Straight Lines and Pair of Straight Lines — Page 5 | JEE Mathematics

Q41

Let

a, b, c

and

d

be non-zero numbers. If the point of intersection of the lines

4ax + 2ay + c = 0

and

5bx + 2by + d = 0

lies in the fourth quadrant and is equidistant from the two axes then :

A

3bc - 2ad = 0

B

3bc + 2ad = 0

C

2bc - 3ad = 0

D

2bc + 3ad = 0

Correct Answer

Option A

Solution

Since the point of intersection lies on fourth quadrant and equidistant from the two axes, i.e., let the point be (k, $-$ k) and this point satisfies the two equations of the given lines. $\therefore$ 4ak $-$ 2ak + c = 0 ......... (1) and 5bk $-$ 2bk + d = 0 .....

(2) From (1) we get,

k = {{ - c} \over {2a}}

Putting the value of k in (2) we get,

5b\left( { - {c \over {2a}}} \right) - 2b\left( { - {c \over {2a}}} \right) + d = 0

or,

- {{5bc} \over {2a}} + {{2bc} \over {2a}} + d = 0

or,

- {{3bc} \over {2a}} + d = 0

or,

- 3bc + 2ad = 0

or,

3bc - 2ad = 0

Q42

Consider the set of all lines px + qy + r = 0 such that 3p + 2q + 4r = 0. Which one of the following statements is true?

A The lines are not concurrent

B The lines are concurrent at the point

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

C The lines are all parallel

D Each line passes through the origin

Correct Answer

Option B

Solution

Equation of lines; px + qy + r = 0 . . . . .

(1) Also given 3p + 2q + 4r = 0 . . . . . . (2) divide equation (2) by 4, we get

{3 \over 4}P + {2 \over 4}q + r = 0

. . . . (3) By comparing (1) and (3) we get, x =

{3 \over 4}

and y =

{2 \over 4}

=

{1 \over 2}

For any value of p,q and r, the equation of set of lines will pan through

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

Q43

Let

PS

be the median of the triangle with vertices

P(2, 2)

,

Q(6, -1)

and

R(7, 3)

. The equation of the line passing through

(1, -1)

band parallel to PS is :

A

4x + 7y + 3 = 0

B

2x - 9y - 11 = 0

C

4x - 7y - 11 = 0

D

2x + 9y + 7 = 0

Correct Answer

Option D

Solution

Let

P,Q,R,

be the vertices of

\Delta PQR

Since

PS

is the median,

S

is mid-point of

QR

So,

S = \left( {{{7 + 6} \over 2},{{3 - 1} \over 2}} \right) = \left( {{{13} \over 2},1} \right)

Now, slope of

PS

= {{2 - 1} \over {2 - {{13} \over 2}}} = - {2 \over 9}

Since, required line is parallel to

PS

therefore slope of required line $=$ slope of

PS

Now, equation of line passing through

(1, -1)

and having slope

- {2 \over 9}

is

y - \left( { - 1} \right) = - {2 \over 9}\left( {x - 1} \right)

9y + 9 = - 2x + 2

\Rightarrow 2x + 9y + 7 = 0

Q44

The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices

(0, 0)

(0, 41)

and

(41, 0)

is :

A 820

B 780

C 901

D 861

Correct Answer

Option B

Solution

The number of integral points lie inside the triangle are 1.

If x = 1, then y may be 1, 2, 3, ....., 39 2.

If x = 2, then y may be 1, 2, 3, ....., 38 3.

If x = 3, then y may be 1, 2, 3, ....., 37 $\vdots$ 39.

If x = 39, then the value of y is 1.

Hence, the number of interior points are

1 + 2 + 3 + .... + 39 = {{39 \times 40} \over 2} = 780

Q45

Two sides of a rhombus are along the lines,

x - y + 1 = 0

and

7x - y - 5 = 0

. If its diagonals intersect at

(-1, -2)

, then which one of the following is a vertex of this rhombus?

A

\left( {{{ 1} \over 3}, - {8 \over 3}} \right)

B

\left( - {{{ 10} \over 3}, - {7 \over 3}} \right)

C

\left( { - 3, - 9} \right)

D

\left( { - 3, - 8} \right)

Correct Answer

Option A

Solution

Let other two sides of rhombus are

x - y + \lambda = 0

and

7x - y + \mu = 0

then

O

is equidistant from

AB

and

DC

and from

AD

and

BC

$\therefore$

\left| { - 1 + 2 + 1} \right| = \left| { - 1 + 2 + \lambda } \right| \Rightarrow \lambda = - 3

and

\left| { - 7 + 2 - 5} \right| = \left| { - 7 + 2 + \mu } \right| \Rightarrow \mu = 15

$\therefore$ Other two sides are

x-y-3=0

and

7x-y+15=0

On solving the equations of sides pairwise, we get the vertices as

\left( {{1 \over 3},{{ - 8} \over 3}} \right),\left( {1,2} \right),\left( {{{ - 7} \over 3},{{ - 4} \over 3}} \right),\left( { - 3, - 6} \right)

Q46

If a variable line drawn through the intersection of the lines

{x \over 3} + {y \over 4} = 1

and

{x \over 4} + {y \over 3} = 1,

meets the coordinate axes at A and B, (A

\ne

B), then the locus of the midpoint of AB is :

A 6xy = 7(x + y)

B 4(x + y)2 − 28(x + y) + 49 = 0

C 7xy = 6(x + y)

D 14(x + y)2 − 97(x + y) + 168 = 0

Correct Answer

Option C

Solution

L1 : 4x + 3y $-$ 12 = 0 L2 : 3x + 4y $-$ 12 = 0 Equation of line passing through the intersection of these two lines L1 and L2 is L1 + $\lambda$ L2 = 0 $\Rightarrow$

\,\,\,

(4x + 3y $-$ 12) + $\lambda$ (3x + 4y $-$ 12) = 0 $\Rightarrow$

\,\,\,

x(4 + 3 $\lambda$ ) + y(3 + 4 $\lambda$ ) $-$ 12(1 + $\lambda$ ) = 0 this line meets x coordinate at point A and y coordinate at point B.

\therefore\,\,\,

Point A =

\left( {{{12\left( {1 + \lambda } \right)} \over {4 + 3\lambda }},0} \right)

and Point B =

\left( {0,\,\,{{12\left( {1 + \lambda } \right)} \over {3 + 4\lambda }}} \right)

Let coordinate of midpoint of line AB is (h, k).

\therefore\,\,\,

h =

{{6\left( {1 + \lambda } \right)} \over {4 + 3\lambda }}

. . . . . (1) and k =

{{6\left( {1 + \lambda } \right)} \over {3 + 4\lambda }}

. . . . (2) Eliminate $\lambda$ from (1) and (2), then we get 6(h + k) = 7 hk

\therefore\,\,\,

Locus of midpoint of line AB is , 6(x + y) = 7xy

Q47

Let k be an integer such that the triangle with vertices (k, – 3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point :

A

\left( {1,{3 \over 4}} \right)

B

\left( {1, - {3 \over 4}} \right)

C

\left( {2,{1 \over 2}} \right)

D

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

Correct Answer

Option C

Solution

Given, vertices of triangle are (k, – 3k), (5, k) and (–k, 2).

{1 \over 2}\left| \begin{array}{lll}k & { - 3k} & 1 \\ 5 & k & 1 \\ { - k} & 2 & 1 \end{array} \right| = \pm 28

$\Rightarrow$ k(k - 2) + 3k(5 + k) + 1(10 + k2) = $\pm$ 56 $\Rightarrow$ 5k2 + 13k + 10 = $\pm$ 56 $\Rightarrow$ 5k2 + 13k - 66 = 0 $\Rightarrow$ k =

{{ - 13 \pm \sqrt { - 1151} } \over {10}}

So no real solution exist. or 5k2 + 13k - 46 = 0 $\therefore$ k =

{{ - 23} \over 5}

or k = 2 since k is an integer $\therefore$ k = 2 Thus, the coordinate of vertices of triangle are A(2, -6), B(5, 2) and C(-2, 2).

Now, equation of altitude from vertex A is y - (-6) =

{{ - 1} \over {\left( {{{2 - 2} \over { - 2 - 5}}} \right)}}\left( {x - 2} \right)

$\Rightarrow$ x = 2 .......(1) Equation of altitude from vertex B is y - 2 =

{{ - 1} \over {\left( {{{2 + 6} \over { - 2 - 2}}} \right)}}\left( {x - 5} \right)

$\Rightarrow$ 2y - 4 = x - 5 $\Rightarrow$ x - 2y = 1 .......(

2) Point H( $\alpha$ , $\beta$ ) lies on both (1) and (2), $\therefore$ $\alpha$ = 2 .........(

3) $\alpha$ - 2 $\beta$ = 1 ......(

4) Solving (3) and (4), we get $\alpha$ = 2 , $\beta$ =

{1 \over 2}

$\therefore$ Orthocentre is

\left( {2,{1 \over 2}} \right)

.

Q48

A square, of each side 2, lies above the x-axis and has one vertex at the origin. If one of the sides passing through the origin makes an angle 30o with the positive direction of the x-axis, then the sum of the x-coordinates of the vertices of the square is :

A

2\sqrt 3 - 1

B

2\sqrt 3 - 2

C

\sqrt 3 - 2

D

\sqrt 3 - 1

Correct Answer

Option B

Solution

Let, coordinate of point A = (x, y).

\therefore\,\,\,

For point A,

{x \over {\cos {{30}^ \circ }}}

=

{y \over {\sin {{30}^ \circ }}}

= 2 $\Rightarrow$ x =

\sqrt 3

and y = 1 Similarly, For point B,

{x \over {\cos {{75}^ \circ }}}

=

{y \over {\sin {{75}^ \circ }}}

= 2

\sqrt 2

\therefore\,\,\,

x =

\sqrt 3 - 1

y =

\sqrt 3 + 1

For point C,

{x \over {cos{{120}^ \circ }}}

=

{y \over {sin{{120}^ \circ }}}

= 2 $\Rightarrow$

\,\,\,

x = $-$ 1 y =

\sqrt 3

\therefore\,\,\,

Sum of the x - coordinate of the vertices = 0 +

\sqrt 3

+

\sqrt 3

$-$ 1 + ( $-$ 1) = 2

\sqrt 3

$-$ 2

Q49

A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is :

A 3x + 2y = 6xy

B 3x + 2y = 6

C 2x + 3y = xy

D 3x + 2y = xy

Correct Answer

Option D

Solution

Let coordinate of point R = (h, k).

Equation of line PQ, (y $-$ 3) = m (x $-$ 2).

Put y = 0 to get coordinate of point p, 0 $-$ 3 = (x $-$ 2) $\Rightarrow$ x = 2 $-$

{3 \over m}

\therefore\,\,\,

p = (2 $-$

{3 \over m}

, 0) As p = (h, 0) then h = 2 $-$

{3 \over m}

$\Rightarrow$

{3 \over m}

= 2 $-$ h $\Rightarrow$ m =

{3 \over {2 - h}}

. . . . . . (1) Put x = 0 to get coordinate of point Q, y $-$ 3 $=$ m (0 $-$ 2) $\Rightarrow$ y = 3 $-$ 2m

\therefore\,\,\,

point Q = (0, 3 $-$ 2m) And From the graph you can see Q = (0, k).

\therefore\,\,\,

k = 3 $-$ 2m $\Rightarrow$ m =

{{3 - k} \over 2}

. . . . (2) By comparing (1) and (2) get

{3 \over {2 - h}} = {{3 - k} \over 2}

$\Rightarrow$ (2 $-$ h)(3 $-$ k) = 6 $\Rightarrow$ 6 $-$ 3h $-$ 2K + hk = 6 $\Rightarrow$ 3 h + 2K = hk

\therefore\,\,\,

locus of point R is 3x + 2y= xy

Q50

A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of 60o with the line x + y = 0. Then an equation of the line L is :

A x +

\sqrt 3

y = 8

B

\sqrt 3

x + y = 8

C (

\sqrt 3

+ 1)x + (

\sqrt 3

– 1)y = 8

\sqrt 2

D (

\sqrt 3

- 1)x + (

\sqrt 3

+ 1)y = 8

\sqrt 2

Correct Answer

Option D

Solution

The equation of line is x cos $\theta$ + y sin $\theta$ = p $\Rightarrow$ x cos (75o) + y sin (75o) = 4 $\Rightarrow$

x\left( {{{\sqrt 3 - 1} \over {2\sqrt 2 }}} \right) + y\left( {{{\sqrt 3 + 1} \over {2\sqrt 2 }}} \right) = 4

$\Rightarrow$

x\left( {\sqrt 3 - 1} \right) + y\left( {\sqrt 3 + 1} \right) = 8\sqrt 2