Find any possible values for x and y if the point (x,y) lies on the line 4y-16x+12=0

1 answer

To find the possible values for \(x\) and \(y\) that lie on the line described by the equation \(4y - 16x + 12 = 0\), we can start by rearranging this equation into the slope-intercept form \(y = mx + b\).

  1. Begin with the original equation: \[ 4y - 16x + 12 = 0 \]

  2. Isolate \(4y\): \[ 4y = 16x - 12 \]

  3. Divide everything by 4 to solve for \(y\): \[ y = 4x - 3 \]

Now, we have expressed \(y\) in terms of \(x\). This means that for any value of \(x\), we can find a corresponding value of \(y\).

Example values:

  • If \(x = 0\): \[ y = 4(0) - 3 = -3 \quad \Rightarrow \quad (0, -3) \]

  • If \(x = 1\): \[ y = 4(1) - 3 = 1 \quad \Rightarrow \quad (1, 1) \]

  • If \(x = 2\): \[ y = 4(2) - 3 = 5 \quad \Rightarrow \quad (2, 5) \]

  • If \(x = -1\): \[ y = 4(-1) - 3 = -7 \quad \Rightarrow \quad (-1, -7) \]

Conclusion:

The point \((x, y)\) can take on infinitely many values of \(x\) and the corresponding \(y\) can be calculated using the equation \(y = 4x - 3\). Thus, any point \((x, 4x - 3)\) will lie on the line defined by the equation \(4y - 16x + 12 = 0\).