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Co-ordinate Geometry: Locus and Equation

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Co-ordinate Geometry:

Locus and Equation:

The path that is traced by a moving point, under a certain condition or conditions, is known as its locus.
For example, let P be the point which moves in such away that the distance between it and the fixed point at the center O always remains same and is considered equal to the radius r. Then obviously, it traces out a circle of radius r as shown below.

 

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In Co-ordinate geometry, the locus is represented by an equation. The equation of a locus is the relation between x & y that satisfies the co-ordinates of any point on the locus. A point’s co-ordinates must satisfy the equation of the locus & vice versa, if the point lies on a locus.

 

 

Methods for Finding the Equation of a locus:

  1. Take a point; be P(x, y) as a moving point whose locus is yet to be determined.
  2. Then, form an equation of the variables x & y, translating into the given algebraic condition, according to which the point moves.
  3. Finally, simplify the equation to have the required equation.

Example I: Find the locus of a point that is equidistant from the point (1, 3) and (2, 4).

Solution:

Let P(x, y) be any point on the locus. It is known that the distance of P from A(1, 3) and B(2, 4) is always same i.e. PA = PB.

Now,

PA = \(\sqrt{(x-1)2+(y-3)2}\)

PB = \(\sqrt{(x-2)2+(y-4)2}\)

Thus,

PA2 = PB2

or, (x-1)2 + (y-3)2 = (x-2)2 + (y-4)2
or, x2 – 2x + 1 + y2 – 6y + 9 = x2 – 4x + 4 + y2 – 8y + 16
or, 2x + 2y = 10

Hence, x + y = 5, which is the required equation.

Example II: Find the equation of the locus of a point which moves in such a way that its distance from (-1, 4) is always 9.

 

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Solution:

Let P(x, y) be a point moving in such a way that its distance from (-1, 4) is always 9 i.e. PA = 9.

Therefore,

(PA)2 = 81
or, (x+1)2 + (y-4)2 = 81
or, x2 + 2x + 1 + y2 – 8y + 16 = 81

Thus, x2 + y2 + 2x – 8y – 64 = 0, which is the required equation.

 

 

Equation of a Straight line:

In Co-ordinate geometry, the equation of a straight line means the mathematical representation of the line, in which, the relation between x and y is found, that is satisfied by the co-ordinates of each and every point on the line.

Equation of a straight line parallel to the axes:

Let PQ be the line parallel to the y-axis on its right side, at a distance a from it. So, the abscissa of any point on the line PQ will be a i.e. x = a for each and every point on the line.
Therefore, the equation of the line PQ is x = a.

 

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If the line was on the left side of the y-axis, its equation would be x = -a. Similarly, the equation of a straight line parallel to the x-axis would be y = b, where b is the distance of a straight line from the x-axis and the line is above the x-axis. If the straight line is below the x-axis, then y = -b.

Note: The equation of x-axis is y = 0 and the equation of y-axis is x = 0.

Example:

Find the equation of a straight line parallel to the x-axis and passing through the point (3, 2).

Solution:

We know,

The equation of a straight line parallel to the x-axis is y = b..... (i)

Since, the line passes through the (3, 2), we get 2 = b

Hence, from (i), y = 2, which is the required equation.

 

 

Equation of a Straight line passing through a given Point with a given Slope:

 

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Let PR be the line passing through the given point P(x1, y1) with a given slope m = tanθ.

Again, let Q (x, y) be any point on the line. Then, draw perpendiculars PL, QM, and PN from P & Q on the x-axis and PN from P on QM.

Then,

m = tanθ = \(\frac {QN}{PN}\)
= \(\frac {QM-NM}{LM}\)
= \(\frac {QM-PL}{OM-OL}\)
= \(\frac {y-y1}{x-x1}\)

Thus, y-y1 = m (x-x1), which is the required equation of the line in point-slope form.

Example:

Find the equation of a straight line passing through the point (3, -2) and having slope -½ .

Solution:

We have,

x1 = 3, y1 = -2, m =-½

We know that,

y-y1 = m (x-x1)
or, y+2 = -½(x-3)
or, 2y+4 = -x+3

Thus, x+2y+1 = 0 is the required equation.

 

 

Equation of a Straight line joining two Points:

Let the two points be (x1, y1) and (x2, y2). Since, a line passes through these points, its slope m is given as ......... m = \(\frac {y2-y1}{x2-x1}\).

We know, y-y1 = m (x-x1)

Thus, y-y1 = \(\frac {y2-y1}{x2-x1}\) (x-x1), which is the required equation of the straight line in two points form.

Example:

Find the equation of a straight line passing through the points (4, 7) and (7, 9).

Solution: Given,

(x1, y1) = (4, 7) & (x2, y2) = (7, 9)

We know,

y-y1 = \(\frac {y2-y1}{ x2-x1}\) (x-4)
or, y-7 = \(\frac {9-7}{7-4 }\) (x-4),
or, y-7 = \(\frac {2}{3}\) (x-4)
or, 3y-21 = 2x-8

Thus, 2x-3y+13 = 0 is the required equation of the straight line.

 

 

Intercepts:

 

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Let a line PQ cut the x-axis and y-axis at P and Q respectively. Then OP is said to be x-intercept and OQ is said to be y-intercept.

Equation of a straight line which cuts off a given intercept on the y-axis and is inclined at a given angle to the y-axis:

Let the straight line AB cut y-axis at L; such as OL = c (given y-intercept) and BQX = θ(say) to the x-axis.

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Let P(x, y) be any point on the line AB. Then, draw the perpendiculars PN and LM from P and L on the x-axis and on PN respectively.

Then,

PN = PM + MN
or, y = LM tanθ + OL
or y = x tanθ+ c

Thus, y = mx + c, which is the required equation of the straight line.

Note: If a line passes through the origin making an angle θ with the x-axis, then its equation is y = mx.

Example:

Find the equation of a straight line having the slope as 2 and y-intercept -5.

Solution: Here,

m = 2
y-intercept (c) = -5

We know,

y = mx + c
or, y = 2x – 5

Thus, 2x – y – 5 = 0 is the required equation.

 

 

Equation of a Straight line which cuts off given Intercepts from the Axes:

Let the line PQ cut off x-intercept, OP = a and y-intercept, OQ = b.

 

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Let A(x, y) be any point on the line. Then draw perpendicular AL from A on the x-axis.

Then, from similar triangles ALP and QOP, we get,

\(\frac {LA}{OQ}\) = \(\frac {LP}{OP}\) = \(\frac {OP-OL}{OP}\)
\(\frac {y}{b}\) = \(\frac {a-x}{a}\) = 1 - \(\frac {x}{a}\)

Thus, \(\frac {x}{a}\) + \(\frac {y}{b}\) = 1, which is the required equation of the straight line.

Example:

Find the equation of a straight line through the points (3, 4) and making equal intercepts on the axes.

Solution: We know,

\(\frac {x}{a}\) + \(\frac {y}{b}\) = 1 ............................. (i)

Since, a = b and the line passes through the point (3, 4), from (i), we get,

\(\frac {3}{a}\) + \(\frac {4}{a}\) = 1

Thus, a = 7 = b

Again, from (i) \(\frac {x}{7}\) + \(\frac {y}{7}\) = 1

Thus, x + y – 7 = 0 is the required equation.

 

 

The Point of Intersection of two Straight lines:

 

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Let the two straight lines be:

  • ax + by + c = 0 .......................... (i)
  • a’x + b’y + c’ = 0 ....................... (ii),

If (x1, y1) be the co-ordinates of the point of intersection, then we have,

ax1 + by1 + c = 0

a’x1 + b’y1 + c’ = 0

On solving, we get:

\(\frac {x1}{bc’-b’c}\) = \(\frac {y1}{a’c-ac’}\) = \(\frac {1}{ab’-a’b}\)

So, \(\frac {x1}{bc’-b’c}\) = \(\frac {1}{ab’-a’b}\)

Thus, x1 = \(\frac {bc’-b’c }{ab’-a’b}\)

Again,

\(\frac {y1}{a’c-ac’}\) = \(\frac {1}{ab’-a’b}\)

Thus, y1 = \(\frac {a’c-ac’}{ab’-a’b}\)

Therefore,

Point of Intersection = (\(\frac {bc’-b’c }{ab’-a’b}\) , \(\frac {a’c-ac’}{ab’-a’b}\))

Example: Find the point of intersection of the straight angle lines x – 4y + 3 = 0 and 6x + 3y + 4 = 0.

Solution:

Here, the straight lines are;

x – 4y + 3 = 0 ................(i)

6x + 3y + 4 = 0 ..............(ii)

Solving (i) and (ii), we get,

\(\frac {x}{-16-9}\) = \(\frac {y}{18-4}\) = \(\frac {1}{3+24}\)
or, \(\frac {x}{-25}\) = \(\frac {1}{27}\)

Thus, x = - \(\frac {25}{27}\)

Again,

\(\frac {y}{14}\) = \(\frac {1}{27}\)

Thus, y = \(\frac {14}{27}\)

Hence, the required co-ordinates are (- \(\frac {25}{27}\) , \(\frac {14}{27}\))

Note: The equations can be solved in any method.

(Tamang, Pant, & G.C, 2016)

 

 

 


Bibliography

Tamang, G., Pant, N., & G.C, P. B. (2016). Business Mathematics. Putalisadak: Asmita Publication.



Co-ordinate Geometry:

  • Locus and Equation
  • Methods for Finding the Equation of a locus
  • Equation of a Straight line
  • Equation of a Straight line passing through a given Point with a given Slope
  • Equation of a Straight line joining two Points
  • Intercepts
  • Equation of a Straight line which cuts off given Intercepts from the Axes
  • The Point of Intersection of two Straight lines

 

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