Question:
If \(P(\alpha,\alpha+1)\) is the foot of the perpendicular drawn from the origin to the line \(L\) and the \(x\)-intercept of \(L\) is
\[
\left(-\frac52,0\right),
\]
then the sum of the squares of the distances from the origin to all such possible points \(P\) is
If \(P(\alpha,\alpha+1)\) is the foot of the perpendicular drawn from the origin to the line \(L\) and the \(x\)-intercept of \(L\) is
\[
\left(-\frac52,0\right),
\]
then the sum of the squares of the distances from the origin to all such possible points \(P\) is
Show Hint
If \((x_1,y_1)\) is the foot of the perpendicular from the origin to a line, then the line can be written directly as
\[
x_1x+y_1y=x_1^2+y_1^2.
\]
This avoids finding slopes and makes intercept conditions easy to apply.
Updated On: Jul 29, 2026
- \(\dfrac{45}{8}\)
- \(\dfrac{15}{2}\)
- \(\dfrac{12}{5}\)
- \(\dfrac{10}{7}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
If \(P(x_1,y_1)\) is the foot of the perpendicular from the origin to a line, then the line is perpendicular to the vector
\[
\overrightarrow{OP}.
\]
Hence the equation of the line is
\[
x_1x+y_1y=x_1^2+y_1^2.
\]
We use the given \(x\)-intercept condition to determine the possible values of \(\alpha\).
Step 1: Write the equation of the line whose foot of perpendicular from the origin is \(P(\alpha,\alpha+1)\). Since \[ P(\alpha,\alpha+1), \] the required line is \[ \alpha x+(\alpha+1)y = \alpha^2+(\alpha+1)^2. \] \[ \alpha x+(\alpha+1)y = 2\alpha^2+2\alpha+1. \]
Step 2: Use the given \(x\)-intercept. The \(x\)-intercept is \[ \left(-\frac52,0\right). \] Substituting \[ x=-\frac52,\qquad y=0, \] into the line equation, \[ -\frac52\alpha = 2\alpha^2+2\alpha+1. \] Multiplying by \(2\), \[ -5\alpha = 4\alpha^2+4\alpha+2. \] \[ 4\alpha^2+9\alpha+2=0. \]
Step 3: Find the possible values of \(\alpha\). Factorizing, \[ 4\alpha^2+9\alpha+2 = (4\alpha+1)(\alpha+2). \] Hence, \[ \alpha=-\frac14 \] or \[ \alpha=-2. \]
Step 4: Find the square of the distance \(OP\) for each point. Since \[ P(\alpha,\alpha+1), \] \[ OP^2 = \alpha^2+(\alpha+1)^2. \] For \[ \alpha=-\frac14, \] \[ OP^2 = \frac1{16}+\frac9{16} = \frac{10}{16} = \frac58. \] For \[ \alpha=-2, \] \[ OP^2 = (-2)^2+(-1)^2 = 5. \]
Step 5: Find the required sum. \[ \frac58+5 = \frac58+\frac{40}{8} = \frac{45}{8}. \]
Step 6: Write the final answer. \[ \boxed{\frac{45}{8}} \]
Step 1: Write the equation of the line whose foot of perpendicular from the origin is \(P(\alpha,\alpha+1)\). Since \[ P(\alpha,\alpha+1), \] the required line is \[ \alpha x+(\alpha+1)y = \alpha^2+(\alpha+1)^2. \] \[ \alpha x+(\alpha+1)y = 2\alpha^2+2\alpha+1. \]
Step 2: Use the given \(x\)-intercept. The \(x\)-intercept is \[ \left(-\frac52,0\right). \] Substituting \[ x=-\frac52,\qquad y=0, \] into the line equation, \[ -\frac52\alpha = 2\alpha^2+2\alpha+1. \] Multiplying by \(2\), \[ -5\alpha = 4\alpha^2+4\alpha+2. \] \[ 4\alpha^2+9\alpha+2=0. \]
Step 3: Find the possible values of \(\alpha\). Factorizing, \[ 4\alpha^2+9\alpha+2 = (4\alpha+1)(\alpha+2). \] Hence, \[ \alpha=-\frac14 \] or \[ \alpha=-2. \]
Step 4: Find the square of the distance \(OP\) for each point. Since \[ P(\alpha,\alpha+1), \] \[ OP^2 = \alpha^2+(\alpha+1)^2. \] For \[ \alpha=-\frac14, \] \[ OP^2 = \frac1{16}+\frac9{16} = \frac{10}{16} = \frac58. \] For \[ \alpha=-2, \] \[ OP^2 = (-2)^2+(-1)^2 = 5. \]
Step 5: Find the required sum. \[ \frac58+5 = \frac58+\frac{40}{8} = \frac{45}{8}. \]
Step 6: Write the final answer. \[ \boxed{\frac{45}{8}} \]
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