Coordinate Geometry - Locus

Find the equation of the locus of a point such that the sum of the squares of whose distances from the points $\left(2, 4\right)$ and $\left(-3, -1\right)$ is $30$.

Let $\;$ $A = \left(2, 4\right)$, $\;$ $B = \left(-3, -1\right)$

Let $P \left(x, y\right)$ be a point on the required locus.

Distance $\;$ $PA = \sqrt{\left(x - 2\right)^2 + \left(y - 4\right)^2}$

i.e. $\;$ $PA = \sqrt{x^2 + y^2 - 4x - 8y + 20}$

$\therefore \;$ $\left(PA\right)^2 = x^2 + y^2 - 4x - 8y + 20$

Distance $\;$ $PB = \sqrt{\left(x + 3\right)^2 + \left(y + 1\right)^2}$

i.e. $\;$ $PB = \sqrt{x^2 + y^2 + 6x + 2y + 10}$

$\therefore \;$ $\left(PB\right)^2 = x^2 + y^2 + 6x + 2y + 10$

As per question, $\;$ $\left(PA\right)^2 + \left(PB\right)^2 = 30$

i.e. $\;$ $x^2 + y^2 - 4x - 8y + 20 + x^2 + y^2 + 6x + 2y + 10 = 30$

i.e. $\;$ $2 x^2 + 2 y^2 + 2x - 6y = 0$

i.e. $\;$ $x^2 + y^2 + x - 3y = 0$ $\;\;\;$ is the equation of the required locus.

Coordinate Geometry - Locus

Find the equation of the locus of a point whose distance from $\left(a, 0\right)$ is always four times its distance from the $Y$ axis.

Let $\;$ $A = \left(a, 0\right)$

Let $P \left(x, y\right)$ be a point on the required locus.

Distance $\;$ $PA = \sqrt{\left(x - a\right)^2 + \left(y - 0\right)^2}$

i.e. $\;$ $PA = \sqrt{x^2 + y^2 + a^2 - 2ax}$

$\therefore \;$ $\left(PA\right)^2 = x^2 + y^2 + a^2 - 2ax$

Distance of point $P \left(x, y\right)$ from the $Y$ axis is $= x$

As per question, $\;$ $PA = 4 \times x$

i.e. $\;$ $\left(PA\right)^2 = 16 \times x^2$

i.e. $\;$ $x^2 + y^2 + a^2 - 2ax = 16x^2$

i.e. $\;$ $15x^2 - y^2 + 2ax - a^2 = 0$ $\;\;\;$ is the equation of the required locus.

Coordinate Geometry - Locus

Find the equation of the locus of a point whose distance from $\left(2, -3\right)$ is double its distance from $\left(1, 2\right)$.

Let the given points be $\;$ $A \left(2, -3\right)$ $\;$ and $\;$ $B \left(1, 2\right)$.

Let $P \left(x, y\right)$ be a point on the given locus.

Distance $\;$ $PA = \sqrt{\left(x - 2\right)^2 + \left(y + 3\right)^2}$

Distance $\;$ $PB = \sqrt{\left(x - 1\right)^2 + \left(y - 2\right)^2}$

As per question, $\;$ $PA = 2 \times PB$

i.e. $\;$ $\left(PA\right)^2 = 4 \times \left(PB\right)^2$

i.e. $\;$ $\left(x - 2\right)^2 + \left(y + 3\right)^2 = 4 \times \left[\left(x - 1\right)^2 + \left(y - 2\right)^2\right]$

i.e. $\;$ $x^2 - 4x + 4 + y^2 + 6y + 9 = 4 \times \left[x^2 - 2x + 1 + y^2 - 4y + 4\right]$

i.e. $\;$ $x^2 + y^2 -4x + 6y + 13 = 4x^2 + 4y^2 -8x - 16y + 20$

i.e. $\;$ $3x^2 + 3y^2 -4x -22y +7 = 0$ $\;\;\;$ is the equation of the required locus.

Coordinate Geometry - Locus

Find the equation of the locus of points which are equidistant from the points $\left(2, 3\right)$ and $\left(5, 7\right)$.

Let the given points be $\;$ $A \left(2, 3\right)$ $\;$ and $\;$ $B \left(5, 7\right)$.

Let $P \left(x, y\right)$ be a point on the given locus.

As per question, $\;$ $PA = PB$

i.e. $\;$ $\left(PA\right)^2 = \left(PB\right)^2$

i.e. $\;$ $\left(x - 2\right)^2 + \left(y - 3\right)^2 = \left(x - 5\right)^2 + \left(y - 7\right)^2$

i.e. $\;$ $x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 - 10x + 25 + y^2 - 14y + 49$

i.e. $\;$ $6x + 8y - 61 = 0$ $\;\;\;$ is the equation of the required locus.

Locus

Construct $\triangle ABC$ in which $BC = 6 \; cm$, $\angle ABC = 120^\circ$ and $AB = 7.5 \; cm$

1. Construct the locus of a point equidistant from $B$ and $C$.
2. Draw the locus of a point $P$ such that area $\triangle PBC = $ area $\triangle ABC$.

$\triangle ABC$ is constructed with the given dimensions (red construction arcs).

Locus of a point equidistant from points $B$ and $C$ lies on the perpendicular bisector of $BC$ (line $DE$) (blue construction arcs and lines).

Locus of point $P$ [such that area $\triangle PBC = $ area $\triangle ABC$] lies on line $m$ which is a line through point $A$ and parallel to side $BC$ of $\triangle ABC$ (violet construction arcs and lines).

Locus

Construct $\triangle PQR$ where $QR = 5 \; cm$, $\angle Q =45^\circ$ and $PQ = 6.5 \; cm$.
Find by construction points $X$ and $Y$ which are equidistant from $Q$ and $R$, and are $4 \; cm$ from $Q$.

$\triangle PQR$ is constructed with $QR = 5 \; cm$, $\angle Q =45^\circ$ and $PQ = 6.5 \; cm$.

Locus $\left(L_1\right)$ of points which are equidistant from $Q$ and $R$ lie on the perpendicular bisector $AB$ of $QR$.

Locus $\left(L_2\right)$ of points which are at a distance of $4 \; cm$ from $Q$ lie on the circle $C$ with center at $Q$ and radius $4 \; cm$.

Points $X$ and $Y$ lie at the intersection of $L_1$ and $L_2$.

Locus

Taking a scale of $2 \; cm$ as $2$ units on both the axes, on a graph paper plot the points $A \left(1, 2\right)$, $B \left(8, 3\right)$, $C \left(7, 8\right)$ and $D \left(3, 7\right)$.

Complete the quadrilateral $ABCD$.

Find the point $P$ inside the quadrilateral such that $AP = BP$ and $P$ is also equidistant from $AB$ and $AD$.

Record the length $PB$.

The points $A \left(1, 2\right)$, $B \left(8, 3\right)$, $C \left(7, 8\right)$ and $D \left(3, 7\right)$ are plotted on a graph paper and the quadrilateral $ABCD$ is drawn.

Given: $\;$ $AP = BP$

$\implies$ $P$ lies on the perpendicular bisector of the line $AB$.

[The locus of a point, which is equidistant from two fixed points, is the perpendicular bisector of the line segment joining the two fixed points.]

$EF$ is the perpendicular bisector of the line segment $AB$.

Given: $\;$ $P$ is equidistant from $AB$ and $AD$.

$\implies$ $P$ lies on the angle bisector of lines segments $AB$ and $AD$.

[The locus of a point equidistant from two intersecting lines is the bisector of the angle between the lines.]

$AG$ is the angle bisector of $\angle DAB$.

$EF$ and $AG$ intersect at point $P$, which is the the point inside the quadrilateral such that $AP = BP$ and is equidistant from $AB$ and $AD$.

Length of $PB = 6.5 \; cm$

Locus

Find the points on the locus of points that are 3 units from X axis and 5 units from the point $(5,1)$.

Let $P \left(h,k\right)$ be a point on the required locus.

$P$ is 3 units from the X axis.

$\implies$ Equation of locus is: $\hspace{1em}$ $k = \pm 3$ $\;\;\; \cdots \; (1)$

Also, $P$ is 5 units from the point $(5,1)$.

i.e. $\sqrt{\left(h - 5\right)^2 + \left(k - 1\right)^2} = 5$

$\therefore$ $\;$ Equation of locus is: $\hspace{1em}$ $\left(h - 5\right)^2 + \left(k -1\right)^2 = 5^2$ $\;\;\; \cdots \; (2)$

Substituting $k = + 3$ in equation $(2)$ gives

$\left(h - 5\right)^2 + \left(3 - 1\right)^2 = 25$

i.e. $h^2 - 10 h + 25 + 4 = 25$

i.e. $h^2 - 10 h + 4 = 0$

i.e. $h = \dfrac{10 \pm \sqrt{100 - 16}}{2} = 5 \pm \sqrt{21}$

Substituting $k = - 3$ in equation $(2)$ gives

$h^2 - 10 h + 25 + \left(-3 - 1\right)^2 = 25$

i.e. $h^2 - 10 h + 16 = 0$

i.e. $\left(h - 2\right) \left(h - 8\right) = 0$

$\implies$ $h = 2$, $h = 8$

$\therefore$ $\;$ The points on the required locus are: $\hspace{1em}$ $\left(5 + \sqrt{21}, 3\right)$, $\left(5 - \sqrt{21}, 3\right)$, $\left(2, -3\right)$ and $\left(8, -3\right)$

Locus

If the points $P \left(6,2\right)$, $Q \left(-2,1\right)$ and $R$ are the vertices of $\triangle PQR$ and $R$ is the point on the locus $y = x^2 - 3x + 4$, then find the equation of locus of centroid of $\triangle PQR$.

Let $R \left(p,q\right)$ be the point on the locus.

$R$ satisfies the equation $y = x^2 - 3x + 4$

$\therefore$ $\;$ We have $\hspace{1em}$ $q = p^2 - 3p + 4$ $\;\;\; \cdots \; (1)$

Let $C \left(h,k\right)$ be a point on the equation of the locus of centroid of $\triangle PQR$.

Given: $P = \left(6,2\right)$ and $Q = \left(-2,1\right)$

$\therefore$ $\;$ Centroid of $\triangle PQR$ is

$\left(\dfrac{6 - 2 + p}{3}, \dfrac{2 + 1 + q}{3}\right) = \left(h, k\right)$

$\implies$ $\dfrac{4 + p}{3} = h$ $\implies$ $p = 3 h - 4$ $\;\;\; \cdots \; (2)$

and $\dfrac{3 + q}{3} = k$ $\implies$ $q = 3 k - 3$ $\;\;\; \cdots \; (3)$

Substituting the values of $p$ and $q$ from equations $(2)$ and $(3)$ in equation $(1)$ gives

$3 k - 3 = \left(3 h - 4\right)^2 - 3 \left(3 h - 4\right) + 4$

i.e. $3 k - 3 = 9 h^2 - 24 h + 16 - 9 h + 12 + 4$

i.e. $9 h^2 - 33 h - 3 k + 35 = 0$

$\therefore$ $\;$ The equation of locus of centroid of $\triangle PQR$ is: $\hspace{1em}$ $9x^2 - 33x - 3y + 35 = 0$

Locus

If $P \left(2,-7\right)$ is a given point and Q is a point on $2x^2 + 9y^2 = 18$, then find the equation of the locus of the midpoint of PQ.

Let $Q \left(a, b\right)$ be a point on $2x^2 + 9y^2 = 18$

Then, $2a^2 + 9b^2 = 18$ $\;\;\; \cdots \; (1)$

Coordinates of midpoint of $P \left(2,-7\right)$ and $Q \left(a,b\right)$ are $\left(\dfrac{2 + a}{2}, \dfrac{b - 7}{2}\right)$

Let $Z \left(h,k\right)$ be a point on the required locus.

Then, $h = \dfrac{a + 2}{2}$, $k = \dfrac{b - 7}{2}$

i.e. $a = 2 h - 2$ $\;\;\; \cdots \; (2a)$ and $b = 2 k + 7$ $\;\;\; \cdots \; (2b)$

Substituting the values of $a$ and $b$ from equations $(2a)$ and $(2b)$ in equation $(1)$ gives

$2 \left(2h - 2\right)^2 + 9 \left(2 k + 7\right)^2 = 18$

i.e. $2 \left(4 h^2 - 8h + 4\right) + 9 \left(4 k^2 + 28 k + 49\right) = 18$

i.e. $8 h^2 - 16 h + 8 + 36 k^2 + 252 k + 441 = 18$

i.e. $8 h^2 + 36 k^2 - 16 h + 252 k + 431 = 0$

$\therefore$ $\;$ The equation of the required locus is: $\hspace{1em}$ $8x^2 + 36y^2 - 16 x + 252 y + 431 = 0$

Locus

The coordinates of a moving point P are $\left(\dfrac{a}{2} \left(\text{cosec } \theta + \sin \theta\right), \dfrac{b}{2} \left(\text{cosec } \theta - \sin \theta\right)\right)$, where $\theta$ is a variable parameter. Find the equation of locus of P.

Let $P \left(h, k\right)$ be the moving point.

Then as per question,

$h = \dfrac{a}{2} \left(\text{cosec } \theta + \sin \theta\right)$, $k = \dfrac{b}{2} \left(\text{cosec } \theta - \sin \theta\right)$

i.e. $\dfrac{2h}{a} = \text{cosec } \theta + \sin \theta$ $\;\;\; \cdots (1)$

and $\dfrac{2k}{b} = \text{cosec } \theta - \sin \theta$ $\;\;\; \cdots \; (2)$

Squaring and subtracting equations $(1)$ and $(2)$ gives

$\dfrac{4h^2}{a^2} - \dfrac{4k^2}{b^2} = \left(\text{cosec } \theta + \sin \theta\right)^2 - \left(\text{cosec } \theta - \sin \theta\right)^2$

i.e. $4\left(\dfrac{h^2}{a^2} - \dfrac{k^2}{b^2}\right) = \text{cosec}^2 \theta + \sin^2 \theta + 2 \text{cosec } \theta \sin \theta - \text{cosec}^2 \theta - \sin^2 \theta + 2 \text{cosec } \theta \sin \theta$

i.e. $4\left(\dfrac{h^2}{a^2} - \dfrac{k^2}{b^2}\right) = 4 \text{cosec } \theta \sin \theta$

i.e. $\dfrac{h^2}{a^2} - \dfrac{k^2}{b^2} = 1$

i.e. $h^2 b^2 - k^2 a^2 = a^2 b^2$

$\therefore$ $\;$ The required equation of locus is: $\hspace{1em}$ $x^2 b^2 - y^2 a^2 = a^2 b^2$

Locus

Find the equation of the locus of point P such that the line segment AB, joining the points $A \left(1,-6\right)$ and $B \left(4,-2\right)$, subtends a right angle at P.

Let $P \left(h, k\right)$ be a point on the required locus.

Since the line segment AB subtends a right angle at P, we have

$PA^2 + PB^2 = AB^2$ $\;\;\; \cdots \; (1)$

Given: $A = \left(1,-6\right)$, $B = \left(4,-2\right)$

$\begin{aligned} \therefore \; PA^2 & = \left(h - 1\right)^2 + \left(k + 6\right)^2 \\\\ & = h^2 + k^2 - 2h + 12 k + 1 + 37 \\\\ & = h^2 + k^2 - 2h + 12 k + 37 \;\;\; \cdots \; (2a) \end{aligned}$

$\begin{aligned} PB^2 & = \left(h - 4\right)^2 + \left(k + 2\right)^2 \\\\ & = h^2 + k^2 - 8 h + 4 k + 16 + 4 \\\\ & = h^2 + k^2 - 8 h + 4k +20 \;\;\; \cdots \; (2b) \end{aligned}$

and, $AB^2 = \left(1-4\right)^2 + \left(-6 + 2\right)^2 = 25$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2a)$, $(2b)$ and $(2c)$. equation $(1)$ becomes

$h^2 + k^2 - 2 h + 12 k + 37 + h^2 + k^2 - 8 h + 4 k + 20 = 25$

i.e. $2h^2 + 2 k^2 - 10 h + 16 k + 32 = 0$

i.e. $h^2 + k^2 - 5 h + 8k + 16 = 0$

$\therefore$ $\;$ The required equation of locus is: $\hspace{1em}$ $x^2 + y^2 - 5 x + 8 y + 16 = 0$