JEE Mathematics Locus Notes Set 01

Important Notes and Formulas

Locus of a Point

The path traced by a point that moves under a given geometrical condition is called its locus (plural: loci). Three standard loci are used constantly:
1. A point at a constant distance from a fixed point traces a circle. The fixed point is the centre and the constant distance is the radius.
2. A point equidistant from two fixed points traces the perpendicular bisector of the segment joining them.
3. A point equidistant from two given lines traces the pair of angle bisectors of the lines, the internal and the external bisector.

Equation of a Locus

To find the equation of a locus, take a general point \( P(x,y) \) on it, translate the given condition into an equation in \( x \) and \( y \), and simplify. The result is the equation of the locus.
Straight line parallel to the y-axis at distance 5 from it: every point has x-coordinate 5, so the locus is \( x = 5 \).
Straight line parallel to the x-axis at distance 4 from it: every point has y-coordinate 4, so the locus is \( y = 4 \).
Circle with centre \( C(a,b) \) and radius \( r \): the condition \( CP = r \) gives \( (x-a)^2 + (y-b)^2 = r^2 \).

Angle Between Two Lines

Let two lines have slopes \( m_1 \) and \( m_2 \), that is \( y = m_1x + c_1 \) and \( y = m_2x + c_2 \). If \( \theta \) is the angle between them, then
\[ \tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \]
This follows from \( \tan(\alpha - \beta) = \dfrac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta} \), where \( \alpha, \beta \) are the angles the lines make with the x-axis.

Parallel Lines

Two lines are parallel when the angle between them is zero, that is, when their slopes are equal: \( m_1 = m_2 \).
A line parallel to \( ax + by + c = 0 \) can always be written as \( ax + by + k = 0 \), where only the constant term changes.

Perpendicular Lines

Two lines are perpendicular when \( \tan\theta \) is undefined, that is, when the denominator \( 1 + m_1m_2 \) is zero. So the condition is \( m_1 m_2 = -1 \).
A line perpendicular to \( ax + by + c = 0 \) can be written as \( bx - ay + k = 0 \).

Solved Examples

Example 1. Find the equation of the locus of a point that moves so that its distances from the points \( A(0, a) \) and \( B(0, -a) \) are equal.
Answer:
Let \( P(x,y) \) be the moving point. The condition is \( PA = PB \).
Squaring, \( x^2 + (y-a)^2 = x^2 + (y+a)^2 \).
Hence \( (y+a)^2 - (y-a)^2 = 0 \), that is \( 4ay = 0 \), so \( y = 0 \).
The locus is the x-axis, which is the perpendicular bisector of AB.

Example 2. A point moves so that its distance from \( A(2,4) \) is three times its distance from \( B(-3,0) \). Find the equation of its locus.
Answer:
Let \( P(x,y) \) be the point. The condition \( PA = 3\,PB \) gives \( PA^2 = 9\,PB^2 \).
\( (x-2)^2 + (y-4)^2 = 9\left[(x+3)^2 + y^2\right] \).
Left side: \( x^2 + y^2 - 4x - 8y + 20 \). Right side: \( 9x^2 + 9y^2 + 54x + 81 \).
Moving everything to one side, \( 8x^2 + 8y^2 + 58x + 8y + 61 = 0 \), which is a circle.

Example 3. Find the equation of the line through \( (-5, 4) \) parallel to \( 4x + 3y + 5 = 0 \).
Answer:
Any line parallel to the given line has the form \( 4x + 3y + k = 0 \).
It passes through \( (-5,4) \): \( 4(-5) + 3(4) + k = 0 \), so \( -20 + 12 + k = 0 \) and \( k = 8 \).
The required line is \( 4x + 3y + 8 = 0 \).

Example 4. Find the equation of the line through \( (-5, 4) \) perpendicular to \( 4x + 3y + 5 = 0 \).
Answer:
The slope of the given line is \( -\dfrac43 \), so the required slope is \( \dfrac34 \).
Point-slope form: \( y - 4 = \dfrac34 (x + 5) \), that is \( 4y - 16 = 3x + 15 \).
So the line is \( 3x - 4y + 31 = 0 \).
Check by the other method: a perpendicular line has the form \( 3x - 4y + k = 0 \); putting in \( (-5,4) \) gives \( -15 - 16 + k = 0 \), so \( k = 31 \).

Example 5. Find the acute angle between the lines \( y = 3x + 1 \) and \( y = \dfrac12 x - 4 \).
Answer:
Here \( m_1 = 3 \) and \( m_2 = \dfrac12 \).
\( \tan\theta = \left|\dfrac{3 - \frac12}{1 + \frac32}\right| = \dfrac{5/2}{5/2} = 1 \).
So \( \theta = 45^\circ \).