15. Loci (Locus and Its Constructions) Previous Year Questions

Circumcentre is equidistant from all vertices of a triangle (intersection of perpendicular bisectors). Incentre is the point of concurrence of interior angle bisectors. Both statements are true. [1 Mark]

Correct Option: (b) at equal distance from the three vertices of the triangle.
Circumcentre is the concurrence point of perpendicular bisectors of sides, which are equidistant from triangle vertices. [1 Mark]

(a) Perpendicular bisector of BC. [1 Mark]
(b) Perpendicular bisector of AB. [0.5 Mark]
(c) Point O is midpoint of hypotenuse AC. Locus is circumcircle of ΔABC with centre O and radius 5 cm. [1 Mark]
(d) Angle bisector ray of ∠ABC. [1.5 Marks]

(i) Construct circle of radius $4.5\text{ cm}$. Draw chord $AB = 6\text{ cm}$.
The locus of points equidistant from two points $A$ and $B$ is the perpendicular bisector of $AB$.
Construct the perpendicular bisector of $AB$; it passes through the centre $O$ and intersects the circle at $D$. [1.5 Marks]
(ii) Join $AD$. The locus of points equidistant from two intersecting lines $AD$ and $AB$ is the internal angle bisector of $\angle DAB$.
Bisect $\angle DAB$; let the bisector intersect the circle at $C$. [1.5 Marks]
(iii) Join $BC$ and $CD$ to complete cyclic quadrilateral $ABCD$.
Measure the side $CD$:
Length of side $CD \approx 5.2\text{ cm}$ (acceptable tolerance: $5.0\text{ cm}$ to $5.4\text{ cm}$). [1 Mark]

Step 1 Draw line segment $BC = 7\text{ cm}$. Bisect $BC$ to find centre $O$, then construct semicircle with radius $OB = 3.5\text{ cm}$. [1 Mark]
Step 2 Locate point $A$ equidistant from $B$ and $C$:
Locus of points equidistant from two fixed points $B$ and $C$ is the **perpendicular bisector of $BC$**.
The perpendicular bisector meets the circumference at point $A$.
(Here, $\Delta ABC$ is isosceles right-angled at $A$ with $\angle ABC = \angle ACB = 45^\circ$). [1 Mark]
Step 3 Locate point $D$ equidistant from $AB$ and $BC$:
Locus of points equidistant from two intersecting lines $AB$ and $BC$ is the **internal angle bisector of $\angle ABC$**.
Construct the bisector of $\angle ABC = 45^\circ$ to intersect the semicircle at point $D$. Join $CD$ and $AD$ to complete cyclic quadrilateral $ABCD$. [1 Mark]
Step 4 Measure $\angle ADC$:
Since $ABCD$ is a cyclic quadrilateral, opposite angles are supplementary:
$\angle ABC + \angle ADC = 180^\circ \implies 45^\circ + \angle ADC = 180^\circ \implies \mathbf{\angle ADC = 135^\circ}$. [1 Mark]