Advertisements
Advertisements
Question
Given: ∠BAC, a line intersects the arms of ∠BAC in P and Q. How will you locate a point on line segment PQ, which is equidistant from AB and AC? Does such a point always exist?
Advertisements
Solution
Since, locus of points equidistant from AB and AC is the bisector of ∠BAC. Draw the bisector of ∠BAC intersecting PQ at R.
Since,R is on the bisector, so it is equidistant from AB and AC.
Yes, such a point always exists as there will be definitely a point where angular bisector and line will intersect.
Hence, R is the required point.
APPEARS IN
RELATED QUESTIONS
Use ruler and compasses only for this question.
- Construct ΔABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
- Construct the locus of points inside the triangle which are equidistant from BA and BC.
- Construct the locus of points inside the triangle which are equidistant from B and C.
- Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
Describe the locus for questions 1 to 13 given below:
1. The locus of a point at a distant 3 cm from a fixed point.
Describe the locus of points at a distance 2 cm from a fixed line.
Describe the locus of a stone dropped from the top of a tower.
In the given figure, obtain all the points equidistant from lines m and n; and 2.5 cm from O.

In a quadrilateral ABCD, if the perpendicular bisectors of AB and AD meet at P, then prove that BP = DP.
Prove that the common chord of two intersecting circles is bisected at right angles by the line of centres.
In Fig. ABCD is a quadrilateral in which AB = BC. E is the point of intersection of the right bisectors of AD and CD. Prove that BE bisects ∠ABC.
Find the locus of points which are equidistant from three non-collinear points.
