Straight Lines and Pair of Straight Lines — Page 6 | JEE Mathematics

Q51

Lines are drawn parallel to the line 4x – 3y + 2 = 0, at a distance

{3 \over 5}

from the origin. Then which one of the following points lies on any of these lines ?

A

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

B

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

C

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

D

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

Correct Answer

Option B

Solution

Line parallel to 4x – 3y + 2 = 0 is given as 4x – 3y + $\lambda$ = 0 distance from origin is

\left| {{\lambda \over 5}} \right| = {3 \over 5}

\Rightarrow \lambda = \pm 3

$\therefore$ required lines are 4x – 3y + 3 = 0 & 4x – 3y – 3 = 0 By Putting

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

on both lines it satisfy.

Q52

The region represented by| x – y |

\le

2 and | x + y|

\le

2 is bounded by a :

A rhombus of area 8

\sqrt 2

sq. units

B square of side length 2

\sqrt 2

units

C square of area 16 sq. units

D rhombus of side length 2 units

Correct Answer

Option B

Solution

{C_1}{\rm{ }}:{\rm{ }}\left| {y{\rm{ }}-{\rm{ }}x} \right|{\rm{ }} \le {\rm{ }}2

{C_2}{\rm{ }}:{\rm{ }}\left| {y{\rm{ + }}x} \right|{\rm{ }} \le {\rm{ }}2

Now region is square shown figure is square with side length 2

\sqrt 2

.

Q53

If the two lines x + (a – 1) y = 1 and 2x + a2y = 1 (a

\in

R – {0, 1}) are perpendicular, then the distance of their point of intersection from the origin is :

A

{2 \over \sqrt5}

B

{\sqrt2 \over 5}

C

{2 \over 5}

D

\sqrt{2 \over 5}

Correct Answer

Option D

Solution

Line 1 : x + (a – 1) y = 1 Slope of this line (m1) =

{{ - 1} \over {a - 1}}

Line 2 : 2x + a2y = 1 Slope of this line (m2) =

- {2 \over {{a^2}}}

Line 1 and Line 2 are perpendicular to each other. $\therefore$ m1 m2 = -1 $\Rightarrow$

\left( {{{ - 1} \over {a - 1}}} \right)\left( { - {2 \over {{a^2}}}} \right) = - 1

$\Rightarrow$

{a^3} - {a^2} + 2 = 0

$\Rightarrow$

\left( {a + 1} \right)

\left( {{a^2} - 2a + 2} \right)

= 0 $\therefore$

{a = - 1}

So lines are Line 1 : x - 2y + 1 = 0 Line 2 : 2x + y - 1 = 0 Solving these equation we get point of intersection P

\left( {{3 \over 5}, - {1 \over 5}} \right)

Now distance of P from origin OP =

\sqrt {{{\left( {{3 \over 5}} \right)}^2} + {{\left( { - {1 \over 5}} \right)}^2}}

=

\sqrt {{{10} \over {25}}}

=

\sqrt {{2 \over 5}}

Q54

Slope of a line passing through P(2, 3) and intersecting the line, x + y = 7 at a distance of 4 units from P, is :

A

{{\sqrt 7 - 1} \over {\sqrt 7 + 1}}

B

{{\sqrt 5 - 1} \over {\sqrt 5 + 1}}

C

{{1 - \sqrt 5 } \over {1 + \sqrt 5 }}

D

{{1 - \sqrt 7 } \over {1 + \sqrt 7 }}

Correct Answer

Option D

Solution

We know parametric form of straight line is

{{x - {x_1}} \over {\cos \theta }} = {{y - {y_1}} \over {\sin \theta }} = r

$\Rightarrow$

{{x - 2} \over {\cos \theta }} = {{y - 3} \over {\sin \theta }} = 4

$\therefore$ x = 4 + 2cos $\theta$ y = 4 + 3sin $\theta$ $\therefore$ Point A = (4 + 2cos $\theta$ , 4 + 3sin $\theta$ ) Point A lies on line x + y = 7, $\therefore$ (4 + 2cos $\theta$ ) + (4 + 3sin $\theta$ ) = 7 $\Rightarrow$ sin $\theta$ + cos $\theta$ =

{1 \over 2}

$\Rightarrow$

{\sin ^2}\theta + {\cos ^2}\theta

+ 2sin $\theta$ cos $\theta$ =

{1 \over 4}

$\Rightarrow$ 1 + sin 2 $\theta$ =

{1 \over 4}

$\Rightarrow$ sin 2 $\theta$ =

- {3 \over 4}

$\Rightarrow$

{{2\tan \theta } \over {1 + {{\tan }^2}\theta }}

=

- {3 \over 4}

$\Rightarrow$ 3tan2 $\theta$ + 8tan $\theta$ + 3 = 0 $\Rightarrow$ tan $\theta$ =

{{ - 8 \pm 2\sqrt 7 } \over 6}

So slope =

{{ - 8 \pm 2\sqrt 7 } \over 6}

By checking each options, Slope =

{{1 - \sqrt 7 } \over {1 + \sqrt 7 }}

As

{{1 - \sqrt 7 } \over {1 + \sqrt 7 }}

=

{{{{\left( {1 - \sqrt 7 } \right)}^2}} \over {1 - 7}}

=

{{8 - 2\sqrt 7 } \over { - 6}}

=

{{ - 8 + 2\sqrt 7 } \over 6}

Q55

Suppose that the points (h,k), (1,2) and (–3,4) lie on the line L1 . If a line L2 passing through the points (h,k) and (4,3) is perpendicular to L1 , then

k \over h

equals :

A

{1 \over 3}

B 3

C 0

D -

{1 \over 7}

Correct Answer

Option A

Solution

Equation of line L1 passing through points (1, 2) and (–3, 4) is : y - 2 =

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

(x - 1) $\Rightarrow$ y - 2 =

- {1 \over 2}

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

L1) Line L1 passes through point (h, k). $\therefore$ h + 2k = 5 .......(

1) Line L2 is perpendicular to line L1.

$\therefore$ Equation of line L2 : 2x - y = $\lambda$ This line L2 passes through point (4, 3).

$\therefore$ 2(4) - 3 = $\lambda$ $\Rightarrow$ $\lambda$ = 5 $\therefore$ Line L2 : 2x - y = 5 L2 also passes through (h, k) $\therefore$ 2h - k = 5 ............(

2) Solving (1) and (2) we get, h = 3 and k = 1 $\therefore$

{k \over h} = {1 \over 3}

Q56

If the system of linear equations x – 2y + kz = 1 2x + y + z = 2 3x – y – kz = 3 has a solution (x,y,z), z

\ne

0, then (x,y) lies on the straight line whose equation is :

A 4x – 3y – 4 = 0

B 3x – 4y – 1 = 0

C 4x – 3y – 1 = 0

D 3x – 4y – 4 = 0

Correct Answer

Option A

Solution

x – 2y + kz = 1 ......(1) 2x + y + z = 2 .........(2) 3x – y – kz = 3 ........(3) for locus of (x, y) add equation (1) + (3) 4x – 3y = 4 $\Rightarrow$ 4x – 3y - 4 = 0

Q57

Let O(0, 0) and A(0, 1) be two fixed points. Then the locus of a point P such that the perimeter of

\Delta

AOP is 4, is :

A 9x2 + 8y2 – 8y = 16

B 8x2 – 9y2 + 9y = 18

C 8x2 + 9y2 – 9y = 18

D 9x2 – 8y2 + 8y = 16

Correct Answer

Option A

Solution

Let point C(h, k) Given, AB + BC + AC = 4 From graph, AB = 1 $\therefore$ BC + AC = 3 $\Rightarrow$

\sqrt {{h^2} + {k^2}} + \sqrt {{h^2} + {{\left( {k - 1} \right)}^2}} = 3

$\Rightarrow$

{{h^2} + {{\left( {k - 1} \right)}^2}}

= 9 +

{{h^2} + {k^2}}

-

6\sqrt {{h^2} + {k^2}}

$\Rightarrow$

6\sqrt {{h^2} + {k^2}} = 2k + 8

$\Rightarrow$

9\left( {{h^2} + {k^2}} \right) = {k^2} + 8k + 16

$\Rightarrow$

9{h^2} + 8{k^2} - 8k - 16 = 0

$\therefore$ Locus of point C will be, 9x2 + 8y2 – 8y = 16

Q58

If 5, 5r, 5r2 are the lengths of the sides of a triangle, then r cannot be equal to :

A

{7 \over 4}

B

{5 \over 4}

C

{3 \over 4}

D

{3 \over 2}

Correct Answer

Option A

Solution

For three lengths to form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side.

This is known as the triangle inequality theorem.

Let's apply this to our lengths: $5$ , $5r$ , and $5r^2$ .

We have three inequalities to check: $5 + 5r > 5r^2$ $5 + 5r^2 > 5r$ $5r + 5r^2 > 5$ Solving these inequalities will give us a range of values for $r$ .

Any value outside of this range will mean that $r$ cannot form a triangle with the given sides.

1.

$1 + r > r^2$ We can rewrite this as: $r^2 - r - 1 < 0$ $\Rightarrow\left[r-\left(\dfrac{1-\sqrt{5}}{2}\right)\right]\left[r-\left(\dfrac{1+\sqrt{5}}{2}\right)\right] The roots of this quadratic equation are$ \frac{1-\sqrt{5}}{2} $and$ \frac{1+\sqrt{5}}{2} $, so the inequality holds for$ r $in the interval$ \left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right) $. 2.$ 1 + r^2 > r $We can rewrite this as:$ r^2 - r + 1 > 0$

\begin{array}{lc} \Rightarrow r^2-2 \cdot \frac{1}{2} r+\left(\frac{1}{2}\right)^2+1-\left(\frac{1}{2}\right)^2>0 \\\\ \Rightarrow \left(r-\frac{1}{2}\right)^2+\frac{3}{4}>0 \\\\ \Rightarrow r \in R \end{array}

This inequality is true for all real numbers since the left-hand side is always positive, which means the inequality holds for all real $R$ .

3.

$r + r^2 > 1$ We can rewrite this as: $r^2 + r - 1 > 0$

\begin{aligned} & \Rightarrow {\left[r-\left(\frac{-1-\sqrt{5}}{2}\right)\right]\left[r-\left(\frac{-1+\sqrt{5}}{2}\right)\right]>0} \\\\ & {\left[\because r^2+r-1=0 \Rightarrow r=\frac{-1 \pm \sqrt{1+4}}{2}=\frac{-1 \pm \sqrt{5}}{2}\right]} \\\\ & \Rightarrow r \in\left(-\infty, \frac{-1-\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, \infty\right) \end{aligned}

Taking the intersection of the intervals from the three inequalities, we get $\left(\dfrac{1-\sqrt{5}}{2}, \dfrac{1+\sqrt{5}}{2}\right)$ as the valid range for $r$ .

Given the solution of the inequalities $r \in \left(\dfrac{1-\sqrt{5}}{2}, \dfrac{1+\sqrt{5}}{2}\right)$ , we need to determine which of the options are outside this interval.

Option A: $\dfrac{7}{4}$ The decimal representation of $\dfrac{7}{4}$ is 1.75, which is outside the interval $\left(\dfrac{1-\sqrt{5}}{2}, \dfrac{1+\sqrt{5}}{2}\right)$ or approximately (-0.618, 1.618).

Hence, r cannot be equal to $\dfrac{7}{4}$ .

Option B: $\dfrac{5}{4}$ The decimal representation of $\dfrac{5}{4}$ is 1.25, which is inside the interval $\left(\dfrac{1-\sqrt{5}}{2}, \dfrac{1+\sqrt{5}}{2}\right)$ .

Hence, r can be equal to $\dfrac{5}{4}$ .

Option C: $\dfrac{3}{4}$ The decimal representation of $\dfrac{3}{4}$ is 0.75, which is also inside the interval $\left(\dfrac{1-\sqrt{5}}{2}, \dfrac{1+\sqrt{5}}{2}\right)$ .

Hence, r can be equal to $\dfrac{3}{4}$ .

Option D: $\dfrac{3}{2}$ The decimal representation of $\dfrac{3}{2}$ is 1.5, which is also inside the interval $\left(\dfrac{1-\sqrt{5}}{2}, \dfrac{1+\sqrt{5}}{2}\right)$ .

Hence, r can be equal to $\dfrac{3}{2}$ .

Therefore, the only value that r cannot be equal to, among the given options, is $\dfrac{7}{4}$ .

So, the correct answer is: Option A: $\dfrac{7}{4}$

Q59

If a straight line passing through the point P(–3, 4) is such that its intercepted portion between the coordinate axes is bisected at P, then its equation is :

A x – y + 7 = 0

B 4x – 3y + 24 = 0

C 4x + 3y = 0

D 3x – 4y + 25 = 0

Correct Answer

Option B

Solution

Let the line be

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

( $-$ 3, 4) =

\left( {{a \over 2},{b \over 2}} \right)

a = $-$ 6, b = 8 equation of line is 4x $-$ 3y + 24 = 0

Q60

Let L denote the line in the xy-plane with x and y intercepts as 3 and 1 respectively. Then the image of the point (–1, –4) in this line is :

A

\left( {{{11} \over 5},{{28} \over 5}} \right)

B

\left( {{{29} \over 5},{{11} \over 5}} \right)

C

\left( {{{29} \over 5},{8 \over 5}} \right)

D

\left( {{8 \over 5},{{29} \over 5}} \right)

Correct Answer

Option A

Solution

Line is

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

$\Rightarrow$ x + 3y – 3 = 0 Let Image of point (–1, –4) is ( $\alpha$ , $\beta$ ) Hence,

{{\alpha + 1} \over 1} = {{\beta + 4} \over 3} = - 2\left( {{{ - 1 - 12 - 3} \over {10}}} \right)

$\Rightarrow$

{{\alpha + 1} \over 1} = {{\beta + 4} \over 3} = {{16} \over 5}

$\Rightarrow$ $\alpha$ =

{{11} \over 5}

, $\beta$ =

{{28} \over 5}