(a) If K(−3, − x) and L(3,−4) are such that KL is perpendicular to MN with M(− 5, −6) and N(−3,− 5), find the value of x.
(b) The curve y = ax2 + bx + c passes through the points (0, 2) and (1, 3) and its gradient where x = 1 is 6. Find the value of a, of b and of c;
(a) To solve this problem, we need to use the concept of perpendicular gradients. Two lines are perpendicular if and only if the product of their gradients is -1.
Step 1: Find the gradient of line KL.
The gradient of a line between two points (x1, y1) and (x2, y2) is given by the formula: m = (y2 - y1) / (x2 - x1).
Using the given points K(-3, -x) and L(3, -4), we can find the gradient of line KL:
mKL = (-4 - (-x)) / (3 - (-3))
= (-4 + x) / 6
Step 2: Find the gradient of line MN.
Using the given points M(-5, -6) and N(-3, -5), we can find the gradient of line MN:
mMN = (-5 - (-6)) / (-3 - (-5))
= (-5 + 6) / (-3 + 5)
= 1/2
Step 3: Set up the equation for perpendicular lines.
Since KL is perpendicular to MN, the product of their gradients will be -1:
mKL * mMN = -1
((-4 + x) / 6) * (1/2) = -1
Step 4: Solve for x.
Multiply both sides of the equation by 6 to eliminate the denominators:
(-4 + x) * (1/2) = -1 * 6
(-4 + x) / 2 = -6
-4 + x = -12
x = -12 + 4
x = -8
Therefore, the value of x is -8.
(b) To solve this problem, we need to use the given information about the curve and its gradient at a specific point.
Step 1: Set up the equation for the curve.
Given that the curve is represented by the equation y = ax^2 + bx + c, we can substitute the coordinates of the two given points (0, 2) and (1, 3) to form two equations:
Equation 1: 2 = a(0)^2 + b(0) + c (substituting x = 0 and y = 2)
2 = c
Equation 2: 3 = a(1)^2 + b(1) + c (substituting x = 1 and y = 3)
3 = a + b + c
Step 2: Find the gradient at x = 1.
The gradient of the curve at a specific point is given by the derivative of the curve equation with respect to x. Since we are interested in the gradient where x = 1, we need to find the derivative and evaluate it at x = 1.
Differentiating the equation y = ax^2 + bx + c with respect to x gives us:
dy/dx = 2ax + b
Evaluating the gradient at x = 1:
dy/dx (x = 1) = 2a(1) + b = 6
Step 3: Substitute the value of c in Equation 2.
From Equation 1, we found that c = 2. Substituting this value into Equation 2 gives us:
3 = a + b + 2
Step 4: Solve the system of equations.
We now have two equations:
a + b + 2 = 3 (Equation 2)
2a + b = 6 (Gradient equation)
Using these equations, we can solve for the values of a and b.
From Equation 2, we have:
a + b = 1 (subtracting 2 from both sides)
Substituting this into the Gradient equation, we have:
2(1) + 1 = 6
2 + 1 = 6
3 = 6
Since the equation 3 = 6 is not true, this means that there is no solution for the system of equations.
Therefore, we cannot determine the values of a, b, and c for the given conditions.