Mathematics
Draw two intersecting lines to include an angle of 30°. Use ruler and compasses to locate points which are equidistant from these lines and also 2 cm away from their point of intersection. How many such point exist ?
Answer
Let two intersecting lines be AB and CD making angles of 30° and 150°. Let these lines intersect at O.
We know that locus of points equidistant from two lines is the angle bisector of angle between them.
From figure,
EF and GH are angular bisectors of AB and CD.
Construct a circle taking O as centre and radius as 2 cm. This circle meet the bisectors at four points P, Q, R and S.
Hence, there are 4 points that are equidistant from two intersecting lines and 2 cm away from their point of intersection.
Related Questions
Without using set square or protractor, construct the quadrilateral ABCD in which ∠BAD = 45°, AD = AB = 6 cm, BC = 3.6 cm and CD = 5 cm.
(i) Measure ∠BCD.
(ii) Locate the point P on BD which is equidistant from BC and CD.
Without using set square or protractor, construct rhombus ABCD with sides of length 4 cm and diagonal AC of length 5 cm. Measure ∠ABC. Find the point R on AD such that RB = RC. Measure the length of AR.
Points A, B and C represent position of three towers such that AB = 60 m, BC = 73 m and CA = 52 m. Taking a scale of 10 m to 1 cm, make an accurate drawing of △ABC. Find by drawing, the location of a point which is equidistant from A, B and C, and its actual distance from any of the towers.
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°.
Hence,
(i) Construct the locus of points equidistant from BA and BC.
(ii) Construct the locus of points equidistant from B and C.
(iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.