A line goes through points (2�ã3,-3) and (8�ã3,3). Find the equation of the line through these 2 points and the angle of incline (between the line and the x-axis).
The slope of your line is 6/(6√3) = 1/√3
Then, using the point (2√3,-3)
y+3 = (1/√3)(x-2√3)
cross-multiplying, expanding and then simplifying I got
x - (√3)y = 5√3 or if we multiply this by √3
(√3)x - 3y = 15
the slope of the line is equal to the tangent of the angle that the line makes with the x-axis
so tan(angle) = 1/√3
angle = 30º (from the ratios of the 30-60-90 triangle)
To find the equation of a line passing through two given points, we can use the point-slope form of the equation:
y - y₁ = m(x - x₁)
where (x₁, y₁) and (x₂, y₂) are the coordinates of the two points, and m is the slope of the line.
Let's first find the slope (m) using the two given points: (2, -3) and (8, 3).
m = (y₂ - y₁) / (x₂ - x₁)
= (3 - (-3)) / (8 - 2)
= 6 / 6
= 1
So, the slope (m) of the line is 1.
Now, let's choose one of the given points, say (2, -3), and substitute its coordinates along with the slope into the point-slope form:
y - (-3) = 1(x - 2)
y + 3 = x - 2
y = x - 2 - 3
y = x - 5
Therefore, the equation of the line passing through the given points is y = x - 5.
Now, let's find the angle of inclination between the line and the x-axis. The angle of inclination (θ) can be found using the equation:
θ = tan⁻¹(m)
where m is the slope we calculated earlier.
θ = tan⁻¹(1)
θ ≈ 45°
So, the angle of inclination between the line and the x-axis is approximately 45 degrees.