Equation of line solution16
In this page equation of line solution16 we are going to see solution of each problem with detailed explanation of the worksheet slope of the line.
(17) If x + 2 y = 7 and 2 x + 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) lie on the circle.
Solution:
Two diameters are intersecting at the point E
x + 2 y = 7 --- (1)
2 x + y = 8 --- (2)
(1) x 2 = > 2 x + 4 y = 14
(1) - (2) 2 x + y = 8
(-) (-) (-)
--------------
3 y = 6
y = 6/3
y = 2
Substitute y = 2 in the first equation
x + 2 (2) = 7
x + 4 = 7
x = 7 - 4
x = 3
Therefore two diameters are intersecting at the point E (3 , 2)
Now we have to find the measurement of radius for that we have to find the distance between the points F(0 , -2) and E(3 , 2)
d = √(x₂ - x₁)² + (y₂ - y₁)²
= √(3 - 0)² + (2 - (-2))²
= √(3)² + (2 + 2)²
= √9 + 4²
= √9 + 16
= √25
= √5 x 5
= 5 units
(18) Find the equation of the straight line segment whose end points are the point of intersection of the straight lines 2 x – 3 y + 4 = 0, x – 2 y + 3 = 0 and the midpoint of the line joining the points (3 ,-2) and (-5 , 8).
Solution:
2 x – 3 y + 4 = 0 ----- (1)
x – 2 y + 3 = 0 ----- (2)
Now we have to find the point of intersection of those two lines
2 x – 3 y + 4 = 0
(2) x 2 = > 2 x - 4 y + 6 = 0
(-) (+) (-)
------------------
y - 2 = 0
y = 2
Substitute y = 2 in the first equation
2 x - 3 (2) + 4 = 0
2 x - 6 + 4 = 0
2 x - 2 = 0
2 x = 2
x = 2/2
x = 1
Now we have to find the midpoint of the line segment joining the points
(3 ,-2) and (-5 , 8)
Midpoint = (x₁ + x₂)/2 , (y₁ + y₂)/2
= (3 - 5)/2 , (- 2 + 8)/2
= -2/2 , 6/2
= (-1 , 3)
(19) If the isosceles triangle PQR, PQ = PR. The base QR lies on the axis, P lies on the y-axis and 2 x – 3 y + 9 =0 is the equation of PQ. Find the equation of PQ. Find the equation of the straight line along PR.
To find the point point P which lies on the y-axis we have to put 0 instead of x in the given equation PQ
2(0) - 3 y = - 9
- 3 y = -9
y = (-9)/(-3)
y = 3
Therefore the point P is (0,3)
To find the point point Q which lies on the x-axis we have to put 0 instead of y in the given equation QR
2 x - 3 (0) = - 9
2 x - 0 = -9
2 x = - 9
x = -9/2
Therefore 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
P (3 , 0) R (9/2 , 0)
(y-y₁)/(y₂ - y₁) = (x-x₁)/(x₂ - x₁)
(y - 3)/(0 - 3) = [x - (9/2)]/[(9/2) - 3)]
(y - 3)/(- 3) = [2x - 9]/[(9-6)/2]
(y - 3)/(- 3) = [2x - 9]/(3/2)
(y - 3)/(- 3) = 2[2x - 9]/3
-y + 3 = 4 x - 18
4 x + y - 18 - 3 = 0
4 x + y - 21 = 0
equation of line solution16 equation of line solution16 equation of line solution16
