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Given: 

Here, b = -a 

To find: 

The equation of line cutoff intercepts from the axes. 

Explanation:

So, the equation of the line is 

Formula used:   \(\frac{x}{a}+\frac{y}{b}=1\) 

 \(\frac{x}{a}+\frac{y}{b}=1\) 

 \(\frac{x}{a}+\frac{y}{-a}=1\) 

⇒ x – y = a 

The line passes through the point (5, 6) So equation satisfy the points, 

⇒ 5 – 6 = a 

⇒ a = -1 

The equation of the line is x – y = -1

(i) Equal in magnitude and both positive

Given:

a = b

Let us find the equation of line cutoff intercepts from the axes.

By using the formula,
The equation of the line is x/a + y/b = 1

x/a + y/a = 1

x + y = a

The line passes through the point (5, 6)

Hence, the equation satisfies the points.

5 + 6 = a

a = 11

∴ The equation of the line is x + y = 11

(ii) Equal in magnitude but opposite in sign

Given:

b = -a

Let us find the equation of line cutoff intercepts from the axes.

By using the formula,
The equation of the line is x/a + y/b = 1

x/a + y/-a = 1

x – y = a

The line passes through the point (5, 6)

Hence, the equation satisfies the points.

5 – 6 = a

a = -1

∴ The equation of the line is x – y = -1

Given: 

Intercepts cut off on the coordinate axes by the line 

ax + by + 8 = 0 ……(i) 

And are equal in length but opposite in sign to those cut off by the line 

2x – 3y + 6 = 0 ……(ii) 

Explanation: 

The slope of two lines are equal 

The slope of the line (i) is \(-\frac{a}{b}\)

 The slope of the line (ii) is \(\frac{2}{3}\) 

∴ \(-\frac{a}{b}=\frac{2}{3}\) 

a = \(-\frac{2b}{3}\)

The length of the perpendicular from the origin to the line (i) is 

The formula used: d = \(|\frac{ax+by+d}{\sqrt{a^2+b^2}}|\) 

d₁ =   \(|\frac{a(0)+b(0)+8}{\sqrt{a^2+b^2}}|\) 

d₁ = \(\frac{8\times3}{\sqrt{13b^2}}\)

The length of the perpendicular from the origin to the line (ii) is 

The formula used: d =   \(|\frac{ax+by+d}{\sqrt{a^2+b^2}}|\) 

d₂ =   \(|\frac{2(0)-3(0)+6}{\sqrt{2^2+3^2}}|\)  

Given: d₁= d₂

  \(\frac{8\times3}{\sqrt{13b^2}}\)  =   \(\frac{6}{\sqrt{13b}}\) 

⇒ b = 4 

∴ a =   \(\frac{2b}{3}\)  =  \(-\frac{8}{3}\) 

Hence the value of a and b is  \(-\frac{8}{3}\) , 4.

Given: equation is perpendicular to x – 3y + 5 = 0 and passes through (3,-2) 

To find: Equation of required line. 

Explanation: 

The equation of the line perpendicular to x − 3y + 5 = 0 is 

3x + y + λ = 0, 

Where λ is a constant. 

It passes through (3, − 2). 

9 – 2 + λ = 0 

⇒ λ = - 7 

Substituting λ = − 7 in 3x + y + λ = 0, 

Hence, we get 3x + y – 7 = 0, which is the required line.

Let L₁ and L₂ be the lines passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18 respectively.

Slopes of the lines x + 2y = 19 and 3x + y = 18 are -1/2 and -3/1 = -3 respectively.

Since the lines L₁ and L₂ pass through the origin, their equations are

y = 2x and y = 1/3 x

i.e. 2x – y = 0 and x – 3y = 0

∴ their combined equation is

(2x – y)(x – 3y) = 0

∴ 2x² – 6xy – xy + 3y² = 0

∴ 2x² – 7xy + 3y² = 0.

Given, A line segment joining (4, 2) and (3, 5) if it cuts off an intercept 3 from y–axis.

To Find: The equation of that line. 

Formula used: The equation of line is y = mx + C 

Explanation: Here, The required equation of line is y = mx + c 

Now, c = 3 (Given)

Let m be slope of given line = – 1

Slope of line joining (x₁ – x₂) and (y₁ – y₂) ,m = \(\frac{y_2-y_1}{x_1-x_2}\)

So, Slope of line joining (4, 2) and (3, 5),m = \(\frac{5-2}{3-4}\) = \(\frac{2}{-1}\) 

Therefore, m = \(\frac{1}{3}\)

Now, The equation of line is y = mx + c

y = \(\frac{1}{3}x+3\) 

x – 3y + 9 = 0

Hence, The equation of line is 2y + 5x + 6 = 0.

Given:

The equation which makes an angle of tan^{– 1}(3) with the x–axis and cuts off an intercept of 4 units on the negative direction of y–axis

By using the formula,

The equation of the line is y = mx + c

Here, angle θ = tan^{– 1}(3)

So, tan θ = 3

The slope of the line is, m = 3

And, Intercept in the negative direction of y–axis is (0, -4)

The required equation of the line is y = mx + c

Now, substitute the values, we get

y = 3x – 4

∴ The equation of the line is y = 3x – 4.

Equation of a line parallel to Y-axis with x-intercept ‘h’ is x = h. 

Here, x-intercept = 3 

∴ The equation of the required line is x = 3.

Given, A straight line passing through the point (6, 2) and the slope is – 3

By using the formula,
The equation of line is [y – y₁ = m(x – x₁)]

Here, the line is passing through (6, 2)

It is given that, the slope of line, m = –3

Coordinates of line are (x₁, y₁) = (6,2)

The equation of line = y – y₁ = m(x – x₁)

Now, substitute the values, we get

y – 2 = – 3(x – 6)

y – 2 = – 3x + 18

y + 3x – 20 = 0

∴ The equation of line is 3x + y – 20 = 0

 Let, the points be A(x₁, y₁) = (3, – 2) and B(x_2, y₂) = (– 1, – 1).

Then, the distance

AB = \(\sqrt{(x_1 − x_2 )^2 + (y_1 − y_2)^2}\)

= \(\sqrt{(3 + 1)^2 + (-2 + 1)^2}\)

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

= \(\sqrt{17}\) units.

Let the given points be A(x₃, y₃) = (2, – 5).

So, the area of ∆ABC is

∆ = \(\frac{1}{2}\)|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

= \(\frac{1}{2}\)|8 (−4 + 5) + k(−5 − 1) + 2 (1 + 4)|

= \(\frac{1}{2}\)|−6k + 2|

= –3k + 1 or 3k – 1

Since A, B, C, are collinear.

∴ ∆ = 0

⇒ – 3k + 1 = 0 or 3k - 1 = 0

⇒ k = \(\frac{1}{3}\)

Let the given points be

P(x₁, y₁) = (a, 0), Q(x₂, y₂) = (0, b) and

R(x₃, y₃) = (3a – 2b).

∴ Area of ∆PQR = \(\frac{1}{2}\)|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

= \(\frac{1}{2}\)|a(b + 2b) + 0(−2b − 0) + 3a(0 − b)|

= 0

⇒ the points (a, 0), (0, b) and (3a, – 2b) are collinear.

Correct option is : (d) 2x – y = 0

Any line passing through origin is of the form y = mx or ax + by = 0. 

Here in the given option, 2x – y = 0 is in the form ax + by = 0. 

∴ Option (d) is the correct answer.

Given: 

Line a²x + ay + 1 = 0 is perpendicular to the line x – ay = 1

To prove:

The line a²x + ay + 1 = 0 is perpendicular to the line x – ay = 1 for all non-zero real values of a. 

Concept Used: 

Product of slope of perpendicular line is -1.

Explanation: 

The given lines are 

a²x + ay + 1 = 0 … (1) 

x − ay = 1 … (2) 

Let m₁ and m₂ be the slopes of the lines (1) and (2).

m₁m₂ = - \(\frac{a^2}{a}\times\frac{1}{a}\) = -1

Hence proved, line a²x + ay + 1 = 0 is perpendicular to the line x− ay = 1 for all non-zero real values of a.

Given: A (h,3) and B (4,1) be the points intersect At right angle at the line 7x – 9y – 19 = 0 

To find:

The value of h. 

Explanation:

The line 7x − 9y − 19 = 0 can be written as

\(y = \frac{7}{9}x-\frac{19}{9}\)  

So, the slope of this line is \(\frac{7}{9}\) 

It is given that the line joining the points A (h,3) and B (4,1) is perpendicular to the line 7x − 9y − 19 = 0.

\(\frac{7}{9}\times \frac{1-3}{4-h} = -1\)  

⇒ 9h – 36 = -14 

⇒ 9h = 22 

⇒ h = \(\frac{22}{9}\) 

Hence, the value of h is \(\frac{22}{9}\)   

Given \(\frac{2\pi}{3}\)

To Find: Slope of the line \(\frac{2\pi}{3}\)

Angle made with the positive x - axis is 

The Slope of the line is m 

Formula Used: m = tanθ 

So, The slope of Line is m = tan \(\frac{2\pi}{3}\)  

 ⇒  tan \(\Big(\frac{2\pi}{3}\Big)\)  =  tan \(\Big(\pi-\frac{\pi}{3}\Big)\)  

 ⇒  tan \(\Big(\frac{2\pi}{3}\Big)\)  =  tan \(\Big(-\frac{\pi}{3}\Big)\)  

 ⇒  tan \(\Big(\frac{2\pi}{3}\Big)\)  =  \(-\sqrt{3}\)  

Hence, The slope of the line is  \(-\sqrt{3}\) .

As slope is given m = - 3 and line is passing through point ( - 2, 3).Using slope - intercept form of equation of line, we will find intercept first

y = mx + c ……………….(1)

3 = - 3( - 2) + c

3 = 6 + c

c = 3 - 6

c = - 3

Putting the value of c in equation (1), we have

y = - 3x + ( - 3)

y = - 3x - 3

3x + y + 3 = 0

So, the required equation of line is 3x + y + 3 = 0.

Equation of a line in normal form is x cos a + y sin a = p

Given, p = 2, sin a = \(\frac{4}{5}\) ⇒ cos a = \(\sqrt{{1} − (\frac{4}{5})^2}\) = \(\frac{3}{5}\).

⇒ x.\(\frac{3}{5}\) + y.\(\frac{4}{5}\) = 2

⇒ 3x + 4y – 10 = 0, is the required equation of the line.

Equation of a line in intercept form is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 …(1)

Since the line passes through (2, 3) and (4, – 5).

∴ \(\frac{2}{a}\) − \(\frac{3}{b}\) = 1 ⇒ 2p − 3q = 1 …(2)

[let \(\frac{1}{a}\) = p, \(\frac{1}{b}\) = q]

And \(\frac{4}{a}\) − \(\frac{5}{b}\) = 1 ⇒ 4p − 5q = 1 …(3)

(2) and (3) ⇒ p = – 1, q = – 1.

∴ a = – 1, b = – 1.

A line which is passing through (1, 2)

To find: The equation of a straight line.

By using the formula,

The equation of line is [y – y₁ = m(x – x₁)]

Here, sin θ = 3/5

We know, sin θ = perpendicular/hypotenuse

= 3/5

So, according to Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

(5)² = (Base)² + (3)²

(Base) = √(25 – 9)

(Base)² = √16 

Base = 4

Hence, tan θ = perpendicular/base

= 3/4

The slope of the line, m = tan θ

= 3/4
The line passing through (x₁,y₁) = (1,2)

The required equation of line is y – y₁ = m(x – x₁)

Now, substitute the values, we get
y – 2 = (3/4) (x – 1)

4y – 8 = 3x – 3

3x – 4y + 5 = 0

∴ The equation of line is 3x – 4y + 5 = 0

Given:

A line which is passing through (2, 2√3), the angle is 75°.

By using the formula,

The equation of line is [y – y₁ = m(x – x₁)]

Here, angle, θ = 75°

The slope of the line, m = tan θ

m = tan 75°

= 3.73 = 2 + √3

The line passing through (x₁, y₁) = (2, 2√3)

The required equation of the line is y – y₁ = m(x – x₁)

Now, substitute the values, we get

y – 2√3 = 2 + √3 (x – 2)

y – 2√3 = (2 + √3)x – 7.46

(2 + √3)x – y – 4 = 0

∴ The equation of the line is (2 + √3)x – y – 4 = 0

The given equation of the line is x + 3y = 7

⇒ y = − \(\frac{1}{3}\)x + \(\frac{7}{3}\)

∴ Slope of the line, m₁ = – \(\frac{1}{3}\)

Let m₂ be the slope of the Perpendicular.

∴ m₁ m₂ = – 1

⇒ − \(\frac{1}{3}\) × m₂ = − 1

⇒ m₂ = 3.

∴ Equation of the perpendicular line with slope 3 and passing through (3, 8) is

y – 8 = 3 (x – 3)

⇒ 3x – y – 1 = 0

∴ The foot of the perpendicular is the point intersection of the lines x + 3y – 7 = 0 and 3x – y – 1 = 0

Solving these equations, we get

x = 1, y = 2, so (1, 2) is the foot of the perpendicular.

Proof Given a straight line, either it cuts the y-axis, or is parallel to or coincident with it. We know that the equation of a line which cuts the y-axis (i.e., it has y-intercept) can be put in the form y = mx + b; further, if the line is parallel to or coincident with the y-axis, its equation is of the form x = x₁, where x = 0 in the case of coincidence. Both of these equations are of the form given in the problem and hence the proof.

Let the co-ordinates of B are (p, q).

Then, slope of line AB, m₁ = \(\frac{q − 1}{p + 3}\).

And, slope of line 2x – 3y = 4 is \(\frac{2}{3}\) = mL.

Since the lines are perpendicular, so m₁ m₂ = – 1

⇒ \(\frac{q − 1}{p + 3}\) × \(\frac{2}{3}\) = −1

⇒ 2q – 2 = – 3p – 9

⇒ 3p + 2q + 7 = 0 …(1)

The midpoint of AB is \(\big(\frac{p − 3}{2} , \frac{q + 1}{2}\big)\), which lies on the line AB.

∴ 2\(\big(\frac{p − 3}{2}\big)\) − 3. \(\big(\frac{q + 1}{2}\big)\) = 4

⇒ 2p – 6 – 3q – 3 = 8

⇒ 2p – 3q – 17 = 0 …(2)

Solving (1) and (2), we get p = 1, q = – 5.

The distance between two parallel line 3x + 4y + 5 = 0 and 3x + 4y + 2 = 0 is 

\(\frac{|5-2|}{\sqrt{3^2+4^2}}\) = \(\frac{3}{\sqrt{25}}\) = \(\frac{3}{5}\)

The distance between two parallel line 3x + 4y + 2 = 0 and 3x + 4y - 5 = 0 is 

\(\frac{|2-(-5)|}{\sqrt{3^2+4^2}}\) = \(\frac{7}{\sqrt{25}}\) = \(\frac{7}{5}\)

Thus required ratio is = \(\frac{3}{\frac{5}{\frac{7}{5}}}\) = \(\frac{3}{7}\)

We have given Coordinates off two lines. 

Given: (5, 6) and (2, 3); (9, – 2) and 96, – 5) 

To Find: Check whether Given lines are perpendicular to each other or parallel to each other. 

Concept Used: If the slopes of this line are equal the lines are parallel to each other. Similarly, If the product of the slopes of this two line is – 1, then lines are perpendicular to each other. 

The formula used: Slope of a line, m = \(\frac{y_2-y_1}{x_2-x_1}\)

Now, The slope of the line whose Coordinates are (5, 6) and (2, 3)

⇒ m₁ = \(\frac{3-6}{2-5}\) 

 ⇒ m₁ =  \(\frac{-3}{-3}\) 

So, m₁ = 1

Now, The slope of the line whose Coordinates are (9, – 2) and (6, – 5) 

⇒ m₂ = \(\frac{-5-(-2)}{6-9}\) 

⇒ m₂ = \(\frac{-3}{-3}\)

So, m₂ = 1 

Here, m₁ = m₂ = 1 

Hence, The lines are parallel to each other.

We have given Coordinates off two line. 

Given: (9, 5) and (– 1, 1); through (3, – 5) and (8, – 3) 

To Find: Check whether Given lines are perpendicular to each other or parallel to each other. 

Concept Used: If the slopes of this line are equal the the lines are parallel to each other. Similarly, If the product of the slopes of this two line is – 1, then lines are perpendicular to each other. 

The formula used: Slope of a line, m = \(\frac{y_2-y_1}{x_2-x_1}\)

Now, The slope of the line whose Coordinates are (9, 5) and (– 1, 1)

⇒ m₁ = \(\frac{1-5}{-1-9}\) 

⇒ m₁ = \(\frac{-4}{-10}\)  

So, m₁ = \(\frac{2}{5}\)

Now, The slope of the line whose Coordinates are (3, – 5) and (8, – 3)

⇒ m₂ = \(\frac{-3-(-5)}{8-3}\) 

⇒ m₂ = \(\frac{2}{5}\)

So, m₂ = \(\frac{2}{5}\) 

Here, m₁ = m₂ = \(\frac{2}{5}\)

Hence, The lines are parallel to each other.

We have given Coordinates off two line. 

Given: (6, 3) and (1,1) and (– 2, 5) and (2, – 5) 

To Find: Check whether Given lines are perpendicular to each other or parallel to each other. 

Concept Used: If the slopes of this line are equal the the lines are parallel to each other. Similarly, If the product of the slopes of this two line is – 1, then lines are perpendicular to each other. 

The formula used: Slope of a line, m = \(\frac{y_2-y_1}{x_2-x_1}\)

Now, The slope of the line whose Coordinates are (6, 3) and (1, 1)

⇒ m₁ = \(\frac{1-3}{1-6}\) 

 ⇒ m₁ =  \(\frac{-2}{-5}\) 

So, m₁ = \(\frac{2}{5}\)

Now, The slope of the line whose Coordinates are (– 2, 5) and (2, – 5) 

⇒ m₂ = \(\frac{-5-5}{2+2}\) 

⇒ m₂ = \(\frac{-10}{4}\)

So, m₂ = \(\frac{-5}{4}\)

Here, m₁m₂ =  \(\frac{2}{5}\times-\frac{5}{2}\)

m₁m₂ = – 1 

Hence, The line is perpendicular to other. 

Given (– 3, 2) and (1, 4) 

To Find The slope of the line passing through the given points

Here, 

The formula used: Slope of line = \(\frac{y_2-y_1}{x_2-x_2}\)

So, The slope of the line, m = \(\frac{4-2}{1-(-3)}\) 

m = \(\frac{2}{4}\) = \(\frac{1}{2}\)

Hence, The slope of the line is  \(\frac{1}{2}\)

We have given Coordinates off two line. 

Given: (3, 15) and (16, 6) and (– 5, 3) and (8, 2) 

To Find: Check whether Given lines are perpendicular to each other or parallel to each other. 

Now, 

Concept Used: If the slopes of these line are equal the the lines are parallel to each other. Similarly, If the product of the slopes of these two line is – 1, then lines are perpendicular to each other. 

The formula used: Slope of a line, m = \(\frac{y_2-y_1}{x_2-x_1}\)

Now, The slope of the line whose Coordinates are (3, 15) and (16, 6)

⇒ m₁ = \(\frac{6-15}{16-3}\) 

 ⇒ m₁ =  \(\frac{-9}{13}\) 

So, m₁ = \(\frac{-9}{13}\)

⇒ m₂ = \(\frac{2-3}{8-(-5)}\) 

⇒ m₂ = \(\frac{-1}{13}\)

So, m₂ = \(\frac{-5}{2}\)

Here, m₁≠m₂ nor m₁m₂ = – 1 

Hence, The lines are neither perpendicular and nor parallel to each other.

Given: 

Points A (0, 1), B(2, 0) and C(-1, -2). 

Assuming:

 m₁, m₂ and m₃ be the slope of the sides AB, BC and CA, respectively. 

Concept Used: 

The slope of the line passing through the two points (x₁, y₁) and (x₂, y₂). 

The equation of the line passing through the two points (x₁, y₁) and (x₂, y₂). 

To find: 

The equation of sides of the triangle. 

Explanation: 

m₁ = \(\frac{0-1}{2-0}\) , m₂ = \(\frac{-2-0}{-1-2}\)  , m₃ = \(\frac{1+2}{1+0}\) 

m₁ =\(-\frac{1}{2}\)  , m₂ = \(-\frac{2}{3}\) and m₃ = 3 

So, the equation of the sides AB, BC and CA are 

Formula used: y – y₁= m (x – x₁) 

y - 1 = \(\frac{1}{2}\)(x - 0), y - 0 = -\(\frac{2}{3}\) (x - 2)and y + 2 = 3(x+1) 

⇒ x + 2y = 2, 2x – 3y =4 and 3x – y +1 = 0 

Hence, equation of sides are x + 2y = 2, 2x – 3y =4 and 3x – y +1 = 0

(i) Given, Line bisects the first quadrant 

To Find: Find the slope of the line. 

Here, If the line bisects in the first quadrant, then the angle must be between line and the positive direction of x - axis . 

Since, Angle = \(\frac{90}{2}\) = 45°

The formula used: The slope of the line, m = tan θ 

Similarly, The slope of the line for a given angle is m = tan 45°

m = 1 

Hence, The slope of the line is 1.

(ii) To Find: Find the slope of the line. 

Here, The line makes an angle of 30° with the positive direction of y - axis (Given) 

Since Angle between line and positive side of axis = 90° + 30°= 120° 

The formula used: The slope of the line, m = tan θ 

Similarly, The slope of the line for a given angle is m = tan 120°

m = – √3 

Hence, The slope of the line is – √3

Given \(\frac{\pi}{3}\)

To Find: Slope of the line 

Angle made with positive x - axis is \(\frac{\pi}{3}\)

The Slope of the line is m 

Formula Used: m = tan θ 

So, The slope of Line is m = tan \(\Big(\frac{\pi}{3}\Big)\) 

⇒ tan\(\Big(\frac{\pi}{3}\Big)\) = \(\sqrt{3}\)

Hence, The slope of the line is √3

(i) Given: Line bisects the first quadrant

We know that, if the line bisects in the first quadrant, then the angle must be between line and the positive direction of x – axis.

Since, angle = 90/2 = 45°

By using the formula,
The slope of the line, m = tan θ

The slope of the line for a given angle is m = tan 45°

So, m = 1

∴ The slope of the line is 1.

(ii) Given: The line makes an angle of 30° with the positive direction of y – axis.

We know that, angle between line and positive side of axis => 90° + 30° = 120°

By using the formula,
The slope of the line, m = tan θ

The slope of the line for a given angle is m = tan 120°

So, m = – √3

∴ The slope of the line is – √3.

Given: 

Points A (1, 4), B(2, -3) and C(-1, -2). 

Assuming: 

m₁, m₂, and m₃ be the slope of the sides AB, BC and CA, respectively. 

Concept Used:

The slope of the line passing through the two points ( x₁, y₁) and ( x₂, y₂). The equation of the line passing through the two points ( x₁, y₁) and ( x₂, y₂). 

To find: 

The equation of sides of the triangle. 

Explanation:

m₁ = \(\frac{-3-4}{2-1}\) ,  m₂ = \(\frac{-2+3}{-1-2}\) ,​​​​m₃ = \(\frac{4+2}{1+1}\)

m₁ = -7, m₂ =  \(\frac{1}{3}\) and m₃ = 3 

So, the equation of the sides AB, BC and CA are 

Formula used: y – y₁= m (x – x₁) 

y – 4 = -7 (x – 1), y + 3 = 

 \(\frac{1}{3}\) (x - 2) and y + 2 = 3(x+1) 

⇒ 7x + y =11, x+ 3y +7 =0 and 3x – y +1 = 0 

Hence, equation of sides are 7x + y =11, x+ 3y +7 =0 and 3x – y +1 = 0

Given equation can be written as

a(x + y – 1) + b(2x + 3y – 1) = 0

⇒ (x + y − 1) + λ(2x + 3y − 1) = 0, where λ = \(\frac{b}{a}\).

This is the form L₁ + λL₂ = 0. So, it represents that line passing through the intersection of

x + y – 1 = 0 and 2x + 3y – 1 = 0

∴ The straight lines pass through a fixed point.

Let the points be A(4, 1), B(6, 5) and C(2, – 1)

AP ⊥ BC is drawn

Slope of line BC, m = \(\frac{−1\, −\,5}{2\, −\, 6}\) = \(\frac{3}{2}\)

Since AP ⊥ BC

∴ Slope of AP, m′ = − \(\frac{1}{m}\) = − \(\frac{2}{3}\)

∴ Equation of line AP is

y − 1 = − \(\frac{2}{3}\)(x − 4)

⇒ 2x + 3y – 11 = 0 …(1)

Let P divide BC in the ratio m : n

∴ Co-ordinates of P are \(\big(\frac{2m + 6n}{m + n}, \frac{−m + 5n}{m +n}\big)\)

Since P lies on AP, so (1)

⇒ 2.\(\frac{2m + 6n}{m + n}\) + 3.\(\big(\frac{−m + 5n}{m +n}\big)\) − 11 = 0

⇒ 4m + 12n - 3m + 15n - 11m – 11n = 0

⇒ \(\frac{m}{n}\) = \(\frac{16}{10}\)

⇒ m : n = 8 : 5

To Prove: Given points are of Parallelogram. 

Explanation: Let us Assume that we have points, A (– 2, – 1), B(4, 0), C(3, 3) and D(– 3, 2), are joining the sides as AB, BC, CD, and AD. 

The formula used: The slope of the line, m = \(\frac{y_2-y_1}{x_2-x_1}\)

Now, The slope of Line AB, m_AB =\(\frac{0-(-1)}{4-(-2)}\) 

m_AB = \(\frac{1}{6}\)

The slope of BC, m_BC =   \(\frac{3-0}{3-4}\) 

m_BC =   \(\frac{3}{-1}\) 

Now, The slope of Line CD, m_CD =   \(\frac{2-3}{-3-3}\) 

m_CD =   \(\frac{1}{6}\) 

The slope of AD, 

m_AD =   \(\frac{2-(-1)}{-3-(-2)}\) 

m_AD =   \(\frac{3}{-1}\) 

Here, We can see that, m_AB = m_CD and m_BC = m_AD  

i.e, AB||CD and BC||AD 

We know, If opposite side of a quadrilateral are parallel that it is a parallelogram.

Hence, ABCD is a Parallelogram.

The given points (x, – 1), (2, 1) and (4, 5) are collinear. 

To Find: The value of x. 

Concept Used: It is given that points are collinear, SO the area of the triangle formed by the points must be zero. 

Formula used: The area of triangle = x₁(y₁ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) 

Explanation: Let be points of triangle A(x, – 1), B(2, 1) and C(4, 5) 

Now, The points are collinear than, Area of a triangle is zero. 

Here, Put the given values in formula and we get, 

x(1 – 5) + (2)(5 – (– 1)) + 4(– 1 – 1) = 0 

x – 5x + 12 – 8 = 0 

– 4x + 4 = 0 

4x = 4 

x = 1 

Hence, The value of x is 1.

Slope of line through (– 2, 6) and (4, 8). 

m₁ = \(\frac{8-6}{4-(-2)}=\frac{1}{3}\) 

and, slope of line through (8, 12) and (x, 24), 

m₂= \(\frac{24-12}{x-8}=\frac{12}{x-8}\)

Since the lines are perpendicular, So, m₁m₂ = – 1 ⇒ \(\frac{1}{3}.\frac{12}{x-8}=-1\) 

⇒ 4 = – x + 8 

⇒ x = 4

To Find: Find the value of x ?

The concept used: If two line is perpendicular then, the product of their slopes is – 1. 

Explanation: We have two lines having point A(– 2,6) and B(4,8) and other line having points C(8,12) and D(x,24).

The formula used: The slope of the line, m = \(\frac{y_2-y_1}{x_2-x_1}\)

Now, The slope of Line AB is, m_AB = \(\frac{4-(-2)}{8-6}\)

m_AB = \(\frac{6}{2}\)

and, The slope of Line CD is, m_CD = \(\frac{x-8}{24-12}\)

m_CD = \(\frac{x-8}{12}\)

We know the product of the slopes of perpendicular line is always – 1. 

Then, m_AB x m_CD = – 1

\(\frac{6}{2}\times\frac{x-8}{12}\) = -1

\(\frac{x-8}{12}\) = -1

x – 8 = – 4 

x = – 4 + 8 

x = 4 

Hence, The value of x is 4.

 Let the slope of line be m 

∴ Slope of the another line be 2m

tan\(\theta\)=\(|\frac{m_1-m_2}{1+m_1m_2}|\)

\(\frac{1}{3}\) = \(|\frac{2m-m}{1+2m^2}|\)

\(\frac{1}{3}\) = \(|\frac{m}{1+2m^2}|\)

2m² + 1 = 3m

2m² – 3m + 1 = 0 

(m – 1) (2m – 1) = 0 

If m – 1 = 0 ⇒ m = 1 

If 2m – 1 = 0 ⇒ m = \(\frac{1}{2}\) 

Hence, slope of lines (1 and 2) or ( \(\frac{1}{3}\) and 1)

The given lines are

x + y = 0 and y = 0

Here, A₁ = B₁ = 1, A₂ = 0, B₂ = 1.

Angle ‘θ’ between the lines is given by

θ = tan⁻¹\(\big|\frac{A_1B_2 − A_2B_1}{A_1A_2 + B_1B_2}\big|\)

= tan⁻¹\(\big|\frac{1 − 0}{0 + 1}\big|\)

= tan⁻¹ (1)

= 45°

Given equation is

3x + 4y = 5

Here, m = – \(\frac{A}{B}\) = − \(\frac{3}{4}\).

And the slope intercept form is

y = − \(\frac{A}{B}\)x − \(\frac{C}{B}\)

⇒ y = − \(\frac{3}{4}\)x − \(\frac{5}{4}\).

Given equation is 5x + 7y -35 = 0

We can rewrite it as 7y = 35 - 5x

⇒ 7y = - 5x + 35

⇒ y = \(-\frac{5}{7}x+5\)

This equation is in the slope-intercept form i.e. it is the form of

y = m x x + c, where m is the slope of the line and c is y-intercept of the line

Therefore, m = -5/7 and c = 5

Conclusion: Slope is -5/7 and y - intercept is 5

Given:

p = 2, sin α = 1/3

We know that cos α = √(1 – sin² α)

= √(1 – 1/9)

= 2√2/3

The equation of the line in normal form is given by

By using the formula,
x cos α + y sin α = p

Now, substitute the values, we get

x2√2/3 + y/3 = 2

2√2x + y = 6

∴ The equation of line in normal form is 2√2x + y = 6.

Given: p = 2, sinα = 1/3 

We know that, cos α = \(\sqrt{1-sin^2α}\)

⇒   \(\sqrt{1-\frac{1}{9}}\) =   \(\frac{2\sqrt{2}}{3}\)

Concept Used: 

The equation of a line in normal form. 

Explanation: 

So, the equation of the line in normal form is 

Formula Used: x cos α + y sin α = p 

⇒ \(\frac{2\sqrt{2}}{3}x+\frac{y}{3}=2\)

⇒ 2√2x + y = 6 

Hence, the equation of line in normal form is 2√2x + y = 6

Let the slopes of the two lines are m₁ = 2 − \(\sqrt{3}\) and m₂ = (2 + \(\sqrt{3}\)).

Then, angle ‘θ’ between the lines is given by

θ = tan⁻¹\(\frac{m_2 − m_1}{1 + m_1m_2}\)

= tan⁻¹\(\frac{(2 + \sqrt{3}) − (2 − \sqrt{3})}{1 + (2 − \sqrt{3})(2 + \sqrt{3})}\)

= tan⁻¹\(\frac{2\sqrt{3}}{1 + 4 − 3}\)

= tan⁻¹(\(\sqrt{3}\))

= tan^{– 1}(tan 60°)

= 60°

Assuming:

The perpendicular drawn from the origin make acute angle α with the positive x–axis. Then, we have, tanα =  \(\frac{5}{12}\) 

We know that, tan(180^∘ + α) = tanα 

So, there are two possible lines, AB and CD, on which the perpendicular drawn from the origin has a slope equal to 5/12 . 

Given: 

Now tan α =  \(\frac{5}{12}\)

⇒ sin α = \(\frac{5}{13}\) and  cos α = \(\frac{12}{13}\)

Explanation: 

So, the equations of the lines in normal form are 

Formula Used: x cos α + y sin α = p 

⇒ x cos α + y sin α = p and x cos(180° + α) + ysin(180° + α) = p 

⇒ x cos α + y sin α = 2 and –x cos α – ysin α = 2 cos (180° + θ) = – cos θ , sin (180° + θ) = – sin θ

⇒ \(\frac{12x}{13}+\frac{5y}{13}\)

and 12x + 5y = – 26 

Hence, the equation of line in normal form is   \(\frac{12x}{13}+\frac{5y}{13}\) = 26 and 12x + 5y = – 26

Slope of the line passing through (– 3, 8) and (2, – 2),

m₁ = \(\frac{8 + 2}{− 3 − 2}\) = − 2.

If m₂ be the slope of the perpendicular line, then m₁ m₂ = – 1

⇒ m₂ = \(\frac{1}{2}\).