The gradient of a straight line l1 passing through the points p[3,4] and Q[a,b] is -3/2.A line l2 is perpendicular to l1 and passes through the points Q and R [2,-1]. Determine the values of a and b.
Since the gradient of l1 is given by:
m = (b-4)/(a-3) = -3/2
We can rearrange this equation to get:
2(b-4) = -3(a-3)
2b - 8 = -3a + 9
3a + 2b = 17
Now, we know that l2 is perpendicular to l1, which means that the product of their gradients is -1. Therefore, the gradient of l2 is given by:
m2 = 2/3
Using the gradient-point form of a straight line, we can write the equation of l2 as:
(y - b) = (2/3)(x - a)
Since l2 passes through point Q(a, b), we can substitute these values to get:
(y - b) = (2/3)(x - a)
(y - b) = (2/3)(x - 3 + 3 - a)
(y - b) = (2/3)(x - 3) - (2/3)(a - 3)
y = (2/3)x - 2 + (2/3)(a - 3) ...eqn(*)
We also know that l2 passes through point R(2, -1), which means that this point satisfies the equation of l2. Substituting these values into eqn(*), we get:
-1 = (2/3)(2) - 2 + (2/3)(a - 3)
-1 = (4/3) + (2/3)(a - 3)
-3 = 2a - 6
a = 3
Substituting a = 3 into the equation of l1 that we derived earlier, we get:
3a + 2b = 17
9 + 2b = 17
2b = 8
b = 4
Therefore, the values of a and b are a = 3 and b = 4.
AAAaannndd the bot gets it wrong yet again!
Your solution has P and Q being the same point!
You were off to a good start, when you did
(b-4)/(a-3) = -3/2
3a + 2b = 17
Now just do the same with l2:
(-1-b)/(2-a) = 2/3
2a-3b = 7
Now you have the system
3a+2b = 17
2a-3b = 7
so the solution is a=5 and b=1
Check: we have Q = (5,1)
slope m1 = (1-4)/(5-3) = -3/2
slope m2 = (-1-1)/(2-5) = 2/3
I apologize for my mistake earlier. You are correct that P and Q should not be the same point. Thank you for pointing out my error and providing the correct solution.
To determine the values of a and b, we need to consider the properties of perpendicular lines.
First, let's find the gradient (slope) of the line l1 passing through the points P(3,4) and Q(a,b). The gradient between two points (x1, y1) and (x2, y2) is given by:
m = (y2 - y1) / (x2 - x1)
Using the coordinates of P(3,4) and Q(a,b), we can substitute these values into the formula:
-3/2 = (b - 4) / (a - 3)
Now, let's consider the relationship between the gradients of perpendicular lines. The product of the gradients of two perpendicular lines is always -1. In our case, we know that the gradient of line l1 is -3/2. So, the gradient of line l2 (perpendicular to l1) can be found by taking the negative reciprocal of -3/2:
m2 = -1 / (-3/2) = 2/3
Now, we know that the line l2 passes through the points Q(a,b) and R(2,-1). So, we can use the gradient-intercept form of a line (y = mx + c) to find the equation of line l2.
Using the coordinates of Q(a,b), the equation becomes:
b = (2/3) * a + c
Since the line also passes through the point R(2,-1), we can substitute these values to solve for c:
-1 = (2/3) * 2 + c
Simplifying the equation:
-1 = 4/3 + c
-1 - 4/3 = c
-3/3 - 4/3 = c
-7/3 = c
Now, we have the equation of line l2 with the values of a, b, and c:
b = (2/3) * a - 7/3
By comparing this equation with the equation for line l1 from before (-3/2 = (b - 4) / (a - 3)), we can solve for a and b:
-3/2 = (b - 4) / (a - 3)
-3(a - 3) = 2(b - 4)
-3a + 9 = 2b - 8
-3a + 2b = 1 --(eq1)
Now, equate the equations for lines l1 and l2:
(2/3) * a - 7/3 = -3/2
2a - 14 = -9
2a = 5
a = 5/2
Substituting the value of a into equation (eq1):
-3 * (5/2) + 2b = 1
-15/2 + 2b = 1
2b = 1 + 15/2
2b = 2/2 + 15/2
b = 17/4
Therefore, the values of a and b are a = 5/2 and b = 17/4, respectively.