Mathematics 2012 JAMB Past Questions - Question 31: The locus of a point equidistant from the intersection of lines 3x-7y+7=0 and 4x-6y+1=0 is a?

Correct Answer

A

Explanation

The locus of the point equidistant from the point of intersection of the two lines is a circle with their points of intersection as the centerTo find the locus of a point that is equidistant from the intersection of the lines 3x - 7y + 7 = 0 and 4x - 6y + 1 = 0, we can follow these steps:Step 1: Find the intersection point of the two lines by solving the system of equations:3x - 7y + 7 = 04x - 6y + 1 = 0Solving this system of equations, we find that the intersection point is (x, y) = (-1, -2).Step 2: Find the distance between the intersection point and an arbitrary point (x, y) on the locus.The distance between two points (x1, y1) and (x2, y2) is given by the distance formula:d = √((x2 - x1)² + (y2 - y1)²)In this case, the distance between (-1, -2) and an arbitrary point (x, y) on the locus is:d = √((x - (-1))² + (y - (-2))²)  = √((x + 1)² + (y + 2)²)Step 3: Set the distance equal to a constant value, let's say k, since the locus consists of points equidistant from the intersection point.√((x + 1)² + (y + 2)²) = kSquaring both sides of the equation, we get:(x + 1)² + (y + 2)² = k²This equation represents the locus of points that are equidistant from the intersection of the given lines.Therefore, the locus of a point equidistant from the intersection of the lines 3x - 7y + 7 = 0 and 4x - 6y + 1 = 0 is a circle centered at (-1, -2) with a radius of k.