Example 1 :
If x + 2y = 7 and 2x + y = 8 are the equations of the lines of two diameter of the circle, find the radius of the circle if the point (0, -2) lies on the circle.
Solution :
Two diameters are intersecting at the point E
x + 2y = 7 ----(1)
2x + y = 8 ----(2)
2(1) - (2) :
2(x + 2y) - (2x + y) = 14 - 8
2x + 4y - 2x - y = 6
3y = 6
y = 2
Substitute y = 2 in (1).
x + 2(2) = 7
x + 4 = 7
x = 7 - 4
x = 3
Diameters are intersecting at the point E (3, 2).
So, center of the circle is E(3, 2).
To find radius, find the distance between the points F and E.
d = √[(x_{2 }- x_{1})^{2 }+ (y_{2 }- y_{1})^{2}]
Substitute (x_{1}, y_{1}) = E(3, 2) and (x_{2}, y_{2}) = F(0, -2).
= √[(0 - 3)^{2 }+ (-2 - 2)^{2}]
= √(9 + 16)
= √25
= 5
Radius of the circle is 5 units.
Example 2 :
Find the equation of the straight line segment whose end points are the point of intersection of the straight lines
2x – 3y + 4 = 0, x – 2y + 3 = 0
and the midpoint of the line joining (3, -2) and (-5, 8).
Solution :
2x – 3y + 4 = 0 ----(1)
x – 2y + 3 = 0 ----(2)
Let A be the point of intersection of (1) and (2) and B be the midpoint of the line joining (3, -2) and (-5, 8).
Target : Find the equation of the line AB.
(1) - 2(2) :
2x - 3y + 4 - 2(x - 2y + 3) = 0
2x - 3y + 4 - 2x + 4y - 6 = 0
y - 2 = 0
y = 2
Substitute y = 2 in (1).
2x - 3(2) + 4 = 0
2x - 6 + 4 = 0
2x - 2 = 0
x = 1
Point of intersection of the lines (1) and (2) is A(1, 2).
Find the midpoint of line joining (3, -2) and (-5, 8).
Midpoint = ((x_{1 }+ x_{2})/2, (y_{1 }+ y_{2})/2)
= ((3 - 5)/2, (-2 + 8)/2)
= (-2/2, 6/2)
= B(-1, 3)
Find the slope of the line AB :
m = (y_{2} - y_{1})/(x_{2} - x_{1})
Substitute (x_{1}, y_{1}) = A(1, 2) and (x_{2}, y_{2}) = B(-1, 3).
m = (3 - 2)/(-1 - 1)
= 1/(-2)
= -1/2
Equation of the line AB :
y - y_{1} = m(x - x_{1})
Substitute (x_{1}, y_{1}) = A(1, 2) and m = -1/2.
y - 2 = (-1/2)(x - 1)
2(y - 2) = -1(x - 1)
2y - 4 = -x + 1
x + 2y - 5 = 0
Example 3 :
If the isosceles triangle PQR, PQ = PR. The base QR lies on the axis, P lies on the y-axis and
2x – 3y + 9 = 0
is the equation of PQ. Find the equation of PQ. Find the equation of the straight line along PR.
Solution :
To find the point point P which lies on the y-axis, substitute x = 0.
2(0) - 3y = -9
-3y = -9
y = 3
The point P is (0, 3)
To find the point point Q which lies on the x-axis, substitute y = 0.
2x - 3(0) = -9
2x - 0 = -9
2x = -9
x = -9/2
The point Q is (-9/2 , 0)
The length of the sides PQ and PR are same. So the point R is (9/2,0)
Equation of PR :
(y-y_{1})/(y_{2}-y_{1}) = (x-x_{1})/(x_{2}-x_{1})
Substitute (x_{1}, y_{1}) = P(3, 0) and (x_{2}, y_{2}) = R(9/2, 0).
(y - 3)/(0 - 3) = (x - 9/2)/(9/2 - 3)
(y - 3)/(-3) = (2x - 9)/(3/2)
-(y - 3) = 2(2x - 9)
- y + 3 = 4x - 18
4x + y - 21 = 0
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