Prove that the equation to a straight line which passes through the point (h,k) and is
perpendicular to the axis of x is
x + ycosΘ = h + kcosΘ
How can a coordinate (x) have an axis?
To prove that the equation of a straight line passing through the point (h, k) and perpendicular to the x-axis is x + ycosθ = h + kcosθ, we can start by understanding the characteristics of the line.
A line perpendicular to the x-axis means that it is parallel to the y-axis. In other words, the slope of the line should be infinite or undefined.
Let's consider the slope-intercept form of a line, y = mx + b, where m is the slope and b is the y-intercept.
Since the line is parallel to the y-axis, the equation will have no x term, so the x coefficient, m, will be zero. Therefore, the equation becomes y = 0x + b, which simplifies to y = b.
This means the line is a horizontal line with a constant y-value, which could be represented as y = k.
Now, since the line passes through the point (h, k), we substitute these values into the equation to obtain k = k. It is true that k = k, so this confirms that the equation y = k is correct.
To convert this equation into a form involving trigonometric functions, we can express y and k in terms of cosθ.
Let's rewrite y = k as y = k * cos(0°), where 0° represents the angle between the x-axis and the line that is perpendicular to it.
Now, let's substitute this value of y into the original equation y = m x + b, yielding y = 0 x + k * cos(0°), which simplifies to y = k * cosθ.
Substituting this result into the equation y = k, we get k * cosθ = k.
Finally, by rearranging the terms, we obtain x + y * cosθ = h + k * cosθ, which is the equation x + ycosθ = h + kcosθ.
Therefore, we have proven that the equation of a straight line passing through the point (h, k) and perpendicular to the x-axis is x + ycosθ = h + kcosθ.