Math Ncert Exemplar 2019 Solutions for Class 11 Science Maths Chapter 10 - Straight Lines

Answer:

Let a be equal interest made b straight line from both axis
Then, Equation of line $\frac{x}{a}+\frac{y}{b}=1$        .....(1)
Since, the given the passes through (1, –2)
$-\frac{-1}{a}=1$
⇒ a = 1
Hence, the equation of line is x + y = –1.

Answer:

Slope of line passing through (2, 3) and (3, –1)
$\frac{-1-3}{3-2}$
= –4
Slope of line perpendicular to given line
$$\begin{gathered} =\frac{-1}{-4} [\because m_1{m}_2=-1] \\ =\frac{1}{4} \end{gathered}$$
Equation of line passing through (5, 2) is:
y – 2 = 1
= x – 4y + 3 = 0.
Hence, equation of line is x – 4y + 3 = 0.

Answer:

Let slope of y = $(2-\sqrt{3}) (x + 5)$ be n₁,
∴ $n_1=2-\sqrt{3}$
Also, slope of y = $(2+\sqrt{3}) (x-7)$ be n₂
$m_2=2+\sqrt{3}$
Let θ be angle between these lines 
$\begin{array}{rcl}\tan \theta & =& \left|\frac{n_1-n_2}{1+m_1{m}_2}\right|\\ & =& \left|\frac{(2-\sqrt{3})-(2+\sqrt{3})}{1+(2-\sqrt{3}) (2+\sqrt{3})}\right|\\ & =& \left|\frac{-2\sqrt{3}}{1+4-3}\right|\\ & =& \sqrt{3}\end{array}$
∴ θ = 60º

Answer:

Let the equation of line be
$\frac{x}{a}+\frac{y}{b}=1$                           .....(1)
Since a + b = 14
⇒ b = 14 – a.
$$\begin{gathered} \therefore \frac{x}{a}+\frac{y}{14-a}=1 \\ \Rightarrow \frac{x(14-a)+ya}{a(14-a)}=1 .....\left(2\right) \end{gathered}$$
Since, Eq(2) passes through (3, 4) then
$\frac{3(14-a)+4a}{a(14-a)}=1$
⇒ a² – 13a + 42 = 0
⇒ (a – 6) (a – 7) = 0
⇒ a = 6, 7
If a = 6, b = 8. then Equation of line will be x + y = 7.

Answer:

Let the required point be (α, β)
Now, α + β = 4             .....(1)
Also, distance of point (α, β) from 4x + 3y = 10
$\left|\frac{4\alpha + 3\beta -10}{\sqrt{4^2+3^2}}\right|=1$
⇒ 4α + 3β – 10 = +5
⇒ 4λ + 3k = 15           .....(2)
⇒ 4λ + 3k = 5             .....(3)
Solving (1) & (2), we get
α = 3, β = 1  i.e. (3, 1)
Solving (1) and (3), we get
α = –7  β = 11  i.e., (–7, 11)
 

Answer:

Slope of $\frac{x}{a}+\frac{y}{b}=1 is m_1$
i.e., $m_1=\frac{-b}{a}$
Slope of line $\frac{x}{a}-\frac{y}{b}=1$ is m₂
$m_2=\frac{b}{a}.$
Let θ be the angle between given lines
$\begin{array}{rcl}\tan \theta & =& \left|\frac{m_1-m_2}{1+m_1{m}_2}\right|\\ & =& \left|\frac{{\displaystyle \frac{-b}{a}}-{\displaystyle \frac{b}{a}}}{1+\left({\displaystyle \frac{-b}{a}}\right)\left({\displaystyle \frac{b}{a}}\right)}\right|\\ & =& \left|\frac{{\displaystyle \frac{-2b}{a}}}{{\displaystyle \frac{a^2-b^2}{a^2}}}\right|\\ & =& \frac{2ab}{a^2-b^2}\end{array}$
Thus, tan θ is $\frac{2ab}{a^2-b^2}.$

Answer:

Given, line passes from point (1, 2) and makes an angle 30º with y-axis.
∴ Angle with x-axis is 60º.
Slope of line = tan 60º = $\sqrt{3}$.
So, Equation of line
$$\begin{gathered} y-2=\sqrt{3}(x-1) \\ \Rightarrow y-2=\sqrt{3}x-\sqrt{3} \\ \Rightarrow \sqrt{3}x-y=\sqrt{3}-2 \end{gathered}$$
Thus, the equation of line is $\sqrt{3}x-y=\sqrt{3}-2$

Answer:

Let the common point of intersection of lines 2x + y = 5 and x + 3y + 8 = 0 be (h, k)
2h + k = 5                .....(1)
h + 3k = –8              .....(2)
Solving (1) and (2), we get
$\left(h, k\right)=\left(\frac{23}{5}, \frac{-21}{5}\right)$
Let slope of line 3x + 4y = 7 is m
i.e., $m=\frac{-3}{4}.$
∴ Equation of line $y+\frac{21}{5}=\frac{-3}{4}\left(x-\frac{23}{5}\right)$
⇒ 3x + 4y + 3 = 0.

Answer:

Intercepts cuts of one Co-ordinate axis by line
ax + by + 8 = 0              .....(1)
Slope of line 1 = $\frac{-a}{b}$.
Also, slope of line 2x – 3x + 6 = 0 is $\frac{2}{3}.$
$$\begin{gathered} \mathrm{Now}, \frac{-a}{b}=\frac{2}{3} \\ a=\frac{-2b}{3} \end{gathered}$$
Let the length of perpendicular from origin to line be d₁
$\begin{array}{rcl}{d}_1& =& \left|\frac{a\left(0\right)+b\left(0\right)+8}{a^2+b^2}\right|\\ & =& \frac{8}{\sqrt{a^2+b^2}}\\ & =& \frac{8\times 3}{\sqrt{13b^2}}\end{array}$
Also, length of perpendicular from origin to line (2).
$d_2=\left|\frac{2\left(0\right)-3\left(0\right)+6}{\sqrt{2^2+3^2}}\right|$
Since, d₁ = d₂
$\frac{8\times 3}{\sqrt{13}b}=\frac{6}{\sqrt{13}}$
= b = 4
So, $a=\frac{-8}{3}$
Thus, the value of a is $\frac{-8}{3}$ and b is 4.

Answer:

Given, Co-ordinate axis divide by point (–5, 4) in ratio (1 : 2)
$$\begin{gathered} \therefore (-5, 4)=\frac{1 \times 0 + 2 \times a}{1 + 2}, \frac{1 \times b + 2 \times a}{1 + 2} \\ \Rightarrow (-5, 4)=\left(\frac{2a}{3}, \frac{b}{3}\right) \\ \Rightarrow a=\frac{-15}{2} \mathrm{and} b=12 \end{gathered}$$
Equation of line
$y-0=\frac{12-0}{0-\left({\displaystyle \frac{-15}{2}}\right)}\left(x-\left(\frac{-15}{2}\right)\right)$
⇒ 5y = 8x + 60
⇒ 8x – 5y + 60 = 0
Thus, required equation of line is 8x + 5y + 60 = 0.

Answer:

GRAPHICS
$$\begin{gathered} \angle BAO=180°-120 \\ = 60° \\ \therefore \angle MAO+\angle MOA=90° \\ \Rightarrow \theta =30° \end{gathered}$$

We know that, the equation of  AB in its normal form 
$$\begin{gathered} x\cos 30°+y\sin 30°=8 \\ \Rightarrow \sqrt{3}x+y=8 \end{gathered}$$

Thus the equation of AB in its normal form is $\sqrt{3}x+y=8$.

Answer:

GRAPHICS

Now, the slope of AC = $\frac{-3}{4}$
Since ABC is an isosceles right-angle triangle,
$\angle BAC=\angle ACB=45°$
Now, let the slope of the line make an angle of $45°$ be m.
$$\begin{gathered} \therefore \tan 45°=\left|\frac{m-\left({\displaystyle \frac{-3}{4}}\right)}{1+m\left({\displaystyle \frac{-3}{4}}\right)}\right| \\ \Rightarrow \frac{4m+3}{4-3m}=\pm 1 \\ \Rightarrow 4m+3=4-3m \mathrm{or} \\ 4m+3=3m-4 \\ \Rightarrow m=\frac{1}{7} \mathrm{or} -7 \end{gathered}$$
So, if the slope of the line is 1/7, then the slope of line AB is -7.
$$\begin{gathered} \mathrm{Equation} \mathrm{of} \mathrm{BC}: y-2=\frac{1}{7}\left(x-2\right) \\ \Rightarrow x-7y+12=0 \\ \mathrm{Equation} \mathrm{of} \mathrm{AB}: y-2=-7\left(x-2\right) \\ \Rightarrow 7x+y-16=0 \end{gathered}$$


 

Answer:

We have, the perpendicular distance of (2, −1) from line x + y = 2 is nothing but the height of the equilateral triangle.
$$\begin{gathered} =\frac{\left|2-1\right|}{\sqrt{2}} \\ =\frac{1}{\sqrt{2}} \end{gathered}$$
Also, the height of an equilateral triangle be = $\frac{\sqrt{3}}{2}\times \mathrm{side}$
$$\begin{gathered} \Rightarrow \frac{\sqrt{3}}{2}\times \mathrm{side}=\frac{1}{\sqrt{2}} \\ \Rightarrow \mathrm{side}=\sqrt{\frac{2}{3}} \end{gathered}$$

Answer:

Let the variable line be $ax+by=1$
Since, the algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2) and (1, 1) on the line is zero
$$\begin{gathered} \Rightarrow \frac{2a+0.b-1}{\sqrt{a^2+b^2}}+\frac{0.a+2b-1}{\sqrt{a^2+b^2}}+\frac{a+b-1}{\sqrt{a^2+b^2}}=0 \\ \Rightarrow a+b-1=0 \\ \Rightarrow a+b=1 \end{gathered}$$

Comparing $ax+by=1$ with a + b = 1
Therefore, the coordinates of the fixed point are (1, 1).

Answer:

Let the slope of the line be m. Also, the line passes through the point A(1, 2)
∴ Equation of line: $y-2=m\left(x-1\right) .....\left(1\right)$
Also, the equation of the given line: x+y=4 
Let these lines meet at point B
Solving 1 and 2, we get
B = $\left(\frac{m+2}{m=1}, \frac{3m+2}{m+1}\right)$

$$\begin{gathered} A{B}^2=\frac{6}{9} \\ \Rightarrow {\left(\frac{m+2}{m+1}-1\right)}^2+{\left(\frac{3m+2}{m+1}-2\right)}^2=\frac{6}{9} \\ \Rightarrow m^2-4m+1=0 \\ \therefore \tan \theta =2+\sqrt{3} \mathrm{or} 2-\sqrt{3} \\ \Rightarrow \theta =75° or 15° \end{gathered}$$
 

Answer:

Given that,
$\frac{1}{a}+\frac{1}{b}=cons\tan t =\frac{1}{k}$
$\Rightarrow \frac{k}{a}+\frac{k}{b}=1$
So, (k, k) lies on the line $\frac{x}{a}+\frac{y}{b}=1$
Thus, the line passes through the fixed point (k, k).

Answer:

Let the line cut the x-axis at (a, 0) and the y-axis at (0, b)
It is given that the point (–4, 3) divides the line segment internally in the ratio 5 : 3.
$$\begin{gathered} \Rightarrow \frac{3a}{8}=-4 \\ \Rightarrow a=-\frac{32}{3} \\ Also, \\ \frac{5b}{8}=3 \\ \Rightarrow b=\frac{24}{3} \\ \mathrm{Using} \mathrm{slope} \mathrm{intercept} \mathrm{form},\mathrm{we} \mathrm{get} \\ \frac{\mathrm{x}}{{\displaystyle \frac{-32}{3}}}+\frac{\mathrm{y}}{{\displaystyle \frac{24}{3}}}=1 \\ \Rightarrow 9\mathrm{x}-20\mathrm{y}+967=0 \end{gathered}$$

Answer:

Given that, x – y + 1 = 0
and 2x – 3y + 5 = 0

Solving these lines, we get the point of intersection as (2, 3)
Let the slope of the required line be m
So, the equation of the required line: $y-3=m\left(x-2\right)$
$\Rightarrow mx-y+3-2m=0$
The distance of the line from points (3, 2) is $\frac{7}{5}$.
$$\begin{gathered} \therefore \left|\frac{3m-2+3-2m}{\sqrt{1+m^2}}\right|=\frac{7}{5} \\ \Rightarrow \frac{49}{25}=\frac{{\left(m+1\right)}^2}{1+m^2} \\ \Rightarrow \left(3m-4\right)\left(4m-3\right)=0 \\ \Rightarrow m=\frac{4}{3}or\frac{3}{4} \\ \mathrm{So}, \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{line} \mathrm{can} \mathrm{be} \\ \mathrm{y}-3=\frac{4}{3}\left(\mathrm{x}-2\right) \mathrm{or} \\ \mathrm{y}-3=\frac{3}{4}\left(\mathrm{x}-2\right) \end{gathered}$$

Answer:

Let the coordinates of the moving point be (x, y)
Given that the sum of the distance from the axis to the point is always 1.

$$\begin{gathered} \therefore \left|x\right|+\left|y\right|=1 \\ \Rightarrow x+y=1 \\ \Rightarrow -x-y=1 \\ \Rightarrow x-y=1 \end{gathered}$$
Hence, these equations give us the locus of the point which is a square.

Answer:

Given that, $y-\sqrt{3} x=2 and y+\sqrt{3} x=2$ are two lines at a distance of 5 units from their point of intersection.
Solving the given two lines, we get the point of intersection A(0, 2)

Now, $y=\sqrt{3}x+2$ is inclined at an angle with a positive direction of x-axis.
Also,  $y=-\sqrt{3}x+2$ is inclined at an angle with a positive direction of x-axis.

GRAPHICS
Clearly, the angles bisector of a line is the y-axis
Now, AP₁ = 5
$$\begin{gathered} \therefore \mathrm{In} △{\mathrm{ABP}}_1 \\ \frac{\mathrm{AB}}{{\mathrm{AP}}_1}=\cos 30° \\ \Rightarrow \mathrm{AB}=\frac{5\sqrt{3}}{2} \\ \Rightarrow \mathrm{OB}=2+\frac{5\sqrt{3}}{2} \\ \mathrm{So}, \mathrm{the} \mathrm{coordinates} \mathrm{of} \mathrm{the} \mathrm{foot} \mathrm{of} \mathrm{the} \mathrm{perpendicular} \mathrm{are} \left(0,2+\frac{5\sqrt{3}}{2}\right). \end{gathered}$$

Answer:

Since p is the length of the perpendicular from the origin to the given point.
$$\begin{gathered} \therefore p=\left|\frac{{\displaystyle \frac{0}{a}}+{\displaystyle \frac{0}{b}}-1}{\sqrt{{\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{1}{b^2}}}}\right| \\ \Rightarrow p^2=\left|\frac{{\displaystyle 1}}{\frac{1}{a^2}+\frac{1}{b^2}}\right| \\ \Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2} .....\left(1\right) \\ \mathrm{Since} a^2,b^2 and p^2 \mathrm{are} \mathrm{in} \mathrm{AP} \\ \therefore 2{\mathrm{p}}^2={\mathrm{a}}^2+{\mathrm{b}}^2 \\ \Rightarrow {\mathrm{p}}^2=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2} \\ \Rightarrow \frac{1}{{\mathrm{p}}^2}=\frac{2}{{\mathrm{a}}^2+{\mathrm{b}}^2} .....\left(2\right) \\ \mathrm{From} \left(1\right) \mathrm{and} \left(2\right) \\ \frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}=\frac{2}{{\mathrm{a}}^2+{\mathrm{b}}^2} \\ \Rightarrow {\mathrm{a}}^4+{\mathrm{b}}^4=0 \end{gathered}$$

Answer:

The equation of a line in slope intercept form is $y-y_1=m\left(x-x_1\right)$.
$$\begin{gathered} \Rightarrow y-\left(-3\right)=\frac{3}{5}\left(x-0\right) \\ =5y-3x+15=0 \end{gathered}$$

Hence, the correct answer is option (a).
 

Answer:

Since the intercept are equal
$$\begin{gathered} \Rightarrow \frac{x}{a}+\frac{y}{a}=1 \\ \Rightarrow x+y=a \\ \Rightarrow y=-x+a \end{gathered}$$

So, the slope of the line is $-1$.

Hence, the correct answer is option (a).

Answer:

Given that,
The slope of line y = x is 1
The slope of the line perpendicular to y = x will be $-1.$
The required line can be written as $y=-x+c.$

Since the line passes through (3, 2).
$$\begin{gathered} \Rightarrow 3=-2+c \\ \Rightarrow c=5 \end{gathered}$$

Thus, the equation of the line is x + y = 5.

Hence, the correct answer is option (b).

Answer:

Since the required line is perpendicular to x + y + 1 = 0.
So, the slope of the required line = 1

Equation of a line in slope intercept form = $y-2=1\left(x-1\right)$
                                                               = x − y + 1 = 0

Hence, the correct answer is an option (b).

Answer:

The line passes through the point (a, 0) and (0, –b)
$\therefore \mathrm{Slope} \mathrm{of} \mathrm{line} m_1=\frac{-b-0}{0-a}=\frac{b}{a}$

The line passes through the point (b, 0) and (0, –a)
$\therefore \mathrm{Slope} \mathrm{of} \mathrm{line} m_2=\frac{-a-0}{0-b}=\frac{a}{b}$
If $\theta$ is the angle between the lines, then
$$\begin{gathered} \tan \theta =\frac{{\displaystyle \frac{b}{a}}-{\displaystyle \frac{a}{b}}}{1+{\displaystyle \frac{b}{a}}\times {\displaystyle \frac{a}{b}}} \\ =\frac{b^2-a^2}{2ab} \end{gathered}$$

Hence, the correct answer is option (c).

 

Answer:

Given that, $\frac{x}{a}+\frac{y}{b}=1$
The given lines pass through (2, –3) and (4, –5).
$$\begin{gathered} \therefore \frac{2}{a}-\frac{3}{b}=1 \mathrm{and} \\ \frac{4}{a}-\frac{5}{b}=1 \end{gathered}$$

Solving the above two equations, we get
$\left(a,b\right)=\left(-1,-1\right)$

Hence, the correct answer is option (d).

Answer:

Given intersecting lines are 2x – 3y + 5 = 0 and 3x + 4y = 0
Solving, we get the point of intersection $\mathrm{P}\left(\frac{-20}{17},\frac{15}{17}\right)$

So, the length of the perpendicular from P to 5x – 2y = 0 is
$$\begin{gathered} =\frac{\left|5\left({\displaystyle \frac{-20}{17}}\right)-2\left(\frac{15}{17}\right)\right|}{\sqrt{5^2+{\left(-2\right)}^2}} \\ =\frac{130}{17\sqrt{29}} \end{gathered}$$

Hence, the correct answer is option (a).

Answer:

The slope of the line $\sqrt{3} x+y=1$, $m_1=-\sqrt{3}$
Let m₂ be the slope of the required line.
$$\begin{gathered} \therefore \tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right| \\ \Rightarrow \tan 60°=\left|\frac{-\sqrt{3}-m_2}{1+\left(-\sqrt{3}\right)m_2}\right| \\ \Rightarrow \sqrt{3}=\pm \left(\frac{-\sqrt{3}-m_2}{1+\left(-\sqrt{3}\right)m_2}\right) \\ \Rightarrow \sqrt{3}=\frac{-\sqrt{3}-m_2}{1+\left(-\sqrt{3}\right)m_2} \left(\mathrm{Taking} + \mathrm{sign}\right) \\ \Rightarrow m_2=\sqrt{3} \\ \Rightarrow \sqrt{3}=-\left(\frac{-\sqrt{3}-m_2}{1+\left(-\sqrt{3}\right)m_2}\right) \left(\mathrm{Taking} - \mathrm{sign}\right) \\ \Rightarrow m_2=0 \\ \therefore \mathrm{Equation} \mathrm{of} \mathrm{line} \mathrm{passing} \mathrm{through} \left(3,2\right) \mathrm{with} \mathrm{slope} \sqrt{3} \mathrm{is} \\ \Rightarrow \sqrt{3}x-y-2-3\sqrt{3}=0 \\ \therefore \mathrm{Equation} \mathrm{of} \mathrm{line} \mathrm{passing} \mathrm{through} \left(3,2\right) \mathrm{with} \mathrm{slope} 0 \mathrm{is} \\ \Rightarrow y+2=0 \end{gathered}$$

Hence, the correct answer is option (a).

Answer:

The equation of a line passing through the point (1, 0) and slope m is 
$$\begin{gathered} =y-0=m\left(x-1\right) \\ =y=mx-m \\ =mx-y-m=0 \end{gathered}$$

Distance from origin = $\frac{\sqrt{3}}{2}$
$$\begin{gathered} \left|\frac{-m}{\sqrt{1+m^2}}\right|=\frac{\sqrt{3}}{2} \\ \Rightarrow m^2=3 \\ \Rightarrow m=\pm \sqrt{3} \\ \mathrm{Therefore}, \mathrm{equation} \mathrm{of} \mathrm{lines} \mathrm{are} \\ -\sqrt{3}\mathrm{x}-\mathrm{y}+\sqrt{3}=0 \\ \sqrt{3}\mathrm{x}+\mathrm{y}-\sqrt{3}=0 \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

The distance between the lines y = mx + c₁ and y = mx + c₂ is
$\frac{\left|c_1-c_2\right|}{\sqrt{m^2+1}}$

Hence, the correct answer is option (b).

Answer:

Let the foot of the perpendicular from the point P(2, 3) on the line y = 3x + 4 be M(h, k)
M(h, k) lies on the given line.
$\therefore 3h-k+4=0$
Also, the slope of the given line is 3
$$\begin{gathered} \therefore \mathrm{Slope} \mathrm{of} \mathrm{PM} = \frac{-1}{3}=\frac{k-3}{h-2} \\ \Rightarrow h+3k-11=0 .....\left(1\right) \\ \mathrm{Solving} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get} \\ \left(h,k\right)=\left(\frac{-1}{10},\frac{37}{10}\right) \end{gathered}$$

Hence, the correct answer is an option (b).
 

Answer:

Since the middle point is P(2, 3), then the line meet axes at A(6, 0) and B(0, 4)
The equation of a line in intercept form is
$$\begin{gathered} \Rightarrow \frac{x}{6}+\frac{y}{4}=1 \\ \Rightarrow 2x+3y=12 \end{gathered}$$

Hence, the correct answer is an option (a).
 

Answer:

Since the required line is parallel to y = 3x – 1
Therefore, slope of the line is 3.
Now, the equation of line is
$$\begin{gathered} y-2=3\left(x-1\right) \\ \Rightarrow 3x-y-1=0 \end{gathered}$$

Hence, the correct answer is option (c).

Answer:

Coordinates of the vertices of the square are A(0,0),B(0,1),C(1,1) and D(1,0).
Now, the equation of AC is  y = x and

Answer:

For specifying a straight line, 2 geometrical parameters  are required.

Hence, the correct answer is option (b).

Answer:

Reflection od A(4, 1) about y = x is S(1, 4).
Now, translation through a distance 2 units along the positive x-axis will give us the point B(1+2, 4) = B(3,4)

Hence, the correct answer is option (b).

Answer:

Given lines,'
4x + 3y + 10 = 0           .....(1)
5x – 12y + 26 = 0         .....(2)
7x + 24y – 50 = 0         .....(3)
Let (p,q) be the point that is equidistant from the given lines.
Distance of (p, q), from eq(1)
=$\frac{\left|4p+3q+10\right|}{5}$
Similarly, the distance of (p,q) from eq(2) and eq(3) are
$\Rightarrow \frac{\left|5p-12q+26\right|}{13} and \frac{\left|7p+24q-50\right|}{25}$
Let the point (p, q) = (0, 0)
$\Rightarrow \frac{10}{5}=\frac{26}{13}=\frac{50}{25}=2$

Hence, the correct answer is option (c).

Answer:

Since the required line is perpendicular to 3x + y = 3
Therefore, the slope of the required line, $m=\frac{1}{3}.$
Equation of the line is
$$\begin{gathered} y-2=\frac{1}{3}\left(x-2\right) \\ \Rightarrow 3y-6=x-2 \\ \Rightarrow x-3y=-4 \\ \Rightarrow 3y-x=4 \\ \Rightarrow \frac{y}{{\displaystyle \frac{4}{3}}}-\frac{x}{4}=1 \end{gathered}$$

Hence, the correct answer is option (d).

Answer:

Lines 3x + 4y + 2 = 0 & 3x + 4y + 5 = 0 are on the same side of the origin. The distance between these Iines is
$d_1=\frac{\left|2-5\right|}{\sqrt{3^2+4^2}}=\frac{3}{5}$
Lines 3x + 4y + 2 = 0 and 3x + 4y - 5 = 0 are on the opposite sides of the origin. The distance between these lines is
$d_1=\frac{\left|2-\left(-5\right)\right|}{\sqrt{3^2+4^2}}=\frac{7}{5}$

Thus, 3x + 4y + 2 = 0 divides the distance between 3x + 4y + 5 = 0 and 3x + 4y − 5 = 0 in the ratio d1​:d2​, i.e. 3:7.

Hence, the correct answer is option (b)

Answer:

Let ABC be the equilateral triangle with vertex A(h, k).
Also centroid G(0, 0)
Now, $\mathrm{AG}\perp \mathrm{BC}$
The slope of line x + y – 2 = 0 is -1.
$$\begin{gathered} \therefore \mathrm{Slope} \mathrm{of} \mathrm{AG}, \frac{k}{h}=1 \\ Now, dis\tan ce of origin from BC=\frac{\left|0+0-2\right|}{\sqrt{1^2+1^2}}=\sqrt{2} \\ \therefore \mathrm{Distance} \mathrm{of} \mathrm{A} \mathrm{from} \mathrm{BC} \\ \left|\mathrm{h}+\mathrm{k}-2\right|=6 \\ \Rightarrow \mathrm{h}+\mathrm{k}-8=0 \mathrm{or} \mathrm{h}+\mathrm{k}+4=0 \\ \Rightarrow \mathrm{h}=4 \mathrm{or} -2 \\ \therefore \mathrm{Vertex} \mathrm{is} \left(-2,-2\right) \end{gathered}$$

Hence, the correct answer is option (c).
 

Answer:

Given line, 
ax + by + c = 0        .....(1)

a, b, c are in A.P.,
$$\begin{gathered} \Rightarrow 2b=a+c \\ \Rightarrow a-2b+c=0 .....\left(2\right) \end{gathered}$$
Comparing (1) and (2), we get
(x, y) = (1, $-$2)

Thus, the given line always passes through (1, $-$2).
 

Answer:

The equation of line cutting equal intercept from the axes is
$\frac{x}{a}+\frac{y}{a}=1$
Now, the line passes through (1, –2).
$$\begin{gathered} \Rightarrow \frac{1}{a}-\frac{2}{a}=1 \\ \Rightarrow \frac{-1}{a}=1 \\ \Rightarrow a=-1 \end{gathered}$$

Thus, the equation of line is $x+y+1=0$.

Answer:

Let the slope of the line be m.
$$\begin{gathered} \tan 45°=\frac{\left|m_1-m_2\right|}{1+m_1{m}_2} \\ 1=\frac{\left|{\displaystyle \frac{1}{2}}-m\right|}{1+{\displaystyle \frac{m}{2}}} \\ \Rightarrow m=3 or \frac{-1}{3} \\ Case-1 m=3 \\ y-2=3\left(x-3\right) \\ \Rightarrow 3x-y=7 \\ Case-2 m=\frac{-1}{3} \\ y-2=\frac{-1}{3}\left(x-3\right) \\ \Rightarrow x+3y=9 \end{gathered}$$
 

Answer:

For points (3, 4)
$3\left(3\right)-4\left(4\right)-8<0$

For points  (2, –6)
$3\left(2\right)-4\left(-6\right)-8>0$

Thus, the given points lies on the opposite sides of the line.

Answer:

Let the moving point be P(h, k)
Given point A(3, –2)
$A{P}^2={\left(h-3\right)}^2+{\left(k+2\right)}^2={d_1}^2$
Now distance of the point (h, k)  from the line 5x – 12y = 3 is 
$$\begin{gathered} d_2=\frac{\left|5h-12k-3\right|}{\sqrt{25+144}}=\frac{\left|5h-12k-3\right|}{13} \\ \mathrm{Given} \mathrm{that} {{\mathrm{d}}_1}^2={\mathrm{d}}_2 \\ \Rightarrow {\left(\mathrm{h}-3\right)}^2+{\left(\mathrm{k}+2\right)}^2=\frac{5\mathrm{h}-12\mathrm{k}-3}{13} \\ \Rightarrow {\mathrm{h}}^2-6\mathrm{h}+9+{\mathrm{k}}^2+4\mathrm{k}+4=\frac{5\mathrm{h}-12\mathrm{k}-3}{13} \\ \Rightarrow 13{\mathrm{h}}^2+13{\mathrm{k}}^2-83\mathrm{h}+64\mathrm{k}+172=0 \\ \mathrm{So}, \mathrm{the} \mathrm{locus} \mathrm{of} \mathrm{the} \mathrm{point} \mathrm{is}: \\ 13{\mathrm{h}}^2+13{\mathrm{k}}^2-83\mathrm{h}+64\mathrm{k}+172=0 \end{gathered}$$

Answer:

Let the line cut the axes in A and B and if (h,k) be the midpoints of AB, then
$2h=\frac{p}{\cos \theta }, 2k=\frac{p}{\sin \theta }$

Answer:

Without loss of generality, we can choose the co-ordinates as A = (0,0), B = (x, y) and C = (a, b)

Now, the slope of AB = $\frac{y}{x}$​ is the rational slope of AC = $\frac{b}{a}$​ rational.
So, 
$\tan \left(BAC\right)=\left(\frac{{\displaystyle \frac{y}{x}}-{\displaystyle \frac{a}{b}}}{1+{\displaystyle \frac{ay}{bx}}}\right) \mathrm{which} \mathrm{is} \mathrm{rational}$
 

As we cannot find any rational point on the k line at 60o so getting an equilateral triangle is not possible.

Thus, the above statement is True.

Answer:

$$\begin{gathered} \mathrm{Slope} \mathrm{of} \mathrm{AB}: \\ {m}_{AB}=\frac{5-1}{0-\left(-2\right)}=2 \\ \mathrm{Slope} \mathrm{of} BC: \\ {m}_{BC}=\frac{2-5}{-1-\left(0\right)}=3 \\ \mathrm{Slope} \mathrm{of} CA: \\ {m}_{CA}=\frac{1-2}{-2-\left(-1\right)}=1 \end{gathered}$$

Since the slopes are different, therefore the given points are not collinear.
Hence, the given statement is False.

Answer:

Equation of the line passing through the point (a cos^₃θ, a sin^₃θ) and perpendicular to the line x sec θ + y cosec θ = a is x cos θ – y sin θ = a sin 2θ.

Thus, the above statement is True.

Answer:

Given lines are,
 x + 2y – 10 = 0            .....(1)
2x + y + 5 = 0              .....(2)
Solving the above lines, we get
Point = $\left(\frac{-20}{3},\frac{25}{3}\right)$
Now, 
$\Rightarrow 5\times \left(\frac{-20}{3}\right)+4\times \frac{25}{3}=0$

Hence, the above statement is True.

Answer:

Slope of line x + y = 2 is $-1$
Let m be the slope of the other line which makes an angle of $60°$ with the above line.

$$\begin{gathered} \tan 60=\left|\frac{m+1}{1-m}\right| \\ \sqrt{3}=\left|\frac{m+1}{1-m}\right| \\ \Rightarrow 3{\left(1-m\right)}^2={\left(1+m\right)}^2 \\ \Rightarrow m=2\pm \sqrt{3} \end{gathered}$$
The equation of the two sides passing through (2,3) are $y-3=\left(2\pm \sqrt{3}\right) \left(x-2\right)$.

Hence, the given statement is True.
 

Answer:

Given lines are,
4x + y – 1 = 0       .....(1)
7x – 3y – 35 = 0   .....(2)
Solving the above lines, we get
(x, y) = (2, – 7)

Now, we have to find the equation line joining the point (3, 5) and (2, – 7)
$$\begin{gathered} y-5=\frac{-7-5}{2-3}\left(x-3\right) \\ \Rightarrow 12x-y-31=0 \\ \mathrm{Now}, \mathrm{distance} \mathrm{of} \mathrm{the} \mathrm{above} \mathrm{line} \mathrm{from} \mathrm{the} \mathrm{point} (0, 0) \mathrm{is} \\ \mathrm{d}=\left|\frac{12\times 0-0-31}{\sqrt{{12}^2+{\left(-1\right)}^2}}\right| \\ =\frac{31}{\sqrt{145}} \\ \mathrm{Also}, \mathrm{distance} \mathrm{of} \mathrm{the} \mathrm{above} \mathrm{line} \mathrm{from} \mathrm{the} \mathrm{point} (8, 34) \mathrm{is} \\ \mathrm{d}=\left|\frac{12\times 8-34-31}{\sqrt{{12}^2+{\left(-1\right)}^2}}\right| \\ =\frac{31}{\sqrt{145}} \end{gathered}$$

Thus, the above statement is True.
 

Answer:

Equation of a line: $\frac{x}{a}+\frac{y}{b}=1$       .....(1)
And $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$            .....(2)
Also, the equation of line perpendicular to (1)
$\frac{x}{a}-\frac{y}{b}=k$
If it passes through the origin then k = 0.
$\frac{x}{a}-\frac{y}{b}=0 .....\left(3\right)$

The locus of the foot of the ⊥ from origin on (1) 

i.e., locus of the point on the intersection of (1) and (3) is obtained by eliminating the parameters a and b between them.

Squaring (1) and (3) and adding, we get
$$\begin{gathered} \left(\frac{1}{a^2}+\frac{1}{b^2}\right)x^2+\left(\frac{1}{b^2}+\frac{1}{a^2}\right)y^2=1 \\ \frac{x^2}{c^2}+\frac{y^2}{c^2}=1 \end{gathered}$$

Thus, the above statement is True.

Answer:

Lines ax + 2y + 1 = 0, bx + 3y + 1 = 0 and cx + 4y + 1 = 0 are concurrent
$$\begin{gathered} \therefore \left|\begin{array}{ccc}a& 2& 1\\ b& 3& 1\\ c& 4& 1\end{array}\right|=0 \\ \Rightarrow a\left(3-4\right)-\left(b-c\right)2+1\left(4b-3c\right)=0 \\ \Rightarrow -a+2b-c=0 \\ \Rightarrow 2b=a+c \end{gathered}$$
Therefore, a, b, c are in AP

Thus, the above statement is False.

Answer:

Let the slope of the line joining the points (3, –4) and (–2, 6) be m₁
$m_1=\frac{6-\left(-4\right)}{-2-3}=-2$
Also, the slope of the line joining the points (–3, 6) and (9, –18). be m₂
$m_2=\frac{-18-\left(6\right)}{9-\left(-3\right)}=-2$

Now, 
$m_1\times m_2\ne -1$

Thus, the given two lines are not perpendicular.

Hence, the above statement is False.

Answer:

(a) Let $\left(x_1,y_1\right)$ be the point on the line x + 5y = 13.
$\therefore x_1+5y_1=13$
Distance of the line 12x – 5y + 26 = 0 from the point $\left(x_1,y_1\right)$ is
$$\begin{gathered} \frac{\left|12x_1-5y_1+26\right|}{\sqrt{{12}^2+{\left(-5\right)}^2}}=2 \\ \Rightarrow \frac{\left|12x_1-13+x_1+26\right|}{13}=2 \\ \Rightarrow \pm \left(x_1+1\right)=2 \\ \mathrm{When} x_1=1 \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} {\mathrm{x}}_1 \mathrm{in} \left(1\right) \\ {y}_1=\frac{12}{5} \\ \mathrm{When} x_1=-3 \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} {\mathrm{x}}_1 \mathrm{in} \left(1\right) \\ {y}_1=\frac{16}{5} \\ \mathrm{So}, \mathrm{the} \mathrm{required} \mathrm{points} \mathrm{are} \left(1,\frac{12}{5}\right) \mathrm{and} \left(-3,\frac{16}{5}\right) \end{gathered}$$

(b) Let $\left(x_1,y_1\right)$ be the point on the line x + y = 4.
$\therefore x+y=4 .....\left(1\right)$
Distance of the line 4x + 3y – 10 = 0 from the point $\left(x_1,y_1\right)$ is
$$\begin{gathered} \frac{\left|4x_1+3y_1-10\right|}{\sqrt{4^2+{\left(3\right)}^2}}=1 \\ \Rightarrow \frac{\left|4x_1+3\left(4-x_1\right)-10\right|}{5}=1 \\ \Rightarrow \pm \left(\frac{x_1+2}{5}\right)=1 \\ \mathrm{When} x_1=3 \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} {\mathrm{x}}_1 \mathrm{in} \left(1\right) \\ {y}_1=1 \\ \mathrm{When} x_1=-7 \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} {\mathrm{x}}_1 \mathrm{in} \left(1\right) \\ {y}_1=11 \\ \mathrm{So}, \mathrm{the} \mathrm{required} \mathrm{points} \mathrm{are} \left(3,1\right) \mathrm{and} \left(-7,11\right) \end{gathered}$$

(c)
Equation of line joining points A(–2, 5) and B(3, 1) 
$$\begin{gathered} y-5=\frac{1-5}{3+2}\left(x+2\right) \\ \Rightarrow y-5=\frac{-4}{5}\left(x+2\right) \\ \Rightarrow 4x+5y-17=0 \end{gathered}$$
Let $P\left(x_1,y_1\right) and Q\left(x_2,y_2\right)$ be any two points on line AB.
$P\left(x_1,y_1\right)$ divides line AB  in the ratio 1:2.
$$\begin{gathered} x_1=\frac{1.3+2\left(-2\right)}{1+2} \\ =\frac{-1}{3} \\ {y}_1=\frac{1.1+2.5}{1+2} \\ =\frac{11}{3} \\ \mathrm{So}, \mathrm{the} \mathrm{coordinates} \mathrm{of} \mathrm{P}\left(x_1,y_1\right)=\left(\frac{-1}{3},\frac{11}{3}\right). \\ Now, point Q\left(x_2,y_2\right) is the mid-point of PB. \\ {x}_2=\frac{3-{\displaystyle \frac{1}{3}}}{2} \\ =\frac{4}{3} \\ {y}_2=\frac{1+{\displaystyle \frac{11}{3}}}{2} \\ =\frac{7}{3} \\ Hence, \mathrm{the} \mathrm{coordinates} \mathrm{of} \mathrm{P}\left(x_2,y_2\right)=\left(\frac{4}{3},\frac{7}{3}\right). \end{gathered}$$

Thus, (a) → (iii),  (b) → (ii),  (c) → (i)
  

  

  
 

Answer:

(a) Given that,
(2x + 3y + 4) + λ(6x – y + 12) = 0
$$\begin{gathered} \Rightarrow \left(2+6\lambda \right)x+\left(3-\lambda \right)y+4+12\lambda =0 .....\left(1\right) \\ \mathrm{If} \mathrm{equation}\left(1\right) \mathrm{is} \mathrm{parallel} \mathrm{to} \mathrm{y}-\mathrm{axis} \\ \left(3-\mathrm{\lambda }\right)=0 \\ \Rightarrow \mathrm{\lambda }=3 \end{gathered}$$

(b) Given that,
(2x + 3y + 4) + λ(6x – y + 12) = 0           .....(1)
$$\begin{gathered} \Rightarrow \left(2+6\lambda \right)x+\left(3-\lambda \right)y+4+12I=0 \\ \mathrm{Slope}=-\left(\frac{2+6\lambda }{3-\lambda }\right) \\ \mathrm{Second} \mathrm{equation} \mathrm{is} 7x+y-4=0 .....\left(2\right) \\ \mathrm{slope}=-7 \\ \mathrm{If} \mathrm{equation} \left(1\right) \mathrm{and} \left(2\right) \mathrm{are} \mathrm{perpendicular} \mathrm{to} \mathrm{each} \mathrm{other} \\ -\left(\frac{2+6\lambda }{3-\lambda }\right)\times \left(-7\right)=-1 \\ \Rightarrow \lambda =\frac{-17}{41} \end{gathered}$$

(c) Given that,
(2x + 3y + 4) + λ(6x – y + 12) = 0           .....(1)
If equation (1) passes through the point (1, 2)
$$\begin{gathered} \Rightarrow \left(2\times 1+3\times 2+4\right)+\lambda \left(6\times 1-2+12\right)=0 \\ \Rightarrow \lambda =\frac{-3}{4} \end{gathered}$$

(c) Given that,
(2x + 3y + 4) + λ(6x – y + 12) = 0           .....(1)

If equation (1) is parallel to x-axis
$$\begin{gathered} \Rightarrow 2+\lambda 6=0 \\ \Rightarrow \lambda =\frac{-1}{3} \end{gathered}$$

Answer:

(a) Given lines are,
2x – 3y = 0            .....(1)
4x – 5y = 2            .....(2)
Equation of line passing through (1) and (2)
$\left(2x-3y\right)+k\left(4x-5y-2\right)=0 .....\left(3\right)$
If equation (3) passes through (2, 1), we get
$$\begin{gathered} \left(2\times 2-3\times 1\right)+k\left(4\times 2-5\times 1-2\right)=0 \\ \Rightarrow 1+k\left(8-7\right)=0 \\ \Rightarrow k=-1 \\ \mathrm{So}, \mathrm{the} \mathrm{required} \mathrm{equation} \mathrm{is} \\ \left(2x-3y\right)-1\left(4x-5y-2\right)=0 \\ \Rightarrow x-y-1=0 \end{gathered}$$

(b) Equation of the line passing through the intersection of 2x – 3y = 0 and 4x – 5y = 2 is
$\left(2x-3y\right)+k\left(4x-5y-2\right)=0 .....\left(1\right)$
$$\begin{gathered} Slope=\frac{2+4k}{3+5k} \\ \mathrm{Slope} \mathrm{of} \mathrm{line} x+2y+1=0 is \frac{-1}{2}. \\ \mathrm{If} \mathrm{the} \mathrm{line} \mathrm{are} \mathrm{perpendicular} \mathrm{to} \mathrm{each} \mathrm{other} \\ \frac{-1}{2}\left(\frac{2+4k}{3+5k}\right)=-1 \\ \Rightarrow k=\frac{-2}{3} \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{k} \mathrm{in} \mathrm{equation} \left(1\right), \mathrm{we} \mathrm{get} \\ \Rightarrow 2\mathrm{x}-\mathrm{y}=4 \end{gathered}$$

(c) Equation of the line passing through the given line  2x – 3y = 0 and 4x – 5y = 2 is
$\left(2x-3y\right)+k\left(4x-5y-2\right)=0 .....\left(1\right)$
$$\begin{gathered} \mathrm{Slope}=\frac{2+4k}{3+5k} \\ \mathrm{Slope} \mathrm{of} \mathrm{given} \mathrm{line} 3x-4y+5=0 \mathrm{is} \frac{3}{4}. \\ If the two equations are parallel, then \\ \frac{2+4k}{3+5k}=\frac{3}{4} \\ \Rightarrow k=1 \\ \mathrm{So}, \mathrm{equation} \mathrm{of} \mathrm{required} \mathrm{line} \mathrm{is} \\ 3x-4y-1=0 \end{gathered}$$

(d)  Equation of the line passing through the given line  2x – 3y = 0 and 4x – 5y = 2 is
$\left(2x-3y\right)+k\left(4x-5y-2\right)=0 .....\left(1\right)$
$$\begin{gathered} \mathrm{Slope}=\frac{2+4k}{3+5k} \\ \mathrm{Since} \mathrm{the} \mathrm{equaton} \mathrm{is} \mathrm{equally} \mathrm{inclined} \mathrm{with} \mathrm{axes} \\ \therefore -\tan 45°=-1 \\ \frac{2+4k}{3+5k}=-1 \\ \Rightarrow k=\frac{-5}{9} \\ \mathrm{So}, \mathrm{equation} \mathrm{of} \mathrm{required} \mathrm{line} \mathrm{is} \\ x+y-5=0 \end{gathered}$$