In this page concurrency of straightlines question8 we are going to see solution of eighth question on the quiz 'concurrency of straight lines'.
Definition:
If three or more straight lines passing through the same point then that common point is called the point of concurrency
Steps to find whether the three given lines are concurrent:
(i) Solve any two equations of the straight lines and obtain their point of intersection.
(ii) Substitute the co-ordinates of the point of intersection in the third equation.
(iii) Check whether the third equation is satisfied
(iv) If it is satisfied,the point lies on the third line and so the three straight lines are concurrent.
Concurrency of straightlines question8
Show that the straight lines 3 x - 4 y + 5 = 0,7 x - 8 y + 5 = 0 and
4 x + 5 y - 45 = 0 are concurrent.Find the point of concurrency.
Solution:
Let us see how to solve the concurrency of straightlines question8.
The given equations are
3 x - 4 y + 5 = 0 ----------(1)
7 x - 8 y + 5 = 0 ----------(2)
4 x + 5 y - 45 = 0 ----------(3)
First we need to solve any two equations then we have to plug the point into the remaining equation
In
the first equation coefficient of x is 3,in the second equation the
coefficient of x is 7 and we have same signs for both equations.But the
coefficient of y in the first equation is -4 and coefficient of y in the
second equation is -8 and we have same signs.
To make the coefficient of y of the first equation as 8 we need to multiply the the whole equation by 2.
Then we are going to subtract the first equation from the second equation since we have same signs.
(1) x 2 = > 6 x - 8 y + 10 = 0
6 x - 8 y + 10 = 0 ----------(1)
7 x - 8 y + 5 = 0 ----------(2)
(-) (+) (-)
----------------
- x + 5 = 0
- x = -5
x = 5
Substitute x = 5 in the first equation
3 x - 4 y + 5 = 0
3 (5)- 4 y + 5 = 0
15 - 4 y + 5 = 0
20 - 4 y = 0
- 4 y = - 20
y = - 20/(-4)
y = 5
So the point of intersection of the first and second line is (5,5)
Now we have to apply the point (5,5) in the third equation 4 x + 5 y - 45 = 0
4 (5) + 5 (5) - 45 = 0
20 + 25 - 45 = 0
45 - 45 = 0
0 = 0
The third equation is satisfied.So the point (5,5) lies on the lies on the third line.So the straight lines are concurrent. This is the solution for concurrency of straightlines question8.
Questions |
Solution |
(1) Show that the straight lines 2 x + y - 1 = 0, 2 x + 3 y - 3 = 0 and 3 x + 2 y - 2 = 0 are concurrent.Find the point of concurrency. | |
(2) Show that the straight lines x + y - 5 = 0, 3 x - y + 1 = 0 and 5 x - y - 1 = 0 are concurrent.Find the point of concurrency. | |
(3) Show that the straight lines x - y - 2 = 0, 3 x + 4 y + 15 = 0 and 5 x - 4 y - 7 = 0 are concurrent.Find the point of concurrency. | |
(4) Show that the straight lines x - 3 y + 3 = 0, 2 x + y - 8 = 0 and 5 x - 4 y - 7 = 0 are concurrent.Find the point of concurrency. | |
(5) Show that the straight lines x + y - 7 = 0, 2 x + y - 16 = 0 and 3 x + 8 y - 11 = 0 are concurrent.Find the point of concurrency. | |
(6) Show that the straight lines x + y - 3 = 0, x + 2 y - 5 = 0 and x + 3 y - 7 = 0 are concurrent.Find the point of concurrency. | |
(7) Show that the straight lines 3 x + y + 2 = 0, 2 x - y + 3 = 0 and x + 4 y - 3 = 0 are concurrent.Find the point of concurrency. | |
(8) Show that the straight lines 3 x - 4 y +5=0, 7 x - 8 y + 5 = 0 and 4 x + 5 y - 45 = 0 are concurrent.Find the point of concurrency. | |
(10) Show that the straight lines y - x - 1 = 0, 2 x + y + 2 = 0 and 5 x + 3 y + 5 = 0 are concurrent.Find the point of concurrency. |
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