Mathematics
On a graph paper, draw the lines x = 3 and y = -5. Now, on the same graph paper, draw the locus of the point which is equidistant from the given lines.
Answer
Steps of construction :
Draw a line l with equation x = 3 and m having equation y = -5.
Let these lines intersect at point P.
Draw a line n, which is the angle bisector of ∠P.
Since, n is the angle bisector of ∠P so any point on n is equidistant from l and m.
Hence, locus of the point which is equidistant from the given lines is line n..
Related Questions
ABC is a triangle. Point P moves with vertex B as center and radius 2.8 cm. The locus of point P is :
bisector of angle ABC.
a line parallel to BC and at a distance of 2.8 cm from it.
circle with center at point B and radius = 2.8 cm.
perpendicular bisector of BC.
Draw an ∠ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC; and also equidistant from A and B.
On a graph paper, draw the line x = 6. Now on the same graph paper, draw the locus of the point which moves in such a way that its distance from the given line is always equal to 3 units.
Ruler and compasses only may be used in this question. All construction lines and arcs must be clearly shown, and be of sufficient length and clarity to permit the assessment.
(i) Construct a triangle ABC, in which BC = 6 cm, AB = 9 cm, and ∠ABC = 60°.
(ii) Construct the locus of all points, inside △ABC, which are equidistant from B and C.
(iii) Construct the locus of the vertices of the triangles with BC as base, which are equal in area to △ABC.
(iv) Mark the point Q, in your construction, which would make △QBC equal in area to △ABC, and isosceles.
(v) Measure and record the length of CQ.