Maths problem involving Coordimate Geometry

[Sanosuke Sagara]

find the equation of a straight line whose x-intercept and y-intercept are a and b respectively.If this line varies such that $$\frac{1}{a^2}$$ + $$\frac{1}{b^2}$$ = $$\frac{1}{c^2}$$with c as a constant,show that the locus of the foot of the perpendicular from the origin to this line is the curve
$$X^2$$ +$$Y^2$$ = $$C^2$$.


I want to ask what is meant by the phrase 'foot of the perpendicular from the origin to this line ' ?

I hope that somebody will help me to explain the meaning and thanks for anybody that spend some time on this question.

 

[Data]

The 'foot of the perpendicular from the origin to this line' means the curve which the intesection of the line and the perpendicular forms as $$a$$ and $$b$$ are varied. This intersection point changes with $$a$$ and $$b$$, and the path that it traces they move through all their possible values is what the question is looking for.

 

[tacman]

The phrase "foot of the perpendicular from the origin to this line" refers to the point where a line intersects with a perpendicular line drawn from the origin (point (0,0)) to the line. In this problem, we are looking at the locus (or path) of all these points for the given line.

To solve this problem, we first need to find the equation of the line with x-intercept a and y-intercept b. We know that the x-intercept is a, so it must pass through the point (a,0). Similarly, the y-intercept is b, so the line must also pass through the point (0,b). Using the slope-intercept form of a line, we can write the equation as:

y = (-b/a)x + b

Next, we are given that \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}, where c is a constant. We can rewrite this equation as:

\frac{1}{a^2} + \frac{1}{b^2} - \frac{1}{c^2} = 0

Multiplying both sides by (ab)^2, we get:

b^2 + a^2 - (ab)^2 = 0

This equation can be factored as:

(b - ac)(a - bc) = 0

Since c is a constant, we can write this as:

b = ac or a = bc

Substituting these values into the equation for our line, we get two possible equations:

y = (-b/ac)x + b or y = (-a/bc)x + b

Simplifying these equations, we get:

y = (-1/c)x + b or y = (-1/c)x + a

These two equations represent the same line, just in different forms. Now, we need to find the locus of the foot of the perpendicular from the origin to this line. To do this, we can use the distance formula to find the distance from the origin to any point on the line. The distance formula is:

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Since the origin is at (0,0), the distance formula simplifies to:

d = \sqrt{x^2 + y^2}

Substituting the equations for