Straight Lines and Pair of Straight Lines — Page 10 | JEE Mathematics

Q91

Let

A(1,1), B(-4,3), C(-2,-5)

be vertices of a triangle

A B C, P

be a point on side

B C

, and

\Delta_{1}

and

\Delta_{2}

be the areas of triangles

A P B

and

A B C

, respectively. If

\Delta_{1}: \Delta_{2}=4: 7

, then the area enclosed by the lines

A P, A C

and the

x

-axis is :

A

\dfrac{1}{4}

B

\dfrac{3}{4}

C

\dfrac{1}{2}

D 1

Correct Answer

Option C

Solution

{{{\Delta _1}} \over {{\Delta _2}}} = {{{1 \over 2} \times BP \times AH} \over {{1 \over 2} \times BC \times AH}} = {4 \over 7}

P\left( {{{ - 20} \over 7},{{ - 11} \over 7}} \right)

Line

AC:y - 1 = 2(x - 1)

Intersection with x-axis

= \left( {{1 \over 2},0} \right)

Line

AP:y - 1 = {2 \over 3}(x - 1)

Intersection with x-axis

\left( {{{ - 1} \over 2},0} \right)

Vertices are

(1,1),\left( {{1 \over 2},0} \right)

and

\left( {{{ - 1} \over 2},0} \right)

Area

= {1 \over 2}

sq. unit

Q92

The equations of the sides

\mathrm{AB}, \mathrm{BC}

and CA of a triangle ABC are

2 x+y=0, x+\mathrm{p} y=39

and

x-y=3

respectively and

\mathrm{P}(2,3)

is its circumcentre. Then which of the following is NOT true?

A

(\mathrm{AC})^{2}=9 \mathrm{p}

B

(\mathrm{AC})^{2}+\mathrm{p}^{2}=136

C

32<\operatorname{area}\,(\Delta \mathrm{ABC})<36

D

34<\operatorname{area}\,(\triangle \mathrm{ABC})<38

Correct Answer

Option D

Solution

Intersection of

2x + y = 0

and

x - y = 3\,:\,A(1, - 2)

Equation of perpendicular bisector of AB is

x - 2y = - 4

Equation of perpendicular bisector of AC is

x + y = 5

Point B is the image of A in line

x - 2y + 4 = 0

which is obtained as

B\left( {{{ - 13} \over 5},{{26} \over 5}} \right)

Similarly vertex

C:(7,4)

Equation of line

BC:x + 8y = 39

So,

p = 8

AC = \sqrt {{{(7 - 1)}^2} + {{(4 + 2)}^2}} = 6\sqrt 2

Area of triangle

ABC = 32.4

Q93

Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be P(1,1). If the line AP intersects the line BC at the point Q

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

, then

k_{1}+k_{2}

is equal to :

A 2

B

\dfrac{4}{7}

C

\dfrac{2}{7}

D 4

Correct Answer

Option B

Solution

Let D be mid-point of AC, then

{{b + 3} \over 2} = 1 \Rightarrow b = - 1

Let E be mid-point of BC,

{{5 - b} \over {b - a}}\,.\,{{{{(3 + b)} \over 2}} \over {{{a + b} \over 2} - 1}} = - 1

On putting

b = - 1

, we get

a = 5

or

-3

But

a = 5

is rejected as

ab > 0

A( - 3,3),\,B( - 1,5),\,C( - 3, - 1),\,P(1,1)

Line

BC \Rightarrow y = 3x + 8

Line

AP \Rightarrow y = {{3 - x} \over 2}

Point of intersection

\left( {{{ - 13} \over 7},{{17} \over 7}} \right)

Q94

Let

m_{1}, m_{2}

be the slopes of two adjacent sides of a square of side a such that

a^{2}+11 a+3\left(m_{1}^{2}+m_{2}^{2}\right)=220

. If one vertex of the square is

(10(\cos \alpha-\sin \alpha), 10(\sin \alpha+\cos \alpha))

, where

\alpha \in\left(0, \dfrac{\pi}{2}\right)

and the equation of one diagonal is

(\cos \alpha-\sin \alpha) x+(\sin \alpha+\cos \alpha) y=10

, then

72\left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)+a^{2}-3 a+13

is equal to :

A 119

B 128

C 145

D 155

Correct Answer

Option B

Solution

One vertex of square is $(10(\cos \alpha-\sin \alpha), 10(\sin \alpha+\cos \alpha))$ and one of the diagonal is $(\cos \alpha-\sin \alpha) x+(\sin \alpha+\cos \alpha) y=10$ So the other diagonal can be obtained as $(\cos \alpha+\sin \alpha) x-(\cos \alpha-\sin \alpha) y=0$ So, the point of intersection of the diagonal will be (5( $\cos \alpha-\sin \alpha), 5(\cos \alpha+\sin \alpha))$ .

Therefore, the vertex opposite to the given vertex is $(0,0)$ .

So, the diagonal length $=10 \sqrt{2}$ Side length $(a)=10$ It is given that $a^{2}+11 a+3\left(m_{1}^{2}+m_{2}^{2}\right)=220$ $m_{1}^{2}+m_{2}^{2}=\dfrac{220-100-110}{3}=\dfrac{10}{3}$ and $m_{1} m_{2}=-1$ Slopes of the sides are tan $\alpha$ and $-\cot \alpha$ $\tan ^{2} \alpha=3$ or $\dfrac{1}{3}$ $72\left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)+a^{2}-3 a+13$ $=72 \cdot \dfrac{\tan ^{4} \alpha+1}{\left(1+\tan ^{2} \alpha\right)^{2}}+a^{2}-3 a+13=128$

Q95

Let

\mathrm{A}(\alpha,-2), \mathrm{B}(\alpha, 6)

and

\mathrm{C}\left(\dfrac{\alpha}{4},-2\right)

be vertices of a

\triangle \mathrm{ABC}

. If

\left(5, \dfrac{\alpha}{4}\right)

is the circumcentre of

\triangle \mathrm{ABC}

, then which of the following is NOT correct about

\triangle \mathrm{ABC}

?

A area is 24

B perimeter is 25

C circumradius is 5

D inradius is 2

Correct Answer

Option B

Solution

Circumcentre of $\triangle A B C$

\begin{aligned} &=\left(\frac{\alpha+\frac{\alpha}{4}}{2}, \frac{6-2}{2}\right) \\\\ &=\left(\frac{5 \alpha}{8}, 2\right) \\\\ &=\left(5, \frac{\alpha}{4}\right) \\\\ &\Rightarrow \alpha=8 \end{aligned}

$\operatorname{area}(\triangle A B C)=\dfrac{1}{2} \cdot \dfrac{3 \alpha}{4} \times 8=24$ sq. units

\begin{aligned} \text{ Perimeter } &=8+\frac{3 \alpha}{4}+\sqrt{8^{2}+\left(\frac{3 \alpha}{4}\right)^{2}} \\\\ &=8+6+10=24 \end{aligned}

Circumradius $=\dfrac{10}{2}=5$ inradius $(r)=\dfrac{\Delta}{s}=\dfrac{24}{12}=2$

Q96

The combined equation of the two lines

ax+by+c=0

and

a'x+b'y+c'=0

can be written as

(ax+by+c)(a'x+b'y+c')=0

. The equation of the angle bisectors of the lines represented by the equation

2x^2+xy-3y^2=0

is :

A

3{x^2} + xy - 2{y^2} = 0

B

{x^2} - {y^2} - 10xy = 0

C

{x^2} - {y^2} + 10xy = 0

D

3{x^2} + 5xy + 2{y^2} = 0

Correct Answer

Option B

Solution

For pair of straight lines in this form $a x^2+b y^2+2 h x y+2 g x+2 f y+c=0$ Equation of angle bisector is

\frac{x^2-y^2}{a-b}=\frac{x y}{h}

for $2 x^2+x y-3 y^2=0$

a=2, b=-3, h=\frac{1}{2}

Equation of angle bisector is

\begin{aligned} & \frac{x^2-y^2}{5}=\frac{x y}{1 / 2} \\\\ & \Rightarrow \mathrm{x}^2-\mathrm{y}^2-10 \mathrm{xy}=0 \end{aligned}

Q97

If the orthocentre of the triangle, whose vertices are (1, 2), (2, 3) and (3, 1) is

(\alpha,\beta)

, then the quadratic equation whose roots are

\alpha+4\beta

and

4\alpha+\beta

, is :

A

x^2-20x+99=0

B

x^2-22x+120=0

C

x^2-19x+90=0

D

x^2-18x+80=0

Correct Answer

Option A

Solution

\mathrm{m}=-\frac{1}{2}

Here $\mathrm{m_BH} \times \mathrm{m_AC}=-1$

\begin{array}{ll} & \left(\frac{\beta-3}{\alpha-2}\right)\left(\frac{1}{-2}\right)=-1 \\\\ & \beta-3=2 \alpha-4 \\\\ & \beta=2 \alpha-1 \\\\ & \mathrm{~m}_{\mathrm{AH}} \times \mathrm{m}_{\mathrm{BC}}=-1 \\\\ \Rightarrow & \left(\frac{\beta-2}{\alpha-1}\right)(-2)=-1 \\\\ \Rightarrow & 2 \beta-4=\alpha-1 \\\\ \Rightarrow & 2(2 \alpha-1)=\alpha+3 \\\\ \Rightarrow & 3 \alpha=5 \\\\ & \alpha=\frac{5}{3}, \beta=\frac{7}{3} \Rightarrow \mathrm{H}\left(\frac{5}{3}, \frac{7}{3}\right) \\\\ & \alpha+4 \beta=\frac{5}{3}+\frac{28}{3}=\frac{33}{3}=11 \\\\ & \beta+4 \alpha=\frac{7}{3}+\frac{20}{3}=\frac{27}{3}=9 \\\\ & \mathrm{x}^2-20 \mathrm{x}+99=0 \end{array}

Q98

Let

B

and

C

be the two points on the line

y+x=0

such that

B

and

C

are symmetric with respect to the origin. Suppose

A

is a point on

y-2 x=2

such that

\triangle A B C

is an equilateral triangle. Then, the area of the

\triangle A B C

is :

A

\dfrac{10}{\sqrt{3}}

B

2 \sqrt{3}

C

3 \sqrt{3}

D

\dfrac{8}{\sqrt{3}}

Correct Answer

Option D

Solution

Origin $(O)$ is mid-point of $B C(x+y=0)$ . $A$ lies on perpendicular bisector of $B C$ , which is $x-y=0$ A is point of intersection of $x-y=0$ and $y-2 x=2$ $\therefore A \equiv(-2,-2)$ Let $h=A O=\dfrac{-2-2}{\sqrt{1^{2}+1^{2}}}=2 \sqrt{2}$

\text{ Area }=\frac{h^{2}}{\sqrt{3}}=\frac{8}{\sqrt{3}}

Q99

A light ray emits from the origin making an angle 30

^\circ

with the positive

x

-axis. After getting reflected by the line

x+y=1

, if this ray intersects

x

-axis at Q, then the abscissa of Q is :

A

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

B

{2 \over {3 - \sqrt 3 }}

C

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

D

{2 \over {3 + \sqrt 3 }}

Correct Answer

Option D

Solution

Let $Q(h, O)$ $\because$ OP reflected by $x+y=1$ . So, image of $Q$ lies on $y=\dfrac{x}{\sqrt{3}}$

\begin{aligned} & \therefore \quad \frac{x-h}{1}=\frac{y}{1}=\frac{-2(h-1)}{2} \\\\ & \therefore \quad x=1, y=1-h \end{aligned}

It lies on $y=\dfrac{x}{\sqrt{3}}$

\begin{aligned} & \therefore \quad 1-h=\frac{1}{\sqrt{3}} \\\\ & \therefore \quad h=1-\frac{1}{\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{3}}=\frac{2}{3+\sqrt{3}} \end{aligned}

Q100

If

(\alpha, \beta)

is the orthocenter of the triangle

\mathrm{ABC}

with vertices

A(3,-7), B(-1,2)

and

C(4,5)

, then

9 \alpha-6 \beta+60

is equal to :

A 30

B 40

C 25

D 35

Correct Answer

Option C

Solution

\begin{aligned} & \text{ Altitude of BC: } y+7=\frac{-5}{3}(x-3) \\\\ & 3 y+21=-5 x+15 \\\\ & 5 x+3 y+6=0 \\\\ & \text{ Altitude of AC: } y-2=\frac{-1}{12}(x+1) \\\\ & 12 y-24=-x-1 \\\\ & x+12 y=23 \\\\ & \alpha=\frac{-47}{19}, \quad \beta=\frac{121}{57} \\\\ & 9 \alpha-6 \beta+60=25 \end{aligned}