Straight Lines and Pair of Straight Lines — Page 9 | JEE Mathematics

Q81

The distance between the two points A and A' which lie on y = 2 such that both the line segments AB and A' B (where B is the point (2, 3)) subtend angle

{\pi \over 4}

at the origin, is equal to :

A 10

B

{48 \over 5}

C

{52 \over 5}

D 3

Correct Answer

Option C

Solution

Let $A(\alpha, 2) \quad$ Given $B(2,3)$

\begin{aligned} & m_{O A}=\frac{2}{\alpha} \quad\&\quad m_{O B}=\frac{3}{2} \\\\ & \tan \frac{\pi}{4}=\left|\frac{\frac{2}{\alpha}-\frac{3}{2}}{1+\frac{2}{\alpha} \cdot \frac{3}{2}}\right| \Rightarrow \frac{4-3 \alpha}{2 \alpha+6}=\pm 1 \\\\ & 4-3 \alpha=2 \alpha+6 \quad \& 4-3 \alpha=-2 \alpha-6 \\\\ & \alpha=\frac{-2}{5} \& \alpha=10 \\\\ & A\left(-\frac{2}{5}, 2\right) \& A^{\prime}(10,2) \text{ and } B(2,3) \\\\ & A A^{\prime}=10+\frac{2}{5}=\frac{52}{5} \end{aligned}

Q82

The distance of the origin from the centroid of the triangle whose two sides have the equations

x - 2y + 1 = 0

and

2x - y - 1 = 0

and whose orthocenter is

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

is :

A

\sqrt 2

B 2

C 2

\sqrt 2

D 4

Correct Answer

Option C

Solution

For point A, 2x $-$ y $-$ 1 = 0 ...... (1) x $-$ 2y + 1 = 0 ......

(2) Solving (1) and (2), we get x = 1, y = 1.

$\therefore$ Point A = (1, 1) Altitude from B to line AC is perpendicular to line AC.

$\therefore$ Equator of altitude BH is 2x + y + $\lambda$ = 0 ......

(3) It passes through point

H\left( {{7 \over 3},{7 \over 3}} \right)

so it satisfy the equation (3).

{{14} \over 3} + {7 \over 3} + \lambda = 0

$\Rightarrow$ $\alpha$ = $-$ 7 $\therefore$ Altitude BH = 2x + y $-$ 7 = 0 ......

(4) Solving equation (1) and (4), we get x = 2, y = 3.

$\therefore$ Point B = (2, 3) Altitude from C to line AB is perpendicular to line AB.

$\therefore$ Equation of altitude CH is x + 2y + $\lambda$ = 0 ......

(5) It passes through point

H\left( {{7 \over 3},{7 \over 3}} \right)

so it satisfy equation (5).

{7 \over 3} + {{14} \over 3} + \lambda = 0

$\Rightarrow$ $\lambda$ = $-$ 7 $\therefore$ Altitude CH = x + 2y $-$ 7 = 0 ......

(6) Solving equation (2) and (6), we get x = 3, y = 2 $\therefore$ Point C = (3, 2) Centroid G (x, y) of triangle A (1, 1), B (2, 3) and C (3, 2) is

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

,

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

Now, distance of point G (2, 2) from center O (0, 0) is

OG = \sqrt {{2^2} + {2^2}} = 2\sqrt 2

Q83

Let a triangle be bounded by the lines L1 : 2x + 5y = 10; L2 :

-

4x + 3y = 12 and the line L3, which passes through the point P(2, 3), intersects L2 at A and L1 at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to :

A

{{110} \over {13}}

B

{{132} \over {13}}

C

{{142} \over {13}}

D

{{151} \over {13}}

Correct Answer

Option B

Solution

$L_{1}: 2 x+5 y=10$ $L_{2}: -4 x+ 3 y=12$ Solving $L_{1}$ and $L_{2}$ we get

C \equiv\left(\frac{-15}{13}, \frac{32}{13}\right)

Now, Let $A\left(x_{1}, \dfrac{1}{3}\left(12+4 x_{1}\right)\right)$ and

\begin{aligned} &B\left(x_{2}, \frac{1}{5}\left(10-2 x_{2}\right)\right) \\\\ &\therefore \quad \frac{3 x_{1}+x_{2}}{4}=2 \\\\ &\text{ and } \frac{\left(12+4 x_{1}\right)+\frac{10-2 x_{2}}{5}}{4}=3 \end{aligned}

So, $3 x_{1}+x_{2}=8$ and $10 x_{1}-x_{2}=-5$ So, $\left(x_{1}, x_{2}\right)=\left(\dfrac{3}{13}, \dfrac{95}{13}\right)$

\begin{aligned} &A=\left(\frac{3}{13}, \frac{56}{13}\right) \text{ and } B=\left(\frac{95}{13}, \frac{-12}{13}\right) \\\\ &=\left|\frac{1}{2}\left(\frac{3}{13}\left(\frac{-44}{13}\right) \frac{-56}{13}\left(\frac{110}{13}\right)+1\left(\frac{2860}{169}\right)\right)\right| \\\\ &=\frac{132}{13} \text{ sq. units } \end{aligned}

Q84

In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4. Let the point B lie on the line x + 3y = 7. If (

\alpha

,

\beta

) is the centroid of

\Delta

ABC, then 15(

\alpha

+

\beta

) is equal to :

A 39

B 41

C 51

D 63

Correct Answer

Option C

Solution

\left. \begin{array}{ll}2x + y = 4 \\ 2x + 6y = 14 \end{array} \right\}y = 2,\,x = 3

B(1, 2) Let C(k, 4 $-$ 2k) Now

A{B^2} = A{C^2}

{5^2} + {( - 1)^2} = {(6 - k)^2} + {( - 3 + 2k)^2}

\Rightarrow 5{k^2} - 24k + 19 = 0

(5k - 19)(k - 1) = 0 \Rightarrow k = {{19} \over 5}

C\left( {{{19} \over 5}, - {{18} \over 5}} \right)

Centroid ( $\alpha$ , $\beta$ )

\alpha = {{6 + 1 + {{19} \over 5}} \over 3} = {{18} \over 5}

\beta = {{1 + 2 - {{18} \over 5}} \over 3} = - {1 \over 5}

Now

15(\alpha + \beta )

15\left( {{{17} \over 5}} \right) = 51

Q85

Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of

\Delta

PQR is :

A

{{25} \over {4\sqrt 3 }}

B

{{25\sqrt 3 } \over 2}

C

{{25} \over {\sqrt 3 }}

D

{{25} \over {2\sqrt 3 }}

Correct Answer

Option D

Solution

Let, side of triangle = a.

h = {{|3 + 7 - 5|} \over {\sqrt {{1^2} + {1^2}} }}

= {5 \over {\sqrt 2 }}

From figure,

h = a\sin 60^\circ

\Rightarrow a = {{2h} \over {\sqrt 3 }}

= {2 \over {\sqrt 3 }} \times {5 \over {\sqrt 2 }}

= {{10} \over {\sqrt 6 }}

$\therefore$ Area

= {3 \over 4}{\left( {{{10} \over {\sqrt 6 }}} \right)^2}

= {{25} \over {2\sqrt 3 }}

Q86

Let the area of the triangle with vertices A(1,

\alpha

), B(

\alpha

, 0) and C(0,

\alpha

) be 4 sq. units. If the points (

\alpha

,

-

\alpha

), (

-

\alpha

,

\alpha

) and (

\alpha

2,

\beta

) are collinear, then

\beta

is equal to :

A 64

B

-

8

C

-

64

D 512

Correct Answer

Option C

Solution

$\because$ A(1, $\alpha$ ), B( $\alpha$ , 0) and C(0, $\alpha$ ) are the vertices of

\Delta

ABC and area of

\Delta

ABC = 4 $\therefore$

\left| {{1 \over 2}\left| \begin{array}{lll}1 & \alpha & 1 \\ \alpha & 0 & 1 \\ 0 & \alpha & 1 \end{array} \right|} \right| = 4

\Rightarrow \left| {1(1 - \alpha ) - \alpha (\alpha ) + {\alpha ^2}} \right| = 8

\Rightarrow \alpha = \, \pm \,8

Now,

(\alpha ,\, - \alpha ),\,( - \alpha ,\alpha )

and

({\alpha ^2},\beta )

are collinear $\therefore$

\left| \begin{array}{lll}8 & { - 8} & 1 \\ { - 8} & 8 & 1 \\ {64} & \beta & 1 \end{array} \right| = 0 = \left| \begin{array}{lll}{ - 8} & 8 & 1 \\ 8 & { - 8} & 1 \\ {64} & \beta & 1 \end{array} \right|

\Rightarrow 8(8 - \beta ) + 8( - 8 - 64) + 1( - 8\beta - 8 \times 64) = 0

\Rightarrow 8 - \beta - 72 - \beta - 64 = 0

\Rightarrow \beta = - 64

Q87

Let

\alpha

1,

\alpha

2 (

\alpha

1 2) be the values of

\alpha

fo the points (

\alpha

,

-

3), (2, 0) and (1,

\alpha

) to be collinear. Then the equation of the line, passing through (

\alpha

1,

\alpha

2) and making an angle of

{\pi \over 3}

with the positive direction of the x-axis, is :

A

x - \sqrt 3 y - 3\sqrt 3 + 1 = 0

B

\sqrt 3 x - y + \sqrt 3 + 3 = 0

C

x - \sqrt 3 y + 3\sqrt 3 + 1 = 0

D

\sqrt 3 x - y + \sqrt 3 - 3 = 0

Correct Answer

Option B

Solution

Points A( $\alpha$ , $-$ 3), B(2, 0) and C(1, $\alpha$ ) are collinear. $\therefore$ Slope of AB = Slope of BC

\Rightarrow {{0 + 3} \over {2 - \alpha }} = {{\alpha - 0} \over {1 - 2}}

\Rightarrow - 3 = \alpha (2 - \alpha )

\Rightarrow - 3 = 2\alpha - {\alpha ^2}

\Rightarrow {\alpha ^2} - 2\alpha - 3 = 0

\Rightarrow {\alpha ^2} - 3\alpha + \alpha - 3 = 0

\Rightarrow \alpha (\alpha - 3) + 1(\alpha - 3) = 0

\Rightarrow (\alpha + 1)(\alpha - 3) = 0

\Rightarrow \alpha = - 1,\,3

Given,

{\alpha _1} < {\alpha _2}

$\therefore$

{\alpha _1} = -1

and

{\alpha _2} = 3

$\therefore$

\left( {{\alpha _1},\,{\alpha _2}} \right) = ( - 1,\,3)

Now, equation of the line passing through ( $-$ 1, 3) and making angle

{\pi \over 3}

with positive x-axis is

(y - {y_1}) = m(x - {x_1})

\Rightarrow y - 3 = \left( {\tan {\pi \over 3}} \right)(x + 1)

\Rightarrow y - 3 = \sqrt 3 (x + 1)

\Rightarrow \sqrt 3 x - y + \sqrt 3 + 3 = 0

Q88

A line, with the slope greater than one, passes through the point

A(4,3)

and intersects the line

x-y-2=0

at the point B. If the length of the line segment

A B

is

\dfrac{\sqrt{29}}{3}

, then

B

also lies on the line :

A

2 x+y=9

B

3 x-2 y=7

C

x+2 y=6

D

2 x-3 y=3

Correct Answer

Option C

Solution

Let inclination of required line is $\theta$ , So the coordinates of point $B$ can be assumed as $\left(4-\dfrac{\sqrt{29}}{3} \cos \theta, 3-\dfrac{\sqrt{29}}{3} \sin \theta\right)$ Which satisfices $x-y-2=0$ $4-\dfrac{\sqrt{29}}{3} \cos \theta-3+\dfrac{\sqrt{29}}{3} \sin \theta-2=0$ $\sin \theta-\cos \theta=\dfrac{3}{\sqrt{29}}$ By squaring $\sin 2 \theta=\dfrac{20}{29}=\dfrac{2 \tan \theta}{1+\tan ^{2} \theta}$ $\tan \theta=\dfrac{5}{2}$ only (because slope is greater than 1 )

\sin \theta=\frac{5}{\sqrt{29}}, \cos \theta=\frac{2}{\sqrt{29}}

Point $B:\left(\dfrac{10}{3}, \dfrac{4}{3}\right)$ Which also satisfies $x+2 y=6$

Q89

Let the point

P(\alpha, \beta)

be at a unit distance from each of the two lines

L_{1}: 3 x-4 y+12=0

, and

L_{2}: 8 x+6 y+11=0

. If

P

lies below

L_{1}

and above

{ }{L_{2}}

, then

100(\alpha+\beta)

is equal to :

A

-

14

B 42

C

-

22

D 14

Correct Answer

Option D

Solution

{L_1}:3x - 4y + 12 = 0

{L_2}:8x + 6y + 11 = 0

Equation of angle bisector of L1 and L2 of angle containing origin

2(3x - 4y + 12) = 8x + 6y + 11

2x + 14y - 13 = 0

...... (i)

{{3\alpha - 4\beta + 12} \over 5} = 1

\Rightarrow 3\alpha - 4\beta + 7 = 0

...... (ii) Solution of

2x + 14y - 13 = 0

and

3x - 4y + 7 = 0

gives the required point

P(\alpha ,\beta ),\,\alpha = {{ - 23} \over {25}},\beta = {{53} \over {50}}

100(\alpha + \beta ) = 14

Q90

A point

P

moves so that the sum of squares of its distances from the points

(1,2)

and

(-2,1)

is 14. Let

f(x, y)=0

be the locus of

\mathrm{P}

, which intersects the

x

-axis at the points

\mathrm{A}

,

\mathrm{B}

and the

y

-axis at the points C, D. Then the area of the quadrilateral ACBD is equal to :

A

{9 \over 2}

B

{{3\sqrt {17} } \over 2}

C

{{3\sqrt {17} } \over 4}

D 9

Correct Answer

Option B

Solution

Let point

P:(h,\,k)

{(h - 1)^2} + {(k - 2)^2} + {(h + 2)^2} + {(k - 1)^2} = 14

2{h^2} + 2{k^2} + 2h - 6k - 4 = 0

Locus of

P:{x^2} + {y^2} + x - 3y - 2 = 0

Intersection with x-axis,

{x^2} + x - 2 = 0

\Rightarrow x = - 2,\,1

Intersection with y-axis,

{y^2} - 3y - 2 = 0

\Rightarrow y = {{3\, \pm \,\sqrt {17} } \over 2}

Area of the quadrilateral ACBD is

= {1 \over 2}(|{x_1}| + |{x_2}|)(|{y_1}| + |{y_2}|)

= {1 \over 2} \times 3 \times \sqrt {17} = {{3\sqrt {17} } \over 2}