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

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

Concept Used:

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

Given: 

Here a + b = 7, b = 7 – a 

Explanation: 

The line is passing through (-3, 8).

 \(\frac{-3}{a}+\frac{8}{b}=1\) 

Substituting b =7 – a, we get

 \(\frac{x}{a}+\frac{y}{7-b}=1\)  

⇒ -3(7 – a) + = 7a – a²

⇒ a² + 4a – 21 = 0 

⇒ (a – 3)(a + 7)= 0

⇒ a = 3 ( since, a can only be positive) 

Substituting a = 3 in equation (i) we get, b = 7 – 3 = 4 

Hence, the equation of the line is  \(\frac{x}{3}+\frac{y}{4}=1\) 

Correct Answer – x=0, x`-sqrt(3)` y =0
The given line is `x+sqrt(3)y+3sqrt(3) = 0.`
`”or ” y=(-(1)/(sqrt(3)))x-3`
Therefore, the slope of (1) is `-1//sqrt(3).`
Let the slope of the required line be m.
Also, the angle between these line is `60^(@)`. Therefore,
`” tan ” 60^(@) = |(m-(-1//sqrt(3)))/(1+m(1//sqrt(3)))|`
`”or ” sqrt(3) = |(sqrt(3)m+1)/(sqrt(3)-m)|`
`”If “(sqrt(3)m+1)/(sqrt(3)-m) =sqrt(3)`
`”or “m=(1)/(sqrt(3))`
Using y=mx +c, the equation of the required line is
`y=(1)/(sqrt(3))x +0`
`”i.e.,” x-sqrt(3) y = 0` (as the line passes through the origin, c=0)
`(sqrt(3)m-1)/(sqrt(3)-m) = -sqrt(3)`
`”or ” sqrt(3)m+1 =-3+sqrt(3)m`
Therefore, finite m does not exist.
Therefore, the slope of the required line is infinity. Thus, the required line is a vertical line. This line passes through the origin.
Therefore, the equation of the required line is x=0.

Given: 

Lines 3x – y = 5 and x + 3y = 1 

To find: 

The equation of the straight line which passes through the point of intersection of the lines 3x – y = 5 and x + 3y = 1 and makes equal and positive intercepts on the axes.

Explanation: 

The equation of the straight line passing through the point of intersection of 

3x − y = 5 and x + 3y = 1 is 3x − y − 5 + λ(x + 3y − 1) = 0 

⇒ (3 + λ)x + (− 1 + 3λ)y − 5 − λ = 0 … (1) 

⇒ y = \(-\Big(\frac{3+λ}{-1+λ}\Big)x\) + \(\Big(\frac{5+λ}{-1+λ}\Big)\)

The slope of the line that makes equal and positive intercepts on the axis is − 1. From equation (1), we have:

\(-\Big(\frac{3+λ}{-1+3λ}\Big)\) = -1

⇒ λ = 2 

Substituting the value of λ in (1), we get the equation of the required line. 

⇒ 3 + 2x + -1 + 6y – 5 – 2 = 0 

⇒ 5x + 5y – 7 = 0 

Hence, equation of required line is 5x + 5y – 7 = 0

Correct Answer – B

Given: A line passing through (1, -2) 

Assuming: 

The equation of the line cutting equal intercepts at coordinates of length ‘ a ‘ is

Explanation: 

Formula used:

 \(\frac{x}{a}+\frac{y}{b}=1\) 

  \(\frac{x}{a}+\frac{y}{a}=1\) 

⇒ x + y = a 

The line x + y = a passes through (1, -2 ) So the point satisfy the equation 1 -2 = a 

⇒ a = -1 

Hence the equation of the line is x + y = -1