A straight line passes through a fixed point (h,k). The locus of the feet of perpendiculars on it drawn from the origin is ………………

Answer

To find the locus of the feet of perpendiculars drawn from the origin to a straight line that passes through a fixed point \((h, k)\), we can follow these steps:

### Step 1: Equation of the Line
Let the equation of the straight line passing through the point \((h, k)\) be given in point-slope form:
\[ y - k = m(x - h) \]
where \(m\) is the slope of the line.

### Step 2: Rearranging the Equation
Rearranging the equation gives us:
\[ y = mx - mh + k \]
This can be rewritten as:
\[ mx - y + (k - mh) = 0 \]

### Step 3: Distance from the Origin to the Line
The distance \(d\) from the origin \((0, 0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
In our case, \(A = m\), \(B = -1\), and \(C = k - mh\). Therefore, the distance from the origin to the line is:
\[ d = \frac{|m(0) - 1(0) + (k - mh)|}{\sqrt{m^2 + (-1)^2}} = \frac{|k - mh|}{\sqrt{m^2 + 1}} \]

### Step 4: Finding the Locus
Let the foot of the perpendicular from the origin to the line be denoted as \((\alpha, \beta)\). The slope of the line connecting the origin to the point \((\alpha, \beta)\) is:
\[ \text{slope} = \frac{\beta}{\alpha} \]
The slope of the perpendicular line is:
\[ -\frac{\alpha}{\beta} \]

### Step 5: Relating Slopes
Since the slopes are negative reciprocals, we have:
\[ m = -\frac{\alpha}{\beta} \]

### Step 6: Substituting for \(m\)
Substituting \(m\) into the distance equation gives:
\[ d = \frac{|k - h\left(-\frac{\alpha}{\beta}\right)|}{\sqrt{\left(-\frac{\alpha}{\beta}\right)^2 + 1}} = \frac{|k + \frac{h\alpha}{\beta}|}{\sqrt{\frac{\alpha^2}{\beta^2} + 1}} = \frac{|k + \frac{h\alpha}{\beta}|}{\sqrt{\frac{\alpha^2 + \beta^2}{\beta^2}}} = \frac{\beta |k + \frac{h\alpha}{\beta}|}{\sqrt{\alpha^2 + \beta^2}} \]

### Step 7: Setting Up the Equation
We know that the distance \(d\) is equal to the perpendicular distance from the origin to the line, which we can set equal to a constant. This leads us to the equation:
\[ \alpha^2 + \beta^2 = k\beta + h\alpha \]

### Step 8: Final Locus Equation
Replacing \(\alpha\) with \(x\) and \(\beta\) with \(y\), we get the final locus equation:
\[ x^2 + y^2 - hx - ky = 0 \]

### Conclusion
Thus, the locus of the feet of the perpendiculars drawn from the origin to the line passing through the point \((h, k)\) is given by:
\[ x^2 + y^2 - hx - ky = 0 \]

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