Straight Lines and Pair of Straight Lines — Page 2 | JEE Mathematics

Q11

If the line

2x + y = k

passes through the point which divides the line segment joining the points

(1, 1)

and

(2, 4)

in the ratio

3 : 2

, then

k

equals :

A

{{29 \over 5}}

B

5

C

6

D

{{11 \over 5}}

Correct Answer

Option C

Solution

The point which divides the line segment joining the points (1, 1) and (2, 4) in the ratio 3 : 2 is

= \left( {{{3 \times 2 + 2 \times 1} \over {3 + 2}},{{3 \times 4 + 2 \times 1} \over {3 + 2}}} \right)

= \left( {{{6 + 2} \over 5},{{12 + 2} \over 5}} \right) = \left( {{8 \over 5},{{14} \over 5}} \right)

Since the line 2x + y = k passes through this point, $\therefore$

2 \times {8 \over 5} + {{14} \over 5} = k

or

{{30} \over 5} = k

or, k = 6

Q12

If one of the lines given by

6{x^2} - xy + 4c{y^2} = 0

is

3x + 4y = 0,

then

c

equals :

A

-3

B

-1

C

3

D

1

Correct Answer

Option A

Solution

3x+4y=0

is one of the lines of the pair

6{x^2} - xy + 4c{y^2} = 0,

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

Put

y = - {3 \over 4}x,

we get

6{x^2} + {3 \over 4}{x^2} + 4c{\left( { - {3 \over 4}x} \right)^2} = 0

\Rightarrow 6 + {3 \over 4} + {{9c} \over 4} = 0 \Rightarrow c = - 3

Q13

Let

A\left( {2, - 3} \right)

and

B\left( {-2, 1} \right)

be vertices of a triangle

ABC

. If the centroid of this triangle moves on the line

2x + 3y = 1

, then the locus of the vertex

C

is the line :

A

3x - 2y = 3

B

2x - 3y = 7

C

3x + 2y = 5

D

2x + 3y = 9

Correct Answer

Option D

Solution

Let the vertex

C

be

(h,k),

then the centroid of

\Delta ABC

is

\left( {{{2 + (- 2) + h} \over 3},{{ - 3 + 1 + k} \over 3}} \right)

or

\left( {{h \over 3},{{ - 2 + k} \over 3}} \right).

It lies on

2x+3y=1

\Rightarrow {{2h} \over 3} - 2 + k = 1

\Rightarrow 2h + 3k = 9

$\therefore$ Locus of

C

is

2x+3y=9

Q14

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

A 1

B 2

C -2

D -4

Correct Answer

Option D

Solution

Slope of

PQ = {{3 - 4} \over {k - 1}} = {{ - 1} \over {k - 1}}

$\therefore$ Slope of perpendicular bisector of

PQ = \left( {k - 1} \right)

Also mid point of

PQ\left( {{{k + 1} \over 2},{7 \over 2}} \right).

Equation of perpendicular bisector is

y - {7 \over 2} = \left( {k - 1} \right)\left( {x - {{k + 1} \over 2}} \right)

\Rightarrow 2y - 7 = 2\left( {k - 1} \right)x - \left( {{k^2} - 1} \right)

\Rightarrow 2\left( {k - 1} \right)x - 2y + \left( {8 - {k^2}} \right) = 0

$\therefore$

y

-intercept

= {{8 - {k^2}} \over { - 2}} = - 4

$\Rightarrow$

8 - {k^2} = - 8

or

{k^2} = 16 \Rightarrow k = \pm 4

Q15

The shortest distance between the line

y - x = 1

and the curve

x = {y^2}

is :

A

{{2\sqrt 3 } \over 8}

B

{{3\sqrt 2 } \over 5}

C

{{\sqrt 3 } \over 4}

D

{{3\sqrt 2 } \over 8}

Correct Answer

Option D

Solution

Let

\left( {{a^2},a} \right)

be the point of shortest distance on

x = {y^2}

Then distance between

\left( {{a^2},a} \right)

and line

x - y + 1 = 0

is given by

\,\,\,\,\,\,\,\,D = {{{a^2} - a + 1} \over {\sqrt 2 }} = {1 \over {\sqrt 2 }}\left[ {{{\left( {a - {1 \over 2}} \right)}^2} + {3 \over 4}} \right]

It is min when

a = {1 \over 2}

and

D{}_{\min } = {3 \over {4\sqrt 2 }} = {{3\sqrt 2 } \over 8}

Q16

A point on the straight line, 3x + 5y = 15 which is equidistant from the coordinate axes will lie only in :

A 1st and 2nd qudratants

B 4th qudratant

C 1st and 2nd and 4th qudratants

D 1st qudratant

Correct Answer

Option A

Solution

Let the point is P(x, y).

According to the question, the point P(x, y) is equidistance from both x and y axis. $\therefore$ |x| = |y| $\Rightarrow$ x = $\pm$ y So the point P lies on the either x = y or x = - y line.

And point P(x, y) also lies on the straight line 3x + 5y = 15.

Form the graph, you can see the point P can either be on 1st qudratant or 2nd qudratant.

Q17

If the straight line, 2x – 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15,

\beta

), then

\beta

equals :

A

{{35} \over 3}

B

-

5

C

-

{{35} \over 3}

D 5

Correct Answer

Option D

Solution

{{17 - \beta } \over { - 8}} \times {2 \over 3} = - 1

$\beta$ = 5

Q18

The pair of lines represented by

3a{x^2} + 5xy + \left( {{a^2} - 2} \right){y^2} = 0

$ are perpendicular to each other for :

A two values of

a

B

\forall \,a

C for one value of

a

D for no values of

a

Correct Answer

Option A

Solution

3a + {a^2} - 2 = 0 \Rightarrow {a^2} + 3a - 2 = 0;

\Rightarrow a = {{ - 3 \pm \sqrt {9 + 8} } \over 2} = {{ - 3 \pm \sqrt {17} } \over 2}

Q19

If the pair of lines

a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0

intersect on the

y

-axis then :

A

2fgh = b{g^2} + c{h^2}

B

b{g^2} \ne c{h^2}

C

abc = 2fgh

D none of these

Correct Answer

Option A

Solution

Put

x=0

in the given equation

\Rightarrow b{y^2} + 2fy + c = 0.

For unique point of intersection

{f^2} - bc = 0

\Rightarrow a{f^2} - abc = 0.

Since

abc + 2fgh - a{f^2} - b{g^2} - c{h^2} = 0

\Rightarrow 2fgh - b{g^2} - c{h^2} = 0

Q20

The equation of the straight line passing through the point

(4, 3)

and making intercepts on the co-ordinate axes whose sum is

-1

is :

A

{x \over 2} - {y \over 3} = 1

and

{x \over -2} +{y \over 1} = 1

B

{x \over 2} - {y \over 3} = -1

and

{x \over -2} +{y \over 1} = -1

C

{x \over 2} + {y \over 3} = 1

and

{x \over 2} +{y \over 1} = 1

D

{x \over 2} + {y \over 3} = -1

and

{x \over -2} +{y \over 1} = -1

Correct Answer

Option A

Solution

Let the required line be

{x \over a} + {y \over b} = 1.......\left( 1 \right)

then

a+b=-1

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.........\left( 2 \right)

(1)

passes through

(4,3),

\Rightarrow {4 \over a} + {3 \over b} = 1

\Rightarrow 4b + 3a = ab\,\,...............\left( 3 \right)

Eliminating

b

from

(2)

and

(3),

we get

{a^2} - 4 = 0 \Rightarrow a = \pm 2 \Rightarrow b = - 3

,

1

$\therefore$ Equation of straight lines are

{x \over 2} + {y \over { - 3}} = 1

or

{x \over { - 2}} + {y \over 1} = 1