Find the equation of the straight line which makes angle 30 with positive direction of x axis and cuts intercepts +5 on the y axis .

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

The equation is:

y-y1= m(x-x1)

Since the angle for the line is 30, then :

m= tan30 = 1/sqrt3= sqrt3/3

Since it cuts the y axis at +5 , then the point (0,5) belongs to the line:

==> y-5 = (sqrt3/3)(x-0)

==> y= (sqrt3/3)x + 5

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Because, from the enunciation, we have information about the slope and the y intercept, we'll put the equation into the standard form:

y = mx + n

m - slope of the line

n - y intercept

Because we know the inclination of the line, a = 30 degrees, we could calculate the slope:

m = tan 30

m = sqrt 3/3

From enunciation, we know that n = 5.

We'll write the equation of the line:

y = (sqrt 3/3)*x + 5

We'll re-write the equation in the general form:

3y - (sqrt3)*x - 15 = 0

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Th equation o a line with slope m and intercepy c on y axis is given by y = mx+c.

In this case m = slope = tan 30 = 1/sqrt3 = sqrt3/3.

Therefore the equation is of the formĀ  y =( tan30)x+c

y= (sqrt3/3)x+c. Since this makes an intercept of 5 on y axis,

y = ((sqrt3)/3)x+5 is the equation of the line.

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