Reduce the equation of the straight line 3x + 4y + 15 = 0 to normal form and find the perpendicular distance of the line from the origin.
Understand the Problem
The question is asking to reduce a given equation of a straight line to its normal form and to find the perpendicular distance of that line from the origin. This involves algebraic manipulation and the application of formulas for the normal form and distance.
Answer
Normal form: $$ \frac{3}{5}x + \frac{4}{5}y = -3 $$; Perpendicular distance: \( 3 \)
Answer for screen readers
The normal form of the line is
$$ \frac{3}{5}x + \frac{4}{5}y = -3 $$
and the perpendicular distance from the origin is ( 3 ).
Steps to Solve
- Write the equation in standard form
The given equation is
$$ 3x + 4y + 15 = 0 $$
To convert it to the standard form, we rearrange it to isolate the constant term:
$$ 3x + 4y = -15 $$
- Convert to normal form
The normal form of a line is given by
$$ Ax + By + C = 0 $$
where the normal vector is $(A, B)$. Here, $A = 3$, $B = 4$, and $C = -15$.
We calculate the magnitude of the vector $(A, B)$:
$$ \sqrt{A^2 + B^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$
Now, we divide the equation by this magnitude:
$$ \frac{3}{5}x + \frac{4}{5}y + 3 = 0 $$
This is equivalent to
$$ \frac{3}{5}x + \frac{4}{5}y = -3 $$
- Calculate the perpendicular distance from the origin
The formula for the perpendicular distance ( D ) from a point ((x_0, y_0)) to the line
$$ Ax + By + C = 0 $$
is given by:
$$ D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} $$
For this problem, substituting ( (x_0, y_0) = (0, 0) ):
$$ D = \frac{|3(0) + 4(0) - 15|}{\sqrt{3^2 + 4^2}} = \frac{|-15|}{5} = \frac{15}{5} = 3 $$
The normal form of the line is
$$ \frac{3}{5}x + \frac{4}{5}y = -3 $$
and the perpendicular distance from the origin is ( 3 ).
More Information
In geometry, the normal form of a line is useful because it emphasizes the normal vector, which is perpendicular to the line, making it easy to calculate distances. The distance from a point to a line is significant in various applications, such as optimization and physics.
Tips
- Forgetting to take the absolute value when calculating distance can lead to incorrect results.
- Not simplifying the equation properly when converting to normal form.
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