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प्रश्न
State and draw the locus of a point equidistant from two given parallel lines.
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उत्तर
The locus of a point equidistant from two given parallel lines AB and CD is the line EF parallel to AB or CD exactly mid-way between AB and CD.
संबंधित प्रश्न
Use ruler and compasses only for this question:
I. Construct ABC, where AB = 3.5 cm, BC = 6 cm and ABC = 60o.
II. Construct the locus of points inside the triangle which are equidistant from BA and BC.
III. Construct the locus of points inside the triangle which are equidistant from B and C.
IV. Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and records the length of PB.
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 distantce from the given line is always equal to 3 units
Angle ABC = 60° and BA = BC = 8 cm. The mid-points of BA and BC are M and N respectively. Draw and describe the locus of a point which is:
- equidistant from BA and BC.
- 4 cm from M.
- 4 cm from N.
Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.
Draw a straight line AB of 9 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.
Without using set squares or protractor, construct a quadrilateral ABCD in which ∠ BAD = 45°, AD = AB = 6 cm, BC= 3.6 cm and CD=5 cm. Locate the point P on BD which is equidistant from BC and CD.
Construct a ti.PQR, in which PQ=S. 5 cm, QR=3. 2 cm and PR=4.8 cm. Draw the locus of a point which moves so that it is always 2.5 cm from Q.
Describe completely the locus of a point in the following case:
Centre of a ball, rolling along a straight line on a level floor.
Without using set squares or protractor construct a triangle ABC in which AB = 4 cm, BC = 5 cm and ∠ABC = 120°.
(i) Locate the point P such that ∠BAp = 90° and BP = CP.
(ii) Measure the length of BP.
Construct a Δ ABC, with AB = 6 cm, AC = BC = 9 cm; find a point 4 cm from A and equidistant from B and C.
Using only a ruler and compass construct ∠ABC = 120°, where AB = BC = 5 cm.
(i) Mark two points D and E which satisfy the condition that they are equidistant from both ABA and BC.
(ii) In the above figure, join AD, DC, AE and EC. Describe the figures:
(a) AECB, (b) ABD, (c) ABE.
