x + q y + r = 0$
$q x + r y + p = 0$
$r x + p y + q = 0$
These lines are concurrent
then,
$| \begin{array}{ccc} p & q & r \\ q & r & p \\ r & p & q \end{array} | = 0$
$- (p^{3} + q^{3} + r^{3} - 3 p q r) = 0$
$p^{3} + q^{3} + r^{3} - 3 p q r = 0$
$(p + q + r) (p^{2} + q^{2} + r^{2} - pq - qr - rp) = 0$
$∴ (p + q + r) = 0$
and $p_{2}^{2} + q_{2}^{2} + r_{2}^{2} - pq - qr - rp = 0$
Hence $p^{2} + q^{2} + r^{2} = pq + qr + rp$

JCECE-2011

Straight Line

88791 If the line $y = m x$ meets the lines $x + 2 y - 1 = 0$ and $2 x - y + 3 = 0$ at the same point, then $m$ is equal to

1 1

2 2

3 -2

4 -1

Explanation:

(D) :

$x + 2 y = 1$
$2 x - y = - 3$
From equation (i) and (ii),
$x = - 1, y = 1$
$y = mx$
So, $1 = m (- 1)$
$m = - 1$

SCRA-2014

Straight Line

88792 The distance between the points of concurrency of the two families of straight lines given by $x + (5 λ + 1) y + 1 - 3 λ = 0$ and
$(5 μ + 2) x - 3 y + 3 + 6 μ = 0$ is

1 4

2

\frac{2 \sqrt{2}}{5}

3

\frac{\sqrt{2}}{5}

4 6

Explanation:

(B) : Family of straight lines given as-
$x + (5 λ + 1) y + 1 - 3 λ = 0 \dots .(i)$
$(5 μ + 2) x - 3 y + 3 + 6 μ = 0$
From equation (i)
$(x + y + 1) + λ (5 y - 3) = 0$
$\begin{array}{cl} Which gives x + y + 1 = 0 \\ and 5 y - 3 = 0 \end{array}$
Solving equations (iii) and (iv), we get-
Point $P (\frac{- 8}{5}, \frac{3}{5})$
Similarly from equation (ii)
Which gives
$(2 x - 3 y + 3) + μ (5 x + 6) = 0$
$2 x - 3 y + 3 = 0$
$x = \frac{- 6}{5}$
Solving equation (v) and (vi), we get the point $Q (\frac{- 6}{5}, \frac{1}{5})$
Hence, point of concurrency are $P (\frac{- 8}{5}, \frac{3}{5})$ and
$Q (\frac{- 6}{5}, \frac{1}{5})$
$∴ PQ = \sqrt{{(\frac{- 8}{5} + \frac{6}{5})}^{2} + {(\frac{3}{5} - \frac{1}{5})}^{2}}$
$= \frac{\sqrt{8}}{5} = \frac{2 \sqrt{2}}{5}$
$\Rightarrow p \cos α - q \sin α = \sqrt{3}$ and $p \sin α + q \cos α = - 1$
Adding and squaring, we get
$p^{2} + q^{2} = 3 + 1 \Rightarrow p^{2} + q^{2} = 4$

TS EAMCET-2022-19.07.2022

Straight Line

88793 The straight lines $x + 3 y - 9 = 0, 4 x + 5 y - 1 = 0$ , $p x + q y + 10 = 0$ are concurrent, if the line $5 x + 6 y + 10 = 0$ passes through the point

1

(q, - p)

2

(q, p)

3

(p, - q)

4

(p, q)

Explanation:

(A) : It is given that lines $x + 3 y - 9 = 0$ ,
$4 x + 5 y - 1 = 0, px + qy + 10 = 0$ are concurrent
$∴ | \begin{array}{lll} 1 & 3 & - 9 \\ 4 & 5 & - 1 \\ p & q & 10 \end{array} | = 0$
$\Rightarrow 50 + q - 3 (40 + p) - 9 (4 q - 5 p) = 0$
$\Rightarrow 50 + q - 120 - 3 p - 36 q + 45 p = 0$
$\Rightarrow 42 p - 35 q - 70 = 0$
$\Rightarrow - 6 p + 5 q + 10 = 0$
So, $5 x + 6 y + 10 = 0$ passes through $(q, - p)$ .

TS EAMCET-2020-10.09.2020

Straight Line

88789 If the lines $x = a + m, y = - 2$ and $y = m x$ are concurrent, then least value of $| a |$ is

1

2 \sqrt{2}

2 0

3

\sqrt{2}

4 None of the above

Explanation:

(B) : The given lines are,
$x = a + m$
$y = - 2$
$y = m x$
These lines are concurrent
$∴ mx = ma + m^{2} \Rightarrow m^{2} + ma + 2 = 0$
$m = \frac{- a \pm \sqrt{a^{2} - 8}}{2} \Rightarrow a^{2} \geq 8$

[JCECE-2016]

Straight Line

88790 Three lines $p x + q y + r = 0, q x + r y + p = 0$ and $r x + p y + q = 0$ are concurrent, if

1

p_{2} + q + r = 0

2

p_{3}^{2} + q_{3}^{2} + r^{2} = pq + qr + rp

3

p^{3} + q^{3} + r^{3} = 3 pqr

4 None of the above

Explanation:

(B) : Gvien lines are,
$p x + q y + r = 0$
$q x + r y + p = 0$
$r x + p y + q = 0$
These lines are concurrent
then,
$| \begin{array}{ccc} p & q & r \\ q & r & p \\ r & p & q \end{array} | = 0$
$- (p^{3} + q^{3} + r^{3} - 3 p q r) = 0$
$p^{3} + q^{3} + r^{3} - 3 p q r = 0$
$(p + q + r) (p^{2} + q^{2} + r^{2} - pq - qr - rp) = 0$
$∴ (p + q + r) = 0$
and $p_{2}^{2} + q_{2}^{2} + r_{2}^{2} - pq - qr - rp = 0$
Hence $p^{2} + q^{2} + r^{2} = pq + qr + rp$

JCECE-2011

Straight Line

88791 If the line $y = m x$ meets the lines $x + 2 y - 1 = 0$ and $2 x - y + 3 = 0$ at the same point, then $m$ is equal to

1 1

2 2

3 -2

4 -1

Explanation:

(D) :

$x + 2 y = 1$
$2 x - y = - 3$
From equation (i) and (ii),
$x = - 1, y = 1$
$y = mx$
So, $1 = m (- 1)$
$m = - 1$

SCRA-2014

Straight Line

88792 The distance between the points of concurrency of the two families of straight lines given by $x + (5 λ + 1) y + 1 - 3 λ = 0$ and
$(5 μ + 2) x - 3 y + 3 + 6 μ = 0$ is

1 4

2

\frac{2 \sqrt{2}}{5}

3

\frac{\sqrt{2}}{5}

4 6

Explanation:

(B) : Family of straight lines given as-
$x + (5 λ + 1) y + 1 - 3 λ = 0 \dots .(i)$
$(5 μ + 2) x - 3 y + 3 + 6 μ = 0$
From equation (i)
$(x + y + 1) + λ (5 y - 3) = 0$
$\begin{array}{cl} Which gives x + y + 1 = 0 \\ and 5 y - 3 = 0 \end{array}$
Solving equations (iii) and (iv), we get-
Point $P (\frac{- 8}{5}, \frac{3}{5})$
Similarly from equation (ii)
Which gives
$(2 x - 3 y + 3) + μ (5 x + 6) = 0$
$2 x - 3 y + 3 = 0$
$x = \frac{- 6}{5}$
Solving equation (v) and (vi), we get the point $Q (\frac{- 6}{5}, \frac{1}{5})$
Hence, point of concurrency are $P (\frac{- 8}{5}, \frac{3}{5})$ and
$Q (\frac{- 6}{5}, \frac{1}{5})$
$∴ PQ = \sqrt{{(\frac{- 8}{5} + \frac{6}{5})}^{2} + {(\frac{3}{5} - \frac{1}{5})}^{2}}$
$= \frac{\sqrt{8}}{5} = \frac{2 \sqrt{2}}{5}$
$\Rightarrow p \cos α - q \sin α = \sqrt{3}$ and $p \sin α + q \cos α = - 1$
Adding and squaring, we get
$p^{2} + q^{2} = 3 + 1 \Rightarrow p^{2} + q^{2} = 4$

TS EAMCET-2022-19.07.2022

Straight Line

88793 The straight lines $x + 3 y - 9 = 0, 4 x + 5 y - 1 = 0$ , $p x + q y + 10 = 0$ are concurrent, if the line $5 x + 6 y + 10 = 0$ passes through the point

1

(q, - p)

2

(q, p)

3

(p, - q)

4

(p, q)

Explanation:

(A) : It is given that lines $x + 3 y - 9 = 0$ ,
$4 x + 5 y - 1 = 0, px + qy + 10 = 0$ are concurrent
$∴ | \begin{array}{lll} 1 & 3 & - 9 \\ 4 & 5 & - 1 \\ p & q & 10 \end{array} | = 0$
$\Rightarrow 50 + q - 3 (40 + p) - 9 (4 q - 5 p) = 0$
$\Rightarrow 50 + q - 120 - 3 p - 36 q + 45 p = 0$
$\Rightarrow 42 p - 35 q - 70 = 0$
$\Rightarrow - 6 p + 5 q + 10 = 0$
So, $5 x + 6 y + 10 = 0$ passes through $(q, - p)$ .

TS EAMCET-2020-10.09.2020

Straight Line

88789 If the lines $x = a + m, y = - 2$ and $y = m x$ are concurrent, then least value of $| a |$ is

1

2 \sqrt{2}

2 0

3

\sqrt{2}

4 None of the above

Explanation:

(B) : The given lines are,
$x = a + m$
$y = - 2$
$y = m x$
These lines are concurrent
$∴ mx = ma + m^{2} \Rightarrow m^{2} + ma + 2 = 0$
$m = \frac{- a \pm \sqrt{a^{2} - 8}}{2} \Rightarrow a^{2} \geq 8$

[JCECE-2016]

Straight Line

88790 Three lines $p x + q y + r = 0, q x + r y + p = 0$ and $r x + p y + q = 0$ are concurrent, if

1

p_{2} + q + r = 0

2

p_{3}^{2} + q_{3}^{2} + r^{2} = pq + qr + rp

3

p^{3} + q^{3} + r^{3} = 3 pqr

4 None of the above

Explanation:

(B) : Gvien lines are,
$p x + q y + r = 0$
$q x + r y + p = 0$
$r x + p y + q = 0$
These lines are concurrent
then,
$| \begin{array}{ccc} p & q & r \\ q & r & p \\ r & p & q \end{array} | = 0$
$- (p^{3} + q^{3} + r^{3} - 3 p q r) = 0$
$p^{3} + q^{3} + r^{3} - 3 p q r = 0$
$(p + q + r) (p^{2} + q^{2} + r^{2} - pq - qr - rp) = 0$
$∴ (p + q + r) = 0$
and $p_{2}^{2} + q_{2}^{2} + r_{2}^{2} - pq - qr - rp = 0$
Hence $p^{2} + q^{2} + r^{2} = pq + qr + rp$

JCECE-2011

Straight Line

88791 If the line $y = m x$ meets the lines $x + 2 y - 1 = 0$ and $2 x - y + 3 = 0$ at the same point, then $m$ is equal to

1 1

2 2

3 -2

4 -1

Explanation:

(D) :

$x + 2 y = 1$
$2 x - y = - 3$
From equation (i) and (ii),
$x = - 1, y = 1$
$y = mx$
So, $1 = m (- 1)$
$m = - 1$

SCRA-2014

Straight Line

88792 The distance between the points of concurrency of the two families of straight lines given by $x + (5 λ + 1) y + 1 - 3 λ = 0$ and
$(5 μ + 2) x - 3 y + 3 + 6 μ = 0$ is

1 4

2

\frac{2 \sqrt{2}}{5}

3

\frac{\sqrt{2}}{5}

4 6

Explanation:

(B) : Family of straight lines given as-
$x + (5 λ + 1) y + 1 - 3 λ = 0 \dots .(i)$
$(5 μ + 2) x - 3 y + 3 + 6 μ = 0$
From equation (i)
$(x + y + 1) + λ (5 y - 3) = 0$
$\begin{array}{cl} Which gives x + y + 1 = 0 \\ and 5 y - 3 = 0 \end{array}$
Solving equations (iii) and (iv), we get-
Point $P (\frac{- 8}{5}, \frac{3}{5})$
Similarly from equation (ii)
Which gives
$(2 x - 3 y + 3) + μ (5 x + 6) = 0$
$2 x - 3 y + 3 = 0$
$x = \frac{- 6}{5}$
Solving equation (v) and (vi), we get the point $Q (\frac{- 6}{5}, \frac{1}{5})$
Hence, point of concurrency are $P (\frac{- 8}{5}, \frac{3}{5})$ and
$Q (\frac{- 6}{5}, \frac{1}{5})$
$∴ PQ = \sqrt{{(\frac{- 8}{5} + \frac{6}{5})}^{2} + {(\frac{3}{5} - \frac{1}{5})}^{2}}$
$= \frac{\sqrt{8}}{5} = \frac{2 \sqrt{2}}{5}$
$\Rightarrow p \cos α - q \sin α = \sqrt{3}$ and $p \sin α + q \cos α = - 1$
Adding and squaring, we get
$p^{2} + q^{2} = 3 + 1 \Rightarrow p^{2} + q^{2} = 4$

TS EAMCET-2022-19.07.2022

Straight Line

88793 The straight lines $x + 3 y - 9 = 0, 4 x + 5 y - 1 = 0$ , $p x + q y + 10 = 0$ are concurrent, if the line $5 x + 6 y + 10 = 0$ passes through the point

1

(q, - p)

2

(q, p)

3

(p, - q)

4

(p, q)

Explanation:

(A) : It is given that lines $x + 3 y - 9 = 0$ ,
$4 x + 5 y - 1 = 0, px + qy + 10 = 0$ are concurrent
$∴ | \begin{array}{lll} 1 & 3 & - 9 \\ 4 & 5 & - 1 \\ p & q & 10 \end{array} | = 0$
$\Rightarrow 50 + q - 3 (40 + p) - 9 (4 q - 5 p) = 0$
$\Rightarrow 50 + q - 120 - 3 p - 36 q + 45 p = 0$
$\Rightarrow 42 p - 35 q - 70 = 0$
$\Rightarrow - 6 p + 5 q + 10 = 0$
So, $5 x + 6 y + 10 = 0$ passes through $(q, - p)$ .

TS EAMCET-2020-10.09.2020

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Question Bank Series

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