find equation of the straight line upon which the length of perpendicular from origin is 2 and slope of perpendicular is 5/12
Answers
Arrange your equation into slope-intercept form:
4x + 3y = 12
3y = -4x + 12
y = -4/3x + 4
Our new line will take the same form of "y = mx + b". First recognize that a perpendicular line has an inverse slope to its reference.
m(reference) = -4/3
m(perpendicular) = 3/4
To solve for "b" we'll plug in (0, 0) since we know the line passes through the origin.
y = 3/4 x + b
0 = 0 + b
b = 0
Our equation is thus:
y = 3/4 x
*Edit*
I'm sorry, I didn't actually answer your question. These two lines intersect at:
3/4x = -4/3x + 4
9x = -16x + 48
25x = 48
x = 1.92
And:
y = 3/4 x
y = 3/4 (1.92)
y = 1.44
We're finding the hypotenuse length given the above "y" and "x" values, using the Pythagorean Theorem.
h² = x² + y²
h² = 1.92² + 1.44²
h = 2.4
The length of the line is 2.4 (units).
The general equation is : xcosα + ysinα = p
Slope = 5/12 = tanα = p/b
So, p = 5
b = 12
Now, cosα = b/h
sinα = p/h
So, h² = 25 + 144
h = 13
Thus, cosα = b/h = 12/13
sinα = p/h = 5/13
Therefore, the equation is -
x×(12/13) + y×(5/13) = 2
12x + 5y = 26
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