Question

4
the equation of line a is y = 7x + 14.
line b is perpendicular to line a and passes through the point \\(\left(4, 13 \frac{3}{7}\
ight)\\).
what would be the solution to the system of equations represented by line a and line b?
\\(\circ\\) a. the system of equations has no real solutions.
\\(\circ\\) b. \\(x = 14, y = 0\\)
\\(\circ\\) c. \\(x = 0, y = 14\\)
\\(\circ\\) d. the system of equations has an infinite number of real solutions.

Explanation:

Step1: Find slope of line B

Slope of A is 7, so slope of B is $-\frac{1}{7}$.

Step2: Write equation of line B

Use point-slope form: $y - 13\frac{3}{7} = -\frac{1}{7}(x - 4)$ → $y = -\frac{1}{7}x + 14$.

Step3: Solve the system

Set $7x + 14 = -\frac{1}{7}x + 14$ → $7x + \frac{1}{7}x = 0$ → $x = 0$. Substitute $x=0$ into A: $y=14$.

Step4: Verify solution

Solution is $(0,14)$.

Answer:

C. $x = 0, y = 14$