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The equation of a line intercept form is 

\(\frac{x}{a}+\frac{y}{b}=1\) 

Given, b = 2 

∴ (1) ⇒ \(\frac{x}{a}+\frac{y}{2a}=1\) 

⇒ 2x + y = 2α 

Since the line passes though the point 

(1, 2), 2. 1 + 2 = 2a ⇒ a = 2 

∴ Equation of the line is 

2x + y – 4 = 0.

Given, A line which passes through the point ( – 3,5) and perpendicular to the line joining (2,5) and ( – 3,6) 

To Find: Find the equation 

Formula Used: The equation of line is (y – y₁) = m(x – x₁) 

Explanation: Here, The line passes through the point ( – 3,5 ), Given 

So, The coordinate (x₁,y₁) = ( – 3,5) 

Now, The line is perpendicular to the line joining (2,5) and ( – 3,6), 

We know, The slope of the line with two points is, m =  \(\frac{y_2 -y_1}{x_2-x_1}\)  

So, the slope of line joining (2, 5 ) and ( – 3,6) is = \(\frac{6-5}{-3-2}\) 

m = \(-\frac{1}{5}\) 

Therefore, The slope of the required line is, m =  \(\frac{1}{Slope \ of \ joining\ line(2,5) and (-3,6)}\) 

So, m =  \(-\frac{1}{\frac{1}{5}}\) 

m = 5 

Now, The equation of straight line is (y – y₁) = m(x – x₁) 

y – 5 = 5 (x – ( – 3) 

y – 5 = 5x + 15 

5x – y + 20 = 0 

Hence, The equation of line is 5x – y + 20 = 0

The mid-point of A(2, 3) and B(6, – 5), 

= \((\frac{2+6}{2},\frac{3-5}{2})=(4, -1)\) 

And slope of AB, m =\(\frac{-5-3}{6-2}=-2\)

∴ Slope of the Perpendicular bisector of 

   AB, m’ = \(\frac{-1}{m}=\frac{1}{2}\). 

Since the bisector passes through (4, – 1), so the equation of the Perpendicular bisector is 

Y – (– 1) = \(\frac{1}{2}\)( – 4) 

⇒ y + 1 = \(\frac{1}{2}\)(x − 4) 

⇒ x – 4 = 2y + 2 

⇒ x – 2y – 6 = 0, is the required equation.

Let P(x, y) be any point on the required locus. 

Given, A(2,4), B(5, 8) and PA² -PB² = 13 

∴ [(x -2)² + (y – 4)² ] – [(x -5)² + (y- 8)²] = 13 

∴ (x² – 4x + 4 + y² – 8y + 16) – (x² – 10x + 25 + y² – 16y + 64) =13 

∴ x² – 4x+ y² – 8y + 20 – x² + 10x – y² + 16y – 89 = 13 

∴ 6x + 8y- 69 = 13 

∴ 6x + 8y – 82 = 0 

∴ 3x + 4y – 41 = 0 

∴ The required equation of locus is 3x + 4y-41 = 0.

Let P(x, y) be any point on the required locus. P is equidistant from A(- 5, 2) and B(4, 1). 

∴ PA = PB 

∴ PA² = PB² 

∴ (x + 5)² + (y – 2)² = (x – 4)² + (y – 1)² 

∴ x² + 10x + 25 + y² — 4y + 4 = x² – 8x + 16 + y² – 2y + 1 

∴ 10x – 4y + 29 = -8x – 2y + 17 

∴ 18x – 2y + 12 = 0 

∴ 9x – y + 6 = 0 

The required equation of locus is 9x -y + 6 = 0.

Let P(x, y) be any point on the required locus. 

Given, A(2, 0), B(0, 3) and AP = 2BP

∴ AP² = 4BP² 

∴ (x – 2)² + (y – 0)² = 4[(x – 0)² + (y – 3)²] 

∴ x² – 4x + 4 + y² = 4(x² + y² – 6y + 9) x² – 4x + 4 + y² = 4x² + 4y² – 24y + 36 

∴ 3x² + 3 y² + 4x – 24y + 32 = 0 

∴ The required equation of locus is 3x² + 3y² + 4x – 24y + 32 = 0. 

Concept Used:

The equation of the line with intercepts a and b is  \(\frac{x}{a}+\frac{y}{b}=1\) 

To find: 

The equation of the line which cutoff intercepts on the axes. 

Given: 

Here a = b and ab = 25 

Explanation: ∴ a² = 25 

⇒ a = 5 since we are to take only positive value of intercepts

Hence, the equation of the required line is

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

 \(\frac{x}{5}+\frac{y}{5}=1\) 

⇒ x + y = 5 

Hence, the equation of line is x + y = 5

Given, Inclination of line = 0 = 135° 

∴ Slope of the line (m) = tan 0 = tan 135° 

= tan (90° + 45°) 

= – cot 45° = – 1 x-intercept of the required line is 7.

∴ The line passes through (7, 0). Equation of the line in slope point form is y₁ – y = m(x – x₁)

∴ The equation of the required line is y — 0 = – 1 (x – 7) 

∴ y = -x + 7 

∴ x + y – 7 = 0

Let m₁ and m₂ be the slope of the lines 2x − y + 3 = 0 and x + 2y + 3 = 0, respectively. 

Let θ be the angle between them. 

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

∵ m₁m₂ = − 1 

Therefore, the angle between the given lines is 90°.

Given: 

The equations of the lines are 

3x + 4y − 7 = 0 … (1) 

4x − 3y + 5 = 0 … (2) 

To find: 

Angles between two lines. 

Explanation:

Let m₁ and m₂ be the slopes of these lines. 

m₁ = \(-\frac{3}{4}\) ,m₂ = \(\frac{4}{3}\)

∵m₁m₂  = \(-\frac{3}{4}\times\frac{4}{3}=-1\)

Hence, the given lines are perpendicular. Therefore, the angle between them is 90°.

To prove:

The points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram

Let us assume the points, A (2, − 1), B (0, 2), C (2, 3) and D (4, 0) be the vertices.

Now, let us find the slopes

Slope of AB = [(2 + 1) / (0 - 2)]

= -3/2

Slope of BC = [(3 - 2) / (2 - 0)]

= 1/2

Slope of CD = [(0 - 3) / (4 - 2)]

= -3/2

Slope of DA = [(-1 - 0) / (2 - 4)]

= 1/2

Thus, AB is parallel to CD and BC is parallel to DA.

Hence proved, the given points are the vertices of a parallelogram.

Now, let us find the angle between the diagonals AC and BD.

Let m₁ and m₂ be the slopes of AC and BD, respectively.

m₁ = [(3 + 1) / (2 - 2)]

= ∞

m₂ = [(0 - 2) / (4 - 0)]

= -1/2

Thus, the diagonal AC is parallel to the y-axis.

∠ODB = tan^-1 (1/2)

In triangle MND,

∠DMN = π/2 – tan^-1 (1/2)

∴ The angle between the diagonals is π/2 – tan^-1 (1/2).

Equation of a line parallel to X-axis is y = k. 

Since the line is at a distance of 5 units above the X-axis, k = 5 

∴ The equation of the required line is y = 5.

Given:

A line which is passing through (–2, 3), the angle is 45°.

By using the formula,

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

Here, angle, θ = 45°

The slope of the line, m = tan θ

m = tan 45°

= 1

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

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

Now, substitute the values, we get

y – 3 = 1(x – (– 2))

y – 3 = x + 2

x – y + 5 = 0

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

The required line passes through the points P(2, 1) and Q(2,-1). 

Since both the given points have same x co-ordinates 

i.e. 2, the given points lie on the line x = 2. 

∴ The equation of the required line is x = 2.

Equation of a line parallel to the Y-axis is x = h. 

Since the line is at a distance of 5 units to the left of the Y-axis, h = -5 

∴ The equation of the required line is x = -5.

A line which is passing through ( – 2,3), the angle is 45° . 

To Find: The equation of a straight line. 

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

Explanation: Here, angle, θ = 45° 

SO, The slope of the line, m = tan θ 

m = tan 45° 

m = 1 

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

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

y – 3 = 1(x – ( – 2)) 

y – 3 = x + 2 

x – y + 5 = 0 

Hence, The equation of line is x – y + 5 = 0

(i) Equation of line parallel to x - axis is given by y = constant, as the y - coordinate of every point on the line parallel to x - axis is 4,i.e. constant. Now the point lies above x - axis means in positive direction of y - axis,

So, the equation of line is given as y = 4.

(ii) Equation of line parallel to x - axis is given by y = constant, as the y - coordinate of every point on the line parallel to x - axis is - 5 i.e. constant. Now the point lies below x - axis means in negative direction of y - axis,

So, the equation of line is given as y = - 5.

(i) Equation of line parallel to y - axis is given by x = constant, as the x - coordinate of every point on the line parallel to y - axis is 6 i.e. constant. Now the point lies to the right of y - axis means in the positive direction of x - axis,

So, required equation of line is x = 6.

(ii) Equation of line parallel to y - axis is given by x = constant, as the x - coordinate of every point on the line parallel to y - axis is - 3. Now point lies to the left of y - axis means in the negative direction of x - axis,

So, required equation of line is given as x = - 3.

The slope of the equation can be calculated using

m = \(\frac{y_2-y_1}{x_2-x_1}\) ⇒ b-b/-a-a = 0

m = 0 (Horizontal line)

Now using two point form of the equation of a line

y - y₁ = \(\frac{y_2-y_1}{x_2-x_1}\)(x - x₁)

y - b = 0(x - a)

y = b

So, required equation of line is y = b.

To Prove: The given line is parallel to another line. 

Proof: Let Assume the coordinate A(2, – 3) and B(– 5, 1), C(7, – 1) and D(0,3). 

The concept used: Slopes of the parallel lines are equal. 

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

Now, The slope of AB = \(\frac{1-(-3)}{-5-2}\)

The Slope of AB = \(\frac{4}{-7}\)

Now, The slope of CD = \(\frac{3-(-1)}{0-7}\)

The Slope of AB = \(\frac{4}{-7}\)

So, The slope of AB = The slope of CD 

Hence, The given Lines are parallel to each other.

To Prove: The Given line is perpendicular to each other. 

Proof: Let Assume the coordinate A(2, – 5) and B(– 2, 5) joining the line AB, C(6,3) and D(1,1) joining the line CD. 

The concept used: The product of the slopes of lines always – 1. 

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

Now, The slope of AB = \(\frac{5-(-5)}{-2-2}\)  

The Slope of AB = \(\frac{10}{-4}\)

Now, The slope of CD = \(\frac{1-3}{1-6}\)

The Slope of AB = \(\frac{2}{5}\) 

So, AB × CD = \(\frac{10}{-4}\times\frac{2}{5}\)

AB × CD = – 1 

Hence, The given Lines are perpendicular to each other.

Equation of a line parallel to X-axis is y = k. 

Since the line is at a distance of 3 units below X-axis, k = -3 

∴ The equation of the required line is y = -3.

Equation of a line parallel to Y-axis is x = h. 

Since the line is at a distance of 2 units to the left of Y-axis, h = -2 

∴ The equation of the required line is x = -2

Equation of a line parallel to X-axis with y-intercept ‘k’ is y = k. 

Here, y-intercept = 5 

∴ The equation of the required line is y = 5.

If a line passing through (x₁, y₁) & (x₂, y₂) then slope of the line is given by slope = (y₂-y₁/x₂-x₁)

(i) Given points are (0,0) and (4,-2)

slope = (-2-0/4-0) = -1/2

(ii) Given points are (0, -3) and (2, 1)

slope = (1-(-3)/2-0) = 2

(iii) Given points are (2, 5) and (-4, -4)

slope = (-4-5/-4-2) = 3/2 = 1.5

(iv) Given points are (-2, 3) and (4, -6)

slope = (-6-3/4+2) = - 3/2 = -1.5

Equation of line parallel to x - axis is given by y = constant, as x - coordinate of every point on the line parallel to y - axis is - 3 i.e. constant.

So, the required equation of line is y = - 3

Given, A straight line passing through the point (0,0) and slope is m. 

To Find: The equation of the straight line. 

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

Explanation: Here, The line is passing through (0,0) 

The slope of line, m = m (Given) 

Coordinates of line are (x₁,y₁) = (0,0) 

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

By putting the values, we get 

y – 0 = m(x – 0) 

y = mx

Hence, The equation of line is y = mx.

The required line passes through the points A(2, 0) and B(3,4). 

Equation of the line in two point form is \(\frac {y-y_1}{y_2-y_1}= \frac {x-x_1}{x_2-x_1}\)

Here, (x y ) = (2,0) and (x ,y ) = (3,4) 

∴ The equation of the required line is

∴ \(\frac {y-0}{4-0}= \frac {x-2}{3-2}\)

∴ y/4 = x-2/1

∴ y = 4(x – 2) 

∴ y = 4x – 8 

∴ 4x – y – 8 = 0

Equation of a line parallel to Y-axis is of the form x = h. 

Since the line passes through A(2, – 3), h = 2 

∴ The equation of the required line is x = 2.

Equation of line parallel to x - axis (horizontal) is y = constant, as y - coordinate of every point on the line parallel to x - axis is - 2 i.e. constant. Therefore equation of the line parallel to x - axis and passing through (4, - 2) is y = - 2.

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

To Find: The equation of a straight line. 

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

Explanation: Here, angle, θ = 75° 

So, The slope of the line, m = tan θ 

m = tan 75°

m = 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) 

y – 2√3 = 2 + √3 (x – 2) 

y – 2√3 = (2 + √3)x – 7.46 

(2 + √3)x – y – 4 = 0 

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

Let the point on the y-axis be P(0, y)

Given: P is equidistant from A(- 4, 3) and B(5, 2).

i.e., PA = PB

⇒ \(\sqrt{(-4-0)^2+(3-y)^2}\) = \(\sqrt{(5-0)^2+(2-y)^2}\)

Squaring both sides, we get

⇒ (- 4 – 0)² + (3 – y)² = (5 – 0)² + (2 – y)²

⇒ 16 + 9 – 6y + y² = 25 + 4 – 4y + y²

⇒ 25 – 6y = 29 – 4y

⇒ 2y = - 4

⇒ y = - 2

Therefore, the required point on the y-axis is (0, - 2).

A line which is passing through (1,2) 

To Find: The equation of a straight line.

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

Explanation: Here, sin θ = \(\frac{3}{5}\) 

We know, sin θ = \(\frac{perpendicular}{Hypotenues}\) =   \(\frac{3}{5}\)  

According to Pythagoras theorem, 

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

(5)² = (Base)² + (3)² 

(Base) = \(\sqrt{25-9}\) 

(Base)² = \(\sqrt{16}\) 

Base = 4

Hence, tan θ = \(\frac{Perpendicular}{Base}\) = \(\frac{3}{4}\) 

SO, The slope of the line, m = tan θ

m = \(\frac{3}{4}\) 

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

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

y – 2 = \(\frac{3}{4}\)(x – 1) 

4y – 8 = 3x – 3 

3x – 4y + 5 = 0 

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

Answer:

Equation of the line = 3x - 4y + 5 = 0

Step-by-step explanation:

Given:

• The line passes through the point (1,2)
• It makes an angle with the positive direction of x axis whose sin is 3/5

To Find:

• The equation of the straight line

Solution:

Here we have to first find the slope of the line

We know that,

m = tan θ

where m is the slope.

We know that

sin θ = opposite/hypotenuse = 3/5

Hence by Pythagoras theorem finding the adjacent side,

Adjacent side = √(25 - 9)

Adjacent side = √16 = 4 units

Also,

tan θ = opposite/adjacent = 3/4

Hence slope of the line is 3/4.

Now point slope form of a line is given by,

y - y₀ = m (x - x₀)

Here (1,2) = (x₀, y₀)

Substitute the data,

y - 2 = 3/4 (x - 1)

Multiply whole equation by 4

4y - 8 = 3x - 3

3x - 4y + 5 = 0

Hence the equation of the line is 3x - 4y + 5 = 0

Equation of line parallel to y - axis (vertical) is given by x = constant, as x - coordinate is constant for every point lying on line i.e. 6.

So, the required equation of line is given as x = 6.

As slope is given m = 4 and passing through (5, - 7).using slope - intercept form of equation of line, we will find value of intercept first

y = mx + c ………..(1)

- 7 = 4(5) + c

- 7 = 20 + c

c = - 7 - 20

c = - 27

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

y = 4x + ( - 27)

y = 4x - 27

4x - y - 27 = 0

So, the required equation of line is 4x - y - 27 = 0.

Given: A line which is parallel to x–axis and passing through (3, –5)

By using the formula,

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

We know that the parallel lines have equal slopes

And, the slope of x–axis is always 0

Then

The slope of line, m = 0

Coordinates of line are (x₁, y₁) = (3, –5)

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

Now, substitute the values, we get

y – (– 5) = 0(x – 3)

y + 5 = 0

∴ The equation of line is y + 5 = 0

Given: A line which is perpendicular to x–axis and having intercept –2

By using the formula,

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

We know that, the line is perpendicular to the x–axis, then x is 0 and y is –1.

The slope of line is, m = y/x 

= -1/0

It is given that x–intercept is –2, so, y is 0.

Coordinates of line are (x₁, y₁) = (–2, 0)

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

Now, substitute the values, we get

y – 0 = (-1/0) (x – (– 2))

x + 2 = 0

∴ The equation of line is x + 2 = 0

Given: A line which is parallel to x–axis and having intercept –2 on y – axis

By using the formula,

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

The parallel lines have equal slopes,

And, the slope of x–axis is always 0

Then

The slope of line, m = 0

It is given that intercept is –2, on y – axis then

Coordinates of line are (x₁, y₁) = (0, – 2)

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

Now, substitute the values, we get

y – (– 2) = 0 (x – 0)

y + 2 = 0

∴ The equation of line is y + 2 = 0

Answer is (B)

Slope of the given line +1=0 is-1.

So, slope of line perpendicular to above line is 1.

Line passes through the point (1,2).

Therefore, equation of the required linens:

y-2 = 1(x- 1) => y-x-1=0.

Answer is (B)

Slope of the given line y = x is 1.

Thus, slope of line perpendicular to y = x is -1.

Line passes through the point (3, 2).

So, equation of the required line is:y-2=-l (x – 3) => x + y = 5

Given: p = 8, α = 300° 

Concept Used: 

Equation of line in normal form. 

Explanation: 

So, the equation of the line in normal form is 

Formula Used: 

x cos α + y sin α = p 

x cos 300° + y sin 300° = 8 

⇒ x cos (360° – 60°) + y sin (360° – 60°) = 8 

We know, cos (360° – θ) = cos θ, sin (360° – θ) = – sin θ 

⇒ x cos60° – y sin60° = 8 

⇒ \(\frac{x}{2}-\frac{\sqrt{3}y}{2}\)

⇒ x – √3y = 16 

Hence, the equation of line in normal form is x – √3y = 16

Given: a, b, c are in AP

b = (a + c)/2

⇒ 2b = a + c

⇒ a – 2b + c = 0 …(i)

Now, comparing the eq. (i) with the given equation ax + by + c = 0, we get

x = 1, y = -2

So, the line will pass through (1, -2)

Ans. If a, b, c are in A.P., then the straight lines ax + by + c = 0 will always pass through (1, -2).

Given: 

The given lines are 

3x − 4y + a = 0 … (1) 

3x − 4y + 3a = 0 … (2) 

4x − 3y − a = 0 … (3) 

4x − 3y − 2a = 0 … (4) 

To prove: 

The area of the parallelogram formed by the lines 3x – 4y + a = 0, 3x – 4y + 3a = 0, 4x – 3y – a = 0 and 4x – 3y – 2a = 0 is \(\frac{2a^2}{7}\) sq. units.

Explanation: 

From Above solution, We know that 

Area of the parallelogram  = \(\Big|\frac{(c_1-d_1)(c_2-d_2)}{\sqrt{a_1b_2-a_2-b_1}}\Big|\) 

⇒ Area of the parallelogram =  \(\Big|\frac{(a-3a)(2a-a)}{\sqrt{-9+16}}\Big|\)  = \(\frac{2a^2}{7}\) square units 

Hence proved.

The equations of the given lines are
3x + y – 2 = 0 ………………………. (1)
px + 2y – 3 = 0 …………………….. (2)
2x – y – 3 = 0 …………………….… (3)
On solving equations (1) and (3), we obtain
x = 1 and y = –1
Since these three lines may intersect at one point, the point of intersection of lines (1) and (3) will also satisfy line (2).
p (1) + 2 (–1) – 3 = 0 p – 2 – 3 = 0 p = 5
Thus, the required value of p is 5.

The equations of the given lines are
3x + y – 2 = 0 ………………………. (1)
px + 2y – 3 = 0 …………………….. (2)
2x – y – 3 = 0 …………………….… (3)
On solving equations (1) and (3), we obtain
x = 1 and y = –1
Since these three lines may intersect at one point, the point of intersection of lines (1) and (3) will also satisfy line (2).
p (1) + 2 (–1) – 3 = 0 p – 2 – 3 = 0 p = 5
Thus, the required value of p is 5

Given: 2x – 3y + 1 = 0, 

x + y = 3, 

2x – 3y = 2 x + y = 4 are given equation 

To prove: 

The lines 2x – 3y + 1 = 0, x + y = 3, 2x – 3y = 2 and x + y = 4 form a parallelogram. 

Explanation: 

The given lines can be written as

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

y = - x + 3 .....(2)

y = \(\frac{2}{3}\) x - \(\frac{2}{3}\) ...(3)

y = - x + 4  ... (4)

The slope of lines (1) and (3) is \(\frac{2}{3}\) and that of lines (2) and (4) is − 1. 

Thus, lines (1) and (3), and (2) and (4) are two pair of parallel lines. 

If both pair of opposite sides are parallel then, we can say that it is a parallelogram. 

Hence proved, the given lines form a parallelogram.

Given line is x-7y + 5 = 0

Slope = m = - a/b = -(1/-7) = 1/7

Slope of the line perpendicular to the given line = – 7 

(∵ m₁ x m₂ =-1) 

Since x-intercept is 3 then (3, 0) is the point on the required line. 

∴ Required equation the line passing through (3, 0) having slope – 7 is, y – 0 = -7(x - 3) 

i.e., 7x ± y – 21 = 0 

Given: equation parallel to 3x – 4y + 6 = 0 and passing through the middle point of the join of points (2, 3) and (4, -1). 

To find: 

The equation to the straight line parallel to 3x – 4y + 6 = 0 and passing through the middle point of the join of points (2, 3) and (4, -1). 

Explanation: 

Let the Given points be A (2, 3) and B (4, − 1). Let M be the midpoint of AB. 

∴ Coordinates of M = \(\Big(\frac{2+4}{2},\frac{3-1}{2}\Big)\) = (3,1)

The equation of the line parallel to 3x − 4y + 6 = 0 is 3x – 4y + λ = 0 

This line passes through M (3, 1). 

∴ 9 – 4 + λ = 0 

⇒ λ = -5 

Substituting the value of λ in 3x – 4y + λ = 0, we get 3x – 4y – 5 = 0 

Hence, the equation of the required line is 3x – 4y – 5 = 0.

Given: equation is parallel to 3x − 4y + 5 = 0 and pass through (2, 3) 

To find: 

Equation of required line.

Explanation: 

The equation of the line parallel to 3x − 4y + 5 = 0 is

 3x – 4y + λ = 0, 

Where, λ is a constant. 

It passes through (2, 3). 

∴6 – 12 + λ = 0 

⇒ λ = 6 

Hence, the required line is 3x − 4y + 6 = 0.