Answer

Verified

449.1k+ views

Hint: Solve by finding the slope of the line. Then, use the formula for the equation of line passing through the point and having a given slope to find its equation. Solve by finding the slope of the line. Then, use the formula for the equation of line passing through the point and having a given slope to find its equation. The equation of a point that passes through a point (a, b) and has a slope m is given as $y-b = m(x-a)$.

Complete step-by-step answer:

We know that the slope of a line is the tangent of the angle that it makes with the positive direction of x-axis.

If m is the slope of the line and \[\theta \] is the angle made with the positive direction of x-axis, then we have the relation as follows:

\[m = \tan \theta ............(1)\]

It is given that the line makes an angle 60° with the positive direction of the y-axis. We know that the angle between positive direction of y-axis and positive direction of x-axis is 90°, then the angle made by the line with the positive direction of the x-axis is 90° minus the angle made with the positive direction of y-axis.

\[\theta = 90^\circ - 60^\circ \]

\[\theta = 30^\circ .............(2)\]

Now using equation (2) in equation (1), we get:

$\Rightarrow$ \[m = \tan 30^\circ \]

$\Rightarrow$ \[m = \dfrac{1}{{\sqrt 3 }}...........(3)\]

It is given that the line passes through the point (3, -2) and we found the value of slope to be \[\dfrac{1}{{\sqrt 3 }}\].

The equation of a point that passes through a point (a, b) and has a slope m is given as follows:

\[y - b = m(x - a)\]

Substituting the values in the above equation, we have:

$\Rightarrow$ \[y - ( - 2) = \dfrac{1}{{\sqrt 3 }}(x - 3)\]

$\Rightarrow$ \[y + 2 = \dfrac{1}{{\sqrt 3 }}(x - 3)\]

Taking \[\sqrt 3 \] to the other side, we have:

$\Rightarrow$ \[\sqrt 3 (y + 2) = x - 3\]

Simplifying the equation, we obtain:

$\Rightarrow$ \[x - \sqrt 3 y - 2\sqrt 3 - 3 = 0\]

$\Rightarrow$ \[x = \sqrt 3 y + 2\sqrt 3 + 3\]

Hence, the required equation is \[x = \sqrt 3 y + 2\sqrt 3 + 3\].

Note: Do not use the given angle to calculate the tangent to find the slope directly, first, you need to find the angle made by the line with the positive direction of x-axis, then proceed with the solution.

Complete step-by-step answer:

We know that the slope of a line is the tangent of the angle that it makes with the positive direction of x-axis.

If m is the slope of the line and \[\theta \] is the angle made with the positive direction of x-axis, then we have the relation as follows:

\[m = \tan \theta ............(1)\]

It is given that the line makes an angle 60° with the positive direction of the y-axis. We know that the angle between positive direction of y-axis and positive direction of x-axis is 90°, then the angle made by the line with the positive direction of the x-axis is 90° minus the angle made with the positive direction of y-axis.

\[\theta = 90^\circ - 60^\circ \]

\[\theta = 30^\circ .............(2)\]

Now using equation (2) in equation (1), we get:

$\Rightarrow$ \[m = \tan 30^\circ \]

$\Rightarrow$ \[m = \dfrac{1}{{\sqrt 3 }}...........(3)\]

It is given that the line passes through the point (3, -2) and we found the value of slope to be \[\dfrac{1}{{\sqrt 3 }}\].

The equation of a point that passes through a point (a, b) and has a slope m is given as follows:

\[y - b = m(x - a)\]

Substituting the values in the above equation, we have:

$\Rightarrow$ \[y - ( - 2) = \dfrac{1}{{\sqrt 3 }}(x - 3)\]

$\Rightarrow$ \[y + 2 = \dfrac{1}{{\sqrt 3 }}(x - 3)\]

Taking \[\sqrt 3 \] to the other side, we have:

$\Rightarrow$ \[\sqrt 3 (y + 2) = x - 3\]

Simplifying the equation, we obtain:

$\Rightarrow$ \[x - \sqrt 3 y - 2\sqrt 3 - 3 = 0\]

$\Rightarrow$ \[x = \sqrt 3 y + 2\sqrt 3 + 3\]

Hence, the required equation is \[x = \sqrt 3 y + 2\sqrt 3 + 3\].

Note: Do not use the given angle to calculate the tangent to find the slope directly, first, you need to find the angle made by the line with the positive direction of x-axis, then proceed with the solution.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Give 10 examples for herbs , shrubs , climbers , creepers

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE