The locus of point which is equidistant from two non-parallel lines AB and CD is :
perpendicular to AB.
perpendicular to CD.
bisector of angle between AB and CD.
perpendicular bisector of CD.
Answer
Since, AB and CD are two non-parallel lines so they will intersect at a point.
We know that,
The locus of a point equidistant from two intersecting lines is the bisector of angles between the lines.
Hence, Option 3 is the correct option.
Three isosceles triangles PBC, QBC and RBC are on the same base, then :
P, Q and R are collinear.
△PQR is isosceles triangle.
Q lies on the circumference of a circle with BC as diameter.
Q is mid-point of line segment PR.
Answer
Steps of construction :
Draw a line segment BC.
Draw XY, perpendicular bisector of BC.
We know that,
Locus of a point equidistant from two given points is the perpendicular bisector of the line joining the two points.
Thus, any point on the line XY is at equal distance from B and C.
∴ P, Q and R lies on the line XY.
Hence, Option 1 is the correct option.
Locus of the centers of the circles passing through two fixed points A and B is :
a line parallel to line segment AB.
the bisector of the line segment AB.
perpendicular to line segment AB.
perpendicular bisector of line segment AB.
Answer
Let O and O' be the center of the circles, passing through points AB.
Steps of construction :
Join AB.
Draw a perpendicular from O to AB and O' to AB.
We know that,
Perpendicular from center to chord bisects it.
Since, both perpendiculars bisects AB.
∴ OO' is a straight line perpendicular to AB and bisecting it.
Hence, Option 4 is the correct option.
A point is equidistant from the sides of an obtuse angle triangle. The point is called :
circumcenter of the triangle.
incenter of the triangle.
centroid of the triangle.
orthocenter of the triangle.
Answer
Let ABC be the obtuse triangle.
Steps of construction :
Draw angle bisectors AX, BY and CZ of the interior angles A, B and C of a triangle.
Let the angle bisectors meet at point I.
We know that,
Locus of a point equidistant from two intersecting lines is the bisector of the angles between the lines.
∴ BY is equidistant from AB and BC
∴ AX is equidistant from AB and AC
∴ CZ is equidistant from AC and BC
∴ The point (incenter) joining AX, BY and CZ will be equidistant from all the three sides of the triangle.
Hence, Option 2 is the correct option.
In △ABC, point P moves such that it remains at a fixed distance (i.e. radius) of 2.8 cm from vertex B. 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.
Answer
We know that,
Locus of a point, in a plane and at a fixed distance from a given fixed point, is the circumference of the circle with the given fixed point as center and given fixed distance as radius.
∴ The locus of point P is a circle with center at point B and radius = 2.8 cm.
Hence, Option 3 is the correct option.
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.
Answer
Steps of construction :
Draw a line segment BC = 5 cm.
At B, draw a ray BX making an angle of 60° and cut off BA = 4.6 cm.
Draw BY, the angle bisector of ∠ABC.
Draw MN, the perpendicular bisector of AB.
Let MN and BY meet at point P.
Hence, P is the point which is equidistant from AB and BC, as well as from A and B.
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..
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.
Answer
Steps of construction :
Draw a line l, x = 6 which is parallel to y-axis.
At a distance of 3 units on both sides from x = 6 mark points P and Q.
Draw line m and n parallel to l and through P and Q respectively.
From figure,
Equation of m : x = 3 and Equation of n : x = 9.
Hence, locus of the point which moves in such a way that its distance from the given line is always equal to 3 units are lines x = 3 and x = 9.
Ruler and compasses may be used in this question. All construction lines and arcs must be clearly shown and be of sufficient length and clarity to permit assessment.
(i) Construct a △ABC, in which BC = 6 cm, AB = 9 cm and angle ABC = 60°.
(ii) Construct the locus of all points inside triangle ABC, which are equidistant from B and C.
(iii) Construct the locus of the vertices of the triangles with BC as base and which are equal in area to triangle 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
Answer
(i) Steps of construction :
Draw a line segment BC = 6 cm.
At B, draw a ray BX making an angle 60° and cut off BA = 9 cm.
Join AC.
Hence, ABC is the required triangle.
(ii) Steps of construction :
- Draw perpendicular bisector of BC which intersects BA at M and BC at L, then any point on LM is equidistant from B and C.
Hence, LM is the required locus.
(iii) Steps of construction :
Through A, draw a line m || BC.
The perpendicular bisector of BC and the parallel line m intersect each other at Q.
Join QB and QC.
Hence, line m is the locus of the vertices of the triangles with BC as base and which are equal in area to triangle ABC.
(iv) The triangle QBC is equal in area to triangle ABC.
Also, QB = QC as Q lies on perpendicular bisector of BC.
Hence, QBC is an isosceles triangle.
(v) On measuring CQ = 8.4 cm.
Construct an isosceles triangle ABC such that AB = 6 cm, BC = AC = 4 cm. Bisect ∠C internally and mark a point P on this bisector such that CP = 5 cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB.
Answer
Steps of construction :
Draw a line segment AB = 6 cm.
With centers A and B and radius = 4 cm, draw two arcs which intersect each other at C.
Join CA and CB.
Draw the angle bisector of angle C and cut off CP = 5 cm.
Draw a line m parallel to AB at a distance of 5 cm.
With P as center and radius = 5 cm, draw arcs which intersect the line m at Q and R.
Join PQ and QR.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
(i) Complete the rectangle ABCD such that :
(a) P is equidistant from AB and BC.
(b) P is equidistant from C and D.
(ii) Measure and record the length of AB.
Answer
(i) Steps of construction :
Draw a line segment BC = 5 cm.
With B as center draw an arc BX, at an angle of 45°.
From BX cut off BP = 4 cm.
Join PC. Hence, BPC is the required triangle.
Draw perpendicular on B and C.
With P as center and radius PC, cut an arc on the perpendicular on C at D.
D as center, draw a line parallel to BC which intersects the perpendicular on B at A.
Hence, ABCD is the required rectangle.
(ii) On measuring we get,
AB = 5.7 cm.
Use ruler and compasses only for the following question. All construction lines and arcs must be clearly shown.
(i) Construct a △ABC in which BC = 6.5 cm, ∠ABC = 60° and AB = 5 cm.
(ii) Construct the locus of points at a distance of 3.5 cm from A.
(iii) Construct the locus of points equidistant from AC and BC.
(iv) Mark 2 points X and Y which are at a distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY.
Answer
(i) Steps of constrcution :
Draw a line segment BC = 6.5 cm.
With B as center draw an arc BX, at an angle of 60°.
From BX cut off BA = 5 cm.
Join AC. Hence, ABC is the required triangle.
With A as center draw a circle of radius = 3.5 cm.
Draw CZ, the angle bisector of ∠C.
Mark the points as X and Y where CZ intersects circle with center A.
(ii) The locus of points at a distance of 3.5 cm from A is a circle with radius = 3.5 cm and center A.
(iii) The locus of points equidistant from AC and BC is CZ, the angle bisector of ∠C.
(iv) On measuring, XY = 4.8 cm.