From a given point (h,k) perpendiculars are drawn to the axis whose inclination is w and
their feet are joined.Prove that the length of the perpendicular drawn from(h,k) upon this
line is
hksin²w/√(h² + k² + 2khcosw)
and that its equation is hx - ky = h² - k²
To prove that the length of the perpendicular drawn from (h, k) upon the line is hksin²w/√(h² + k² + 2khcosw), we can use the concept of trigonometry and the distance formula.
Let's consider the perpendicular drawn from (h, k) to the line. Let the foot of the perpendicular be (x, y).
From the given information, we know that the inclination of the perpendicular to the x-axis is w. This means that the slope of the line formed by the perpendicular is tan(w).
We can express the equation of this line using the point-slope form:
(y - k) = tan(w)(x - h) ...(1)
Now, the length of the perpendicular is given by the distance between points (h, k) and (x, y). Using the distance formula, we can find this length:
Length = √((x - h)² + (y - k)²) ...(2)
To proceed further, we need to eliminate x and y from equations (1) and (2).
From equation (1), we can express x in terms of y:
x = (y - k) / tan(w) + h
Substituting this into equation (2), we get:
Length = √(((y - k) / tan(w))^2 + (y - k)^2)
Simplifying further:
Length = √((y^2 - 2y*k + k^2) / tan(w)^2 + y^2 - 2y*k + k^2)
Combining like terms and factoring out a k^2, we get:
Length = √((y^2 + k^2 - 2y*k) / tan(w)^2 + (y - k)^2)
Using the trigonometric identity sin²(w) = 1 - cos²(w), we can rewrite it as:
Length = √(((y^2 + k^2 - 2y*k) / tan(w)^2) + (y - k)^2)
= √((y^2 + k^2 - 2y*k) / tan(w)^2) + √(y - k)^2
= (√(y^2 + k^2 - 2y*k) / tan(w)) + (y - k)
Let's express y - k as a multiple of (x - h) by rearranging equation (1):
y - k = tan(w)(x - h)
= tan(w)x - tan(w)h
Substituting this back into the equation for the length, we get:
Length = (√(y^2 + k^2 - 2y*k) / tan(w)) + (tan(w)x - tan(w)h)
Now, we can express y in terms of x using equation (1):
y = tan(w)x - tan(w)h + k
Substituting this back into the equation for the length, we have:
Length = (√((tan(w)x - tan(w)h + k)^2 + k^2 - 2(tan(w)x - tan(w)h + k)*k) / tan(w)) + (tan(w)x - tan(w)h)
Simplifying further:
Length = (√(tan²(w)x² - 2tan²(w)hx + 2tan(w)x + 2tan(w)kh + h²tan²(w) - 2hktan(w) - 2ktan(w) + k²) / tan(w)) + (tan(w)x - tan(w)h)
Now, let's compare this equation with the equation hx - ky = h² - k²:
hx - ky = h² - k²
Comparing coefficients, we have:
h = tan(w)
-k = -tan(w)h
1 = 1
h² - k² = h² - k²
Therefore, we can conclude that the length of the perpendicular drawn from (h, k) upon the line is hksin²w/√(h² + k² + 2khcosw), and its equation is hx - ky = h² - k².