Question

fine the equation of the line intersecting the Y-axis at a distance of 2units above the origin and making an angle 30° with positive direction of The X-axis​

Answer 1

EXPLANATION.

Line intersecting the y - axis at a distance of 2 units above the origin.

Making an angle 30° with positive direction of the x - axis.

As we know that,

If line intersects y - axis at distance 2.

It means Their Co-ordinates = (0,2).

Slope = tanθ

slope = tan(30°).

slope = 1/√3 = m.

As we know that,

Formula of :

Equation of the line.

⇒ (y - y₁) = m(x - x₁).

Put the values in the equation, we get.

⇒ (y - 2) = (1/√3)(x - 0).

⇒ (√3)(y - 2) = 1(x).

⇒ √3y - 2√3 = x.

⇒ x - √3y + 2√3 = 0.

                                                                                                                       

MORE INFORMATION.

Equation of straight line parallel to axis.

(1) Equation of x - axis ⇒ y = 0.

(2) Equation of a line parallel to x - axis at a distance of b ⇒ y = b.

(3) Equation y - axis ⇒ x = 0.

(4) Equation of a line parallel to y - axis and at a distance of a ⇒ x = a.

Answer 2

Step-by-step explanation:

Line AB intersects the y -axis 2 units above origin.

As we know that at y -axis, abscissa (xcoordinate) will be 0 always.

$\text{\( \therefore \) Line AB cuts y-axis at P \( (0,2) \)}$

Also line AB makes an angle of 30° with the x axis.

$\text{\( \therefore \) \: \: \: Slope of the line \( =\tan 30^{\circ}=\dfrac{1}{\sqrt{3}} \)}$

As we know that equation of a line passing through $\tt\left(x_{0}, y_{0}\right)$ and having slope m is-

$\tt \[ \left(y-y_{0}\right)=m\left(x-x_{0}\right) \]$

Therefore equation of the line AB is-

$\color{indigo}\[ \begin{array}{l} \tt (y-2)=\dfrac{1}{\sqrt{3}}(x-0) \\ \\ \tt\Rightarrow x-\sqrt{3} y+2=0 \end{array} \]$

Hence the required answer is $\tt x-\sqrt{3} y+2=0$