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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 60Â° 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 90Â°, then the angle made by the line with the positive direction of the x-axis is 90Â° 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 60Â° 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 90Â°, then the angle made by the line with the positive direction of the x-axis is 90Â° 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.

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