Straight Lines and Pair of Straight Lines — Page 11 | JEE Mathematics

Q101

Let

(\alpha, \beta)

be the centroid of the triangle formed by the lines

15 x-y=82,6 x-5 y=-4

and

9 x+4 y=17

. Then

\alpha+2 \beta

and

2 \alpha-\beta

are the roots of the equation :

A

x^{2}-7 x+12=0

B

x^{2}-13 x+42=0

C

x^{2}-14 x+48=0

D

x^{2}-10 x+25=0

Correct Answer

Option B

Solution

Solve the equations $15x - y = 82$ and $6x - 5y = -4$ Multiply the first equation by 5 and the second by 1 and then subtract the second from the first: $75x - 5y = 410$ $6x - 5y = -4$ Subtracting these gives $69x = 414$ which leads to $x = 6$ Substitute $x = 6$ into the first equation to get $y = 8$ So, the first vertex is $(6,8)$ .

Solve the equations $15x - y = 82$ and $9x + 4y = 17$ Multiply the first equation by 4 and the second by 1 and then add: $60x - 4y = 328$ $9x + 4y = 17$ Adding these gives $69x = 345$ which leads to $x = 5$ Substitute $x = 5$ into the second equation to get $y = -7$ So, the second vertex is $(5,-7)$ .

Solve the equations $6x - 5y = -4$ and $9x + 4y = 17$ Multiply the first equation by 4 and the second by 5 and then add: $24x - 20y = -16$ $45x + 20y = 85$ Adding these gives $69x = 69$ which leads to $x = 1$ Substitute $x = 1$ into the first equation to get $y = 2$ So, the third vertex is $(1,2)$ .

So, the vertices of the triangle are $(6,8)$ , $(5,-7)$ , and $(1,2)$ .

Now that we have the vertices of the triangle, we can find the centroid.

The centroid $(\alpha, \beta)$ of a triangle with vertices $(x_1, y_1)$ , $(x_2, y_2)$ , and $(x_3, y_3)$ is given by :

\alpha=\frac{x_1+x_2+x_3}{3}

,

\beta=\frac{y_1+y_2+y_3}{3}

Substituting the coordinates of the vertices into these equations, we get:

\alpha=\frac{6+5+1}{3}=4

,

\beta=\frac{8-7+2}{3}=1

Then, we find $\alpha+2\beta$ and $2\alpha-\beta$ :

\alpha+2\beta=4+2(1)=6 \ 2\alpha-\beta=2(4)-1=7

So, $\alpha+2\beta=6$ and $2\alpha-\beta=7$ are the roots of the quadratic equation.

We can write the quadratic equation as

x^2-(\alpha+2\beta+2\alpha-\beta)x+(\alpha+2\beta)(2\alpha-\beta)=0

Substituting $\alpha=4$ , $\beta=1$ gives us:

x^2-13x+42=0

Q102

If the point

\left(\alpha, \dfrac{7 \sqrt{3}}{3}\right)

lies on the curve traced by the mid-points of the line segments of the lines

x \cos \theta+y \sin \theta=7, \theta \in\left(0, \dfrac{\pi}{2}\right)

between the co-ordinates axes, then

\alpha

is equal to :

A

-

7

B 7

C

-

7

\sqrt3

D 7

\sqrt3

Correct Answer

Option B

Solution

\begin{gathered} x \cos \theta+y \sin \theta=7 \\\\ x-\text{ intercept }=\frac{7}{\cos \theta} \\\\ y-\text{ intercept }=\frac{7}{\sin \theta} \\\\ \mathrm{A}:\left(\frac{7}{\cos \theta}, 0\right) \mathrm{B}:\left(0, \frac{7}{\sin \theta}\right) \end{gathered}

Locus of mid point M : (h, k)

\begin{aligned} & \mathrm{h}=\frac{7}{2 \cos \theta}, \mathrm{k}=\frac{7}{2 \sin \theta} \\\\ & \frac{7}{2 \sin \theta}=\frac{7 \sqrt{3}}{3} \Rightarrow \sin \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{3} \\\\ & \alpha=\frac{7}{2 \cos \theta}=7 \end{aligned}

Q103

Let

C(\alpha, \beta)

be the circumcenter of the triangle formed by the lines

4 x+3 y=69

4 y-3 x=17

, and

x+7 y=61

. Then

(\alpha-\beta)^{2}+\alpha+\beta

is equal to :

A 15

B 17

C 16

D 18

Correct Answer

Option B

Solution

We have,

\begin{aligned} & 4 x+3 y=69 .......(i) \\\\ & 4 y-3 x=17 .......(ii) \\\\ & x+7 y=61 .......(iii) \end{aligned}

On solving (i) and (iii), we get,

\begin{aligned} & x=12, \text{ and } y=7 \\\\ & \text{ So, } A \equiv(12,7) \end{aligned}

On solving (ii) and (iii), we get,

\begin{aligned} & x=5 \text{ and } y=8 \\\\ & \text{ So, B } \equiv(5,8) \end{aligned}

\begin{aligned} & \text{ Hence, circumcentre } \equiv\left(\frac{12+5}{2}, \frac{7+8}{2}\right) \\\\ & \equiv\left(\frac{17}{2}, \frac{15}{2}\right) \end{aligned}

\begin{aligned} & \therefore \alpha=\frac{17}{2}, \beta=\frac{15}{2} \\\\ & \therefore(\alpha-\beta)^2+(\alpha+\beta)=\left(\frac{17}{2}-\frac{15}{2}\right)^2+\left(\frac{17}{2}+\frac{15}{2}\right) \\\\ & =(1)^2+(16)=17 \end{aligned}

Q104

The straight lines

\mathrm{l_{1}}

and

\mathrm{l_{2}}

pass through the origin and trisect the line segment of the line L :

9 x+5 y=45

between the axes. If

\mathrm{m}_{1}

and

\mathrm{m}_{2}

are the slopes of the lines

\mathrm{l_{1}}

and

\mathrm{l_{2}}

, then the point of intersection of the line

\mathrm{y=\left(m_{1}+m_{2}\right)}x

with L lies on :

A

6 x-y=15

B

6 x+y=10

C

\mathrm{y}-x=5

D

y-2 x=5

Correct Answer

Option C

Solution

Given line $L: 9 x+5 y=45$ ..........(i) Also given that $m_1$ and $m_2$ are the slopes of the lines $l_1$ and $l_2$ , respectively.

Let $A$ be the point of intersection of line $l_1$ and $L$ , then co-ordinates of $A$ are $\left(\dfrac{2 \times 5+1 \times 0}{2+1}, \dfrac{2 \times 0+1 \times 9}{2+1}\right)=\left(\dfrac{10}{3}, 3\right)$ And let $B$ be the point of intersection of line $l_2$ and $L$ , then co-ordinates of $B$ are

\left(\frac{1 \times 5+2 \times 0}{1+2}, \frac{1 \times 0+2 \times 9}{1+2}\right)=\left(\frac{5}{3}, 6\right)

Now, slope of line $l_1,\left(m_1\right)=\dfrac{3-0}{\dfrac{10}{3}-0}=\dfrac{9}{10}$ and slope of lines $l_2,\left(m_2\right)=\dfrac{6-0}{\dfrac{5}{3}-0}=\dfrac{18}{5}$

\begin{aligned} \therefore \text{ line } y & =\left(m_1+m_2\right) x \\\\ & =\left(\frac{9}{10}+\frac{18}{5}\right) x=\frac{45}{10} x=\frac{9}{2} x .......(ii) \end{aligned}

Point of intersection of lines (i) and (ii)

\begin{aligned} & 9 x+5\left(\frac{9}{2} x\right)=45 \\\\ & \Rightarrow x+\frac{5 x}{2}=5 \\\\ & \Rightarrow \frac{7 x}{2}=5 \Rightarrow x=\frac{10}{7} \\\\ & \text{ and } y=\frac{9}{2} \times \frac{10}{7}=\frac{45}{7} \\\\ & \therefore \text{ Point of intersection }=\left(\frac{10}{7}, \frac{45}{7}\right) \\\\ & \text{ lies on line } y-x=5 \end{aligned}

Q105

The portion of the line

4 x+5 y=20

in the first quadrant is trisected by the lines

\mathrm{L}_1

and

\mathrm{L}_2

passing through the origin. The tangent of an angle between the lines

\mathrm{L}_1

and

\mathrm{L}_2

is :

A

\dfrac{30}{41}

B

\dfrac{8}{5}

C

\dfrac{2}{5}

D

\dfrac{25}{41}

Correct Answer

Option A

Solution

Co-ordinates of

\mathrm{A}=\left(\frac{5}{3}, \frac{8}{3}\right)

Co-ordinates of

\mathrm{B}=\left(\frac{10}{3}, \frac{4}{3}\right)

Slope of

\mathrm{OA}=\mathrm{m}_1=\frac{8}{5}

Slope of

\mathrm{OB}=\mathrm{m}_2=\frac{2}{5}

\begin{aligned} & \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\ & \tan \theta=\frac{\frac{6}{5}}{1+\frac{16}{25}}=\frac{30}{41} \\ & \tan \theta=\frac{30}{41} \end{aligned}

Q106

Let

A(a, b), B(3,4)

and

C(-6,-8)

respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point

P(2 a+3,7 b+5)

from the line

2 x+3 y-4=0

measured parallel to the line

x-2 y-1=0

is

A

\dfrac{17 \sqrt{5}}{6}

B

\dfrac{15 \sqrt{5}}{7}

C

\dfrac{17 \sqrt{5}}{7}

D

\dfrac{\sqrt{5}}{17}

Correct Answer

Option C

Solution

\mathrm{A}(\mathrm{a}, \mathrm{b}), \quad \mathrm{B}(3,4), \quad \mathrm{C}(-6,-8)

\Rightarrow \mathrm{a}=0, \mathrm{~b}=0 \quad \Rightarrow \mathrm{P}(3,5)

Distance from

\mathrm{P}

measured along

\mathrm{x}-2 \mathrm{y}-1=0

\Rightarrow x=3+r \cos \theta, \quad y=5+r \sin \theta

\begin{aligned} & \text{ Where } \tan \theta=\frac{1}{2} \\ & \mathrm{r}(2 \cos \theta+3 \sin \theta)=-17 \\ & \Rightarrow \mathrm{r}=\left|\frac{-17 \sqrt{5}}{7}\right|=\frac{17 \sqrt{5}}{7} \end{aligned}

Q107

Let

\alpha, \beta, \gamma, \delta \in \mathbb{Z}

and let

A(\alpha, \beta), B(1,0), C(\gamma, \delta)

and

D(1,2)

be the vertices of a parallelogram

\mathrm{ABCD}

. If

A B=\sqrt{10}

and the points

\mathrm{A}

and

\mathrm{C}

lie on the line

3 y=2 x+1

, then

2(\alpha+\beta+\gamma+\delta)

is equal to

A 8

B 5

C 12

D 10

Correct Answer

Option A

Solution

Let E is mid point of diagonals

\frac{\alpha+\gamma}{2}=\frac{1+1}{2}

\alpha+\gamma=2

&

\frac{\beta+\delta}{2}=\frac{2+0}{2}

\beta+\delta=2

2(\alpha+\beta+\gamma+\delta)=2(2+2)=8

Q108

In a

\triangle A B C

, suppose

y=x

is the equation of the bisector of the angle

B

and the equation of the side

A C

is

2 x-y=2

. If

2 A B=B C

and the points

A

and

B

are respectively

(4,6)

and

(\alpha, \beta)

, then

\alpha+2 \beta

is equal to

A 42

B 39

C 48

D 45

Correct Answer

Option A

Solution

\begin{aligned} & A D: D C=1: 2 \\ & \frac{4-\alpha}{6-\alpha}=\frac{10}{8} \\ & \alpha=\beta \\ & \alpha=14 \text{ and } \beta=14 \end{aligned}

Q109

Let

\mathrm{A}

be the point of intersection of the lines

3 x+2 y=14,5 x-y=6

and

\mathrm{B}

be the point of intersection of the lines

4 x+3 y=8,6 x+y=5

. The distance of the point

P(5,-2)

from the line

\mathrm{AB}

is

A

\dfrac{13}{2}

B 8

C

\dfrac{5}{2}

D 6

Correct Answer

Option D

Solution

Solving lines

\mathrm{L}_1(3 \mathrm{x}+2 \mathrm{y}=14)

and

\mathrm{L}_2(5 \mathrm{x}-\mathrm{y}=6)

to get

\mathrm{A}(2,4)

and solving lines

\mathrm{L}_3(4 \mathrm{x}+3 \mathrm{y}=8)

and

\mathrm{L}_4(6 \mathrm{x}+\mathrm{y}=5)

to get

\mathrm{B}\left(\frac{1}{2}, 2\right)

. Finding Eqn. of

\mathrm{AB}: 4 \mathrm{x}-3 \mathrm{y}+4=0

Calculate distance PM

\Rightarrow\left|\frac{4(5)-3(-2)+4}{5}\right|=6

Q110

The distance of the point

(2,3)

from the line

2 x-3 y+28=0

, measured parallel to the line

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

, is equal to

A

3+4 \sqrt{2}

B

6 \sqrt{3}

C

4+6 \sqrt{3}

D

4 \sqrt{2}

Correct Answer

Option C

Solution

Writing

P

in terms of parametric co-ordinates

2+r

\begin{aligned} & \cos \theta, 3+\mathrm{r} \sin \theta \text{ as } \tan \theta=\sqrt{3} \\ & \mathrm{P}\left(2+\frac{\mathrm{r}}{2}, 3+\frac{\sqrt{3} \mathrm{r}}{2}\right) \end{aligned}

\mathrm{P}

must satisfy

2 \mathrm{x}-3 \mathrm{y}+28=0

So,

2\left(2+\frac{r}{2}\right)-3\left(3+\frac{\sqrt{3} \mathrm{r}}{2}\right)+28=0

We find

r=4+6 \sqrt{3}