On a graph paper plot the points A(-2,-3), B(6,5) and C(-2,8). Then the point of intersection of the locus of a point equidistant from A and B and the locus of a point equidistant from AB and BC lies in
On a graph paper plot the points A(-2,-3), B(6,5) and C(-2,8). Then the point of intersection of the locus of a point equidistant from A and B and the locus of a point equidistant from AB and BC lies in
The correct option is B second quadrant
Plot the points A(-2,-3), B(6,5) and C(-2,8) on a graph paper.
We know that the locus of a point which is equidistant from two fixed points is the perpendicular bisector of the line segment joining the two fixed points.
Hence to find the locus of a point equidistant from A and B, we draw a perpendicular bisector 'm' of AB, a shown above.
Also, we know that the locus of a point which is equidistant from two intersecting straight lines is a pair of straight lines which bisect the angles between the given lines.
Hence to find the locus of a point equidistant from AB and BC, we draw the bisector 'l' of angle B.
From these constructions, we note that the lines 'l' and 'm' intersect at a point(P, in the graph) which lies in the second quadrant.
second quadrant
Plot the points A(-2,-3), B(6,5) and C(-2,8) on a graph paper.
We know that the locus of a point which is equidistant from two fixed points is the perpendicular bisector of the line segment joining the two fixed points.
Hence to find the locus of a point equidistant from A and B, we draw a perpendicular bisector 'm' of AB, a shown above.
Also, we know that the locus of a point which is equidistant from two intersecting straight lines is a pair of straight lines which bisect the angles between the given lines.
Hence to find the locus of a point equidistant from AB and BC, we draw the bisector 'l' of angle B.
From these constructions, we note that the lines 'l' and 'm' intersect at a point(P, in the graph) which lies in the second quadrant.
