Find the equation of the straight line which passes through the point

1.	Find the equation of the straight line which passes through the point (1, 2) and makes such an angle with the positive direction of x – axis whose sine is 3/5.
Answer» 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

Find the equation of the straight line which passes through the point (1, 2) and makes such an angle with the positive direction of x – axis whose sine is 3/5.

Answer»

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