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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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
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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.
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
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