Maths · Conditions for concurrence of three lines, the distance of a point from a line
I: A straight line is such that the algebraic sum of the distance from any no. o
I: A straight line is such that the algebraic sum of the distance from any no. of fixed points is zero. Then that line always passes through a fixed point II: The base of the triangle lie along the line \( x=a \) and is of length \( a \).If the area of the triangle is \( a^{2} \) then the third vertex lies on \( x=-a \) or \( x=3 a \) Then which of the following is true.
- A. only I
- B. only II
- C. both 18 ॥
- D. neither I nor II
Step-by-step solution
Statement I: For a line L, the algebraic sum of signed distances from a set of fixed points is zero if and only if L passes through the centroid of those points. Hence L always passes through the centroid, a fixed point. Thus I is true.
Statement II: Let base endpoints be (a, y1) and (a, y1+a). Third vertex (h, k). Area = (1/2)*a*|h-a| = a^2 ⇒ |h-a| = 2a ⇒ h = -a or 3a. So third vertex lies on x = -a or x = 3a. Thus II is true.
Both statements are true.