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A variable line cuts the lines $x^2-(a+b)x+ab=0$ in such a way that intercept between the lines subtends a right angle at the origin. Find the locus of the foot of perpendicular from origin on the variable line.

Lines are $x=a$ and $x=b$ and I assume variable line to $px+qy=1$. Foot of perpendicular$(h,k)$ can be found in terms of $p$ and $q$ but I am not able to eliminate $p,q$ using the information that at intercept between the lines subtends a right angle at the origin

Could someone help me with this?

MathGeek
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1 Answers1

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If $A(x_1,y_1), B(x_2,y_2)$ are the intersection points, then the condition "subtends a right angle at the origin" can be expressed using Pythagoras theorem: $AB^2 = AO^2 + BO^2$ and from here: $(x_1-x_2)^2+(y_1-y_2)^2={x_1}^2 + {y_1}^2 + {x_2}^2 + {y_2}^2$. You get $x_1x_2 + y_1y_2=0$ etc.