Find the constants a and b so that all the four lines whose equation are given by pass through the same point.
x + y = -1
-x + 3y =-11
ax + by =4
2a x -by =2
This means that there should be some point (x,y) for which all these equations are satisfied.
x + y = -1
-x + 3y =-11
Adding these two, we get
4y = -12
=> y = -3
=> x = 2
Now, we move on to the next two equations, where we put in x = 2, y = -3
2a -3b = 4
4a +3b = 2
Once again, we solve it as two linear equations in two variables.
Multiplying the first equation by two and subtracting the second,
2(2a -3b =4)
- (4a +3b =2)
=> -9b = 6
=> b = -2/3
=> a = 1
To find the constants a and b so that all four lines pass through the same point, we need to find the point of intersection of these lines.
First, let's solve the given system of equations to find the values of x and y at the point of intersection.
1) x + y = -1 ...(Eq.1)
2) -x + 3y = -11 ...(Eq.2)
3) ax + by = 4 ...(Eq.3)
4) 2ax - by = 2 ...(Eq.4)
We can solve this system of equations using any method like substitution or elimination. Let's use the elimination method by multiplying Eq.1 by 2 and adding it to Eq.2:
2 * (x + y) = 2 * (-1) => 2x + 2y = -2 ...(Eq.5)
-x + 3y = -11 ...(Eq.2)
Adding Eq.5 and Eq.2:
(2x + 2y) + (-x + 3y) = -2 + (-11)
x + 5y = -13 ...(Eq.6)
Now we have two equations, Eq.3 and Eq.6, with two unknowns a, b, which we can solve simultaneously.
Multiplying Eq.6 by a:
ax + 5ay = -13a ...(Eq.7)
Multiplying Eq.3 by 5:
5ax + 5by = 20 ...(Eq.8)
Adding Eq.7 and Eq.8:
(ax + 5ay) + (5ax + 5by) = -13a + 20
6ax + 5ay + 5by = -13a + 20
However, we can simplify this equation further by using Eq.4, where we have 2ax - by = 2:
Rearranging Eq.4:
2ax = by + 2 ...(Eq.9)
Substituting Eq.9 into the previous equation:
6ax + 5ay + 5by = -13a + 20
6(ax) + 5ay + 5by = -13a + 20
Substituting the value of 2ax from Eq.9:
6(by + 2) + 5ay + 5by = -13a + 20
Expanding and simplifying:
6by + 12 + 5ay + 5by = -13a + 20
11by + 5ay = -13a + 8 ...(Eq.10)
Now we have two equations, Eq.10 and Eq.6, with two unknowns a and b. We can solve these equations simultaneously to find a and b.
Multiplying Eq.6 by 11:
11x + 55y = -143 ...(Eq.11)
Multiplying Eq.10 by -5:
-55by - 25ay = 65a - 40 ...(Eq.12)
Adding Eq.11 and Eq.12:
(11x + 55y) + (-55by - 25ay) = -143 + (65a - 40)
11x - 25ay - 55by = 65a - 183
Now we have one equation, Eq.11, with one unknown x, and one equation, Eq.12 with the unknowns a and b.
Let's substitute the value of x from Eq.11 into Eq.12:
(11x - 55y) - 55by = 65a - 183
-15ay - 55by = 65a - 183 ...(Eq.13)
Now we have Eq.13, which is an equation with only a and b as unknowns. We can further simplify it:
Taking out common factors:
-5a(3y + 11b) = 65a - 183
Dividing both sides by -5a:
3y + 11b = -13 + (183/a) ...(Eq.14)
Now, we have Eq.14, an equation in terms of y, b, and a. We can further simplify it by looking at the coefficient of y, i.e., 3, and the coefficient of b, i.e., 11.
Since 3y + 11b = -13 + (183/a), we can deduce that the sum of y and b has a constant value, i.e., -13 + (183/a).
So, to ensure that all four lines pass through the same point, the constants a and b should be chosen such that -13 + (183/a) is constant.
In summary, to find the constants a and b so that all four lines pass through the same point, you need to choose a value for a and then calculate the corresponding value of b using the equation -13 + (183/a).