Straight Lines and Pair of Straight Lines — Page 7 | JEE Mathematics

Q61

A ray of light coming from the point (2,

2\sqrt 3

) is incident at an angle 30o on the line x = 1 at the point A. The ray gets reflected on the line x = 1 and meets x-axis at the point B. Then, the line AB passes through the point :

A (3, -

\sqrt 3

)

B (4, -

\sqrt 3

)

C

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

D

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

Correct Answer

Option A

Solution

Equation of reflected Ray P'B : y - 2

\sqrt 3

= tan 120o (x - 0) $\Rightarrow$

\sqrt 3

x + y = 2

\sqrt 3

(3, -

\sqrt 3

) satisfy the line.

Q62

If the perpendicular bisector of the line segment joining the points P(1 ,4) and Q(k, 3) has y-intercept equal to –4, then a value of k is :

A

\sqrt {14}

B -4

C –2

D

\sqrt {15}

Correct Answer

Option B

Solution

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

$\therefore$ Slope of perpendicular bisector of PQ,

{m_ \bot } = k - 1

mid point of PQ =

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

equation of perpendicular bisector

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

for y intercept put x = 0

y = {7 \over 2} - \left( {{{{k^2} - 1} \over 2}} \right) = - 4

{{{k^2} - 1} \over 2} = {{15} \over 2} \Rightarrow k = \pm 4

Q63

Two vertical poles AB = 15 m and CD = 10 m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m) above the line AC is :

A 10/3

B 5

C 20/3

D 6

Correct Answer

Option D

Solution

Equation of AD :

y = {{10} \over a}x

Equation of BC :

{x \over a} + {y \over {15}} = 1

\Rightarrow {{ay} \over {10a}} + {y \over {15}} = 1

\Rightarrow {{3y + 2y} \over {30}} = 1

\Rightarrow y = 6

$\therefore$ y coordinate of point P = 6 = height of point P above the line AC.

Q64

The set of all possible values of

\theta

in the interval (0,

\pi

) for which the points (1, 2) and (sin

\theta

, cos

\theta

) lie on the same side of the line x + y = 1 is :

A

\left( {0,{\pi \over 4}} \right)

B

\left( {0,{{3\pi } \over 4}} \right)

C

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

D

\left( {0,{\pi \over 2}} \right)

Correct Answer

Option D

Solution

Let f(x, y) = x + y - 1

\because f\left( {1,2} \right).f\left( {\sin \theta ,\cos \theta } \right) > 0

\Rightarrow 2\left[ {\sin \theta + \cos \theta - 1} \right] > 0

\Rightarrow \sin \theta + \cos \theta > 1

\Rightarrow \sin \left( {\theta + {\pi \over 4}} \right) > {1 \over {\sqrt 2 }}

\Rightarrow \theta + {\pi \over 4} \in \left( {{\pi \over 4},{{3\pi } \over 4}} \right)

\Rightarrow \theta \in \left( {0,{\pi \over 2}} \right)

Q65

Let C be the centroid of the triangle with vertices (3, –1), (1, 3) and (2, 4). Let P be the point of intersection of the lines x + 3y – 1 = 0 and 3x – y + 1 = 0. Then the line passing through the points C and P also passes through the point :

A (–9, –7)

B (9, 7)

C (7, 6)

D (–9, –6)

Correct Answer

Option D

Solution

Centroid C

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

= (2, 2) Point of intersection of lines x + 3y – 1 = 0 and 3x – y + 1 = 0 is P

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

So, equation of line CP is 8x – 11y + 6 = 0 Point (–9, –6) satisfy this equation.

Q66

Let two points be A(1, –1) and B(0, 2). If a point P(x', y') be such that the area of

\Delta

PAB = 5 sq. units and it lies on the line, 3x + y – 4

\lambda

= 0, then a value of

\lambda

is :

A 4

B 1

C -3

D 3

Correct Answer

Option D

Solution

Point P(x', y') lies on the line 3x + y – 4 $\lambda$ = 0 $\therefore$ 3x' + y' – 4 $\lambda$ = 0 Area of

\Delta

PAB = 5 $\Rightarrow$

{1 \over 2}\left| \begin{array}{lll}1 & 1 & { - 1} \\ 1 & 0 & 2 \\ 1 & {x'} & {y'} \end{array} \right|

= $\pm$ 5 $\Rightarrow$ [(–2x') – (y' – 2) – (x')] = $\pm$ 10 $\Rightarrow$ – 3x' – y' + 2 = $\pm$ 10 $\Rightarrow$ 2 - 4 $\lambda$ = $\pm$ 10 $\Rightarrow$ $\lambda$ = -2 or $\lambda$ = 3

Q67

The locus of the mid-points of the perpendiculars drawn from points on the line, x = 2y to the line x = y is :

A 3x - 2y = 0

B 7x - 5y = 0

C 2x - 3y = 0

D 5x - 7y = 0

Correct Answer

Option D

Solution

Slope of line y = x is 1 Line AB is perpendicular to line y = x so Slope of AB = -1 Also slope of AB =

{{\alpha - \beta } \over {2\alpha - \beta }}

$\therefore$

{{\alpha - \beta } \over {2\alpha - \beta }}

= -1 $\Rightarrow$ 3 $\alpha$ = 2 $\beta$ h =

{{2\alpha + \beta } \over 2}

$\Rightarrow$ 2h =

{{4\alpha + 2\beta } \over 2}

=

{{4\alpha + 3\alpha } \over 2}

=

{{7\alpha } \over 2}

Also k =

{{\alpha + \beta } \over 2}

$\Rightarrow$ 2k =

{{5\alpha } \over 2}

So

{h \over k} = {7 \over 5}

$\Rightarrow$ 5h = 7k $\Rightarrow$ 5x = 7y

Q68

A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this line on the coordinate axes is

{1 \over 4}

. Three stones A, B and C are placed at the points (1, 1), (2, 2) and (4, 4) respectively. Then, which of these stones is / are on the path of the man?

A A only

B All the three

C C only

D B only

Correct Answer

Option D

Solution

Given, position of A = (1, 1) Position of B = (2, 2) Position of C = (4, 4) Let x-intercept be a and y-intercept be b.

Equation of line traced is

{x \over a} + {y \over b} = 1

This is the equation of path, let a point (h, k) lie on this path. Then,

{h \over a} + {k \over b} = 1

Also, AM of reciprocal of a and b =

{1 \over 4}

$\therefore$

{{{1 \over a} + {1 \over b}} \over 2} = {1 \over 4}

{1 \over a} + {1 \over b} = {1 \over 2}

On comparing Eqs. (i) and (ii), we get (h, k) = (2, 2) Hence, the required stone is B(2, 2).

Q69

The intersection of three lines x

-

y = 0, x + 2y = 3 and 2x + y = 6 is a :

A Right angled triangle

B Equilateral triangle

C None of the above

D Isosceles triangle

Correct Answer

Option D

Solution

The given three lines are x $-$ y = 0, x + 2y = 3 and 2x + y = 6 then point of intersection, lines x $-$ y = 0 and x + 2y = 3 is (1, 1) lines x $-$ y = 0 and 2x + y = 6 is (2, 2) and lines x + 2y = 3 and 2x + y = 0 is (3, 0) The triangle ABC has vertices A(1, 1), B(2, 2) and C(3, 0) $\therefore$ AB =

\sqrt 2

, BC =

\sqrt 5

and AC =

\sqrt 5

$\therefore$

\Delta

ABC is isosceles

Q70

Let A(

-

1, 1), B(3, 4) and C(2, 0) be given three points. A line y = mx, m > 0, intersects lines AC and BC at point P and Q respectively. Let A1 and A2 be the areas of

\Delta

ABC and

\Delta

PQC respectively, such that A1 = 3A2, then the value of m is equal to :

A 1

B 3

C 2

D

{4 \over {15}}

Correct Answer

Option A

Solution

{A_1} = \Delta ABC = {1 \over 2}\left| {\left| \begin{array}{lll}{ - 1} & 1 & 1 \\ 2 & 0 & 1 \\ 3 & 4 & 1 \end{array} \right|} \right|

{A_1} = {{13} \over 2}

Equation of line AC is

y - 1 = - {1 \over 3}(x + 1)

Line AC intersect with line y = mx at P, Solving we get

P\left( {{2 \over {3m + 1}},{{2m} \over {3m + 1}}} \right)

Equation of line BC is y $-$ 0 = 4(x $-$ 2) Line BC intersect with line y = mx at Q, Solving we get

Q\left( {{{ - 8} \over {m - 4}},{{ - 8m} \over {m - 4}}} \right)

A2 = Area of

\Delta PQC = {1 \over 2}\left| {\left| \begin{array}{lll}2 & 0 & 1 \\ {{2 \over {3m + 1}}} & {{{2m} \over {3m + 1}}} & 1 \\ {{{ - 8} \over {m - 4}}} & {{{ - 8m} \over {m - 4}}} & 1 \end{array} \right|} \right|\, = {{{A_1}} \over 3} = {{13} \over 6}

$\Rightarrow$

{1 \over 2}\left( {2\left( {{{2m} \over {3m + 1}} + {{8m} \over {m - 4}}} \right) - 1\left( {{{ - 16m} \over {(3m + 1)(m - 4)}} + {{16m} \over {(3m + 1)(m - 4)}}} \right)} \right)

= \pm {{13} \over 6}

$\Rightarrow$

{{26{m^2}} \over {3{m^2} - 11m - 4}} = \pm {{13} \over 6}

\Rightarrow 12{m^2} = \pm (3{m^2} - 11m - 4)

taking +ve sign 9m2 + 11m + 4 = 0 (Rejected $\because$ m is imaginary) taking $-$ ve sign 15m2 $-$ 11m $-$ 4 = 0

m = 1, - {4 \over {15}}

$\therefore$ m = 1 (As given m > 0)