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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.
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:
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,
PA^{2} = PB^{2}
or, (x-1)^{2 }+ (y-3)^{2} = (x-2)^{2 }+ (y-4)^{2}
or, x^{2} – 2x + 1 + y^{2} – 6y + 9 = x^{2} – 4x + 4 + y^{2} – 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.
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, x^{2} + 2x + 1 + y^{2} – 8y + 16 = 81
Thus, x^{2} + y^{2} + 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.
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:
Let PR be the line passing through the given point P(x_{1}, y_{1}) 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 = \(\frac {QN}{PN}\)
= \(\frac {QM-NM}{LM}\)
= \(\frac {QM-PL}{OM-OL}\)
= \(\frac {y-y_{1}}{x-x_{1}}\)
Thus, y-y_{1} = m (x-x_{1}), 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,
x_{1} = 3, y_{1} = -2, m =-½
We know that,
y-y_{1} = m (x-x_{1})
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 (x_{1}, y_{1}) and (x_{2}, y_{2}). Since, a line passes through these points, its slope m is given as ......... m = \(\frac {y_{2}-y_{1}}{x_{2}-x_{1}}\).
We know, y-y_{1} = m (x-x_{1})
Thus, y-y_{1} = \(\frac {y_{2}-y_{1}}{x_{2}-x_{1}}\) (x-x_{1}), 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,
(x_{1}, y_{1}) = (4, 7) & (x_{2}, y_{2}) = (7, 9)
We know,
y-y_{1} = \(\frac {y_{2}-y_{1}}{ x_{2}-x_{1}}\) (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:
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.
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.
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:
Let the two straight lines be:
If (x_{1}, y_{1}) be the co-ordinates of the point of intersection, then we have,
ax_{1} + by_{1} + c = 0
a’x_{1} + b’y_{1} + c’ = 0
On solving, we get:
\(\frac {x_{1}}{bc’-b’c}\) = \(\frac {y_{1}}{a’c-ac’}\) = \(\frac {1}{ab’-a’b}\)
So, \(\frac {x_{1}}{bc’-b’c}\) = \(\frac {1}{ab’-a’b}\)
Thus, x_{1} = \(\frac {bc’-b’c }{ab’-a’b}\)
Again,
\(\frac {y_{1}}{a’c-ac’}\) = \(\frac {1}{ab’-a’b}\)
Thus, y_{1} = \(\frac {a’c-ac’}{ab’-a’b}\)
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:
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