Question:
The straight line passing through \((0,0)\) and the foot of perpendicular from \((2,4)\) onto
\[
x+y-1=0
\]
is:
The straight line passing through \((0,0)\) and the foot of perpendicular from \((2,4)\) onto
\[
x+y-1=0
\]
is:
Show Hint
For a line \(ax+by+c=0\), the foot of the perpendicular from a point can be found directly using the projection formula. This saves time in coordinate geometry problems.
Updated On: Jun 26, 2026
- \(y=-3x\)
- \(y=3x\)
- \(y=\dfrac{1}{3}x\)
- \(y=-\dfrac{1}{3}x\)
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The Correct Option is A
Solution and Explanation
Step 1: Identify the given line and point.
The given line is \[ x+y-1=0. \] The point is \[ P(2,4). \] Let the foot of the perpendicular from \(P\) to the line be \(Q\).
Step 2: Use the foot of perpendicular formula.
For the line \[ ax+by+c=0, \] the foot of perpendicular from \((x_1,y_1)\) is \[ \left( x_1-\frac{a(ax_1+by_1+c)}{a^2+b^2}, \, y_1-\frac{b(ax_1+by_1+c)}{a^2+b^2} \right). \] Here, \[ a=1,\quad b=1,\quad c=-1, \] and \[ (x_1,y_1)=(2,4). \] Now, \[ ax_1+by_1+c =2+4-1 =5. \] Therefore, \[ Q= \left( 2-\frac{5}{2}, \, 4-\frac{5}{2} \right) = \left( -\frac12, \frac32 \right). \]
Step 3: Find the equation of the line through \((0,0)\) and \(Q\).
The slope is \[ m= \frac{\frac32-0}{-\frac12-0} = -3. \] Since the line passes through the origin, \[ y=mx. \] Hence, \[ y=-3x. \]
Step 4: Final conclusion.
Therefore, the required line is \[ \boxed{y=-3x}. \]
The given line is \[ x+y-1=0. \] The point is \[ P(2,4). \] Let the foot of the perpendicular from \(P\) to the line be \(Q\).
Step 2: Use the foot of perpendicular formula.
For the line \[ ax+by+c=0, \] the foot of perpendicular from \((x_1,y_1)\) is \[ \left( x_1-\frac{a(ax_1+by_1+c)}{a^2+b^2}, \, y_1-\frac{b(ax_1+by_1+c)}{a^2+b^2} \right). \] Here, \[ a=1,\quad b=1,\quad c=-1, \] and \[ (x_1,y_1)=(2,4). \] Now, \[ ax_1+by_1+c =2+4-1 =5. \] Therefore, \[ Q= \left( 2-\frac{5}{2}, \, 4-\frac{5}{2} \right) = \left( -\frac12, \frac32 \right). \]
Step 3: Find the equation of the line through \((0,0)\) and \(Q\).
The slope is \[ m= \frac{\frac32-0}{-\frac12-0} = -3. \] Since the line passes through the origin, \[ y=mx. \] Hence, \[ y=-3x. \]
Step 4: Final conclusion.
Therefore, the required line is \[ \boxed{y=-3x}. \]
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