Transform `12x^(2)+7xy-12y^(2)-17x-31y-7=0` to rectangular axes through the point (1, -1) inclined at an angle `tan^(-1)((4)/(3))` to the original axes.

Answer

To transform the equation \(12x^2 + 7xy - 12y^2 - 17x - 31y - 7 = 0\) to rectangular axes through the point \((1, -1)\) inclined at an angle \(\tan^{-1}\left(\frac{4}{3}\right)\) to the original axes, we will follow these steps:

### Step 1: Shift the Origin
We first shift the origin to the point \((1, -1)\). We do this by substituting:
\[ x = x' + 1 \quad \text{and} \quad y = y' - 1 \]
Substituting these into the original equation gives:
\[ 12(x' + 1)^2 + 7(x' + 1)(y' - 1) - 12(y' - 1)^2 - 17(x' + 1) - 31(y' - 1) - 7 = 0 \]

### Step 2: Expand the Equation
Now we expand the equation:
1. Expand \(12(x' + 1)^2\):
\[ 12(x'^2 + 2x' + 1) = 12x'^2 + 24x' + 12 \]
2. Expand \(7(x' + 1)(y' - 1)\):
\[ 7(x'y' - x' + y' - 1) = 7x'y' + 7y' - 7x' - 7 \]
3. Expand \(-12(y' - 1)^2\):
\[ -12(y'^2 - 2y' + 1) = -12y'^2 + 24y' - 12 \]
4. Expand \(-17(x' + 1)\):
\[ -17x' - 17 \]
5. Expand \(-31(y' - 1)\):
\[ -31y' + 31 \]

Combining all these expansions, we have:
\[ 12x'^2 + 7x'y' - 12y'^2 + (24x' + 7y' - 7x' + 24y' - 17 - 31 + 12) = 0 \]

### Step 3: Combine Like Terms
Now, we combine like terms:
- For \(x'\): \(12x'^2 + (24 - 7 - 17)x' = 12x'^2 + 0x' = 12x'^2\)
- For \(y'\): \((7 + 24 - 31)y' = 0y' = 0\)
- For \(y'^2\): \(-12y'^2\)

Thus, the equation simplifies to:
\[ 12x'^2 - 12y'^2 - 7 = 0 \]

### Step 4: Rotate the Axes
Next, we need to rotate the axes by the angle \(\theta = \tan^{-1}\left(\frac{4}{3}\right)\). The sine and cosine of this angle can be determined from a right triangle:
- Opposite = 4, Adjacent = 3, Hypotenuse = 5
- \(\sin \theta = \frac{4}{5}\), \(\cos \theta = \frac{3}{5}\)

Using the rotation formulas:
\[ x' = x \cos \theta - y \sin \theta = \frac{3}{5}x - \frac{4}{5}y \]
\[ y' = x \sin \theta + y \cos \theta = \frac{4}{5}x + \frac{3}{5}y \]

### Step 5: Substitute Back into the Equation
Substituting \(x'\) and \(y'\) into the simplified equation \(12x'^2 - 12y'^2 - 7 = 0\):
\[ 12\left(\frac{3}{5}x - \frac{4}{5}y\right)^2 - 12\left(\frac{4}{5}x + \frac{3}{5}y\right)^2 - 7 = 0 \]

### Step 6: Expand and Simplify
Expanding both squares:
1. For \(x'^2\):
\[ \left(\frac{3}{5}x - \frac{4}{5}y\right)^2 = \frac{9}{25}x^2 - \frac{24}{25}xy + \frac{16}{25}y^2 \]
Thus,
\[ 12\left(\frac{9}{25}x^2 - \frac{24}{25}xy + \frac{16}{25}y^2\right) = \frac{108}{25}x^2 - \frac{288}{25}xy + \frac{192}{25}y^2 \]
2. For \(y'^2\):
\[ \left(\frac{4}{5}x + \frac{3}{5}y\right)^2 = \frac{16}{25}x^2 + \frac{24}{25}xy + \frac{9}{25}y^2 \]
Thus,
\[ -12\left(\frac{16}{25}x^2 + \frac{24}{25}xy + \frac{9}{25}y^2\right) = -\frac{192}{25}x^2 - \frac{288}{25}xy - \frac{108}{25}y^2 \]

### Step 7: Combine and Set to Zero
Combining these results:
\[ \left(\frac{108}{25} - \frac{192}{25}\right)x^2 + \left(-\frac{288}{25} - \frac{288}{25}\right)xy + \left(\frac{192}{25} - \frac{108}{25}\right)y^2 - 7 = 0 \]
This simplifies to:
\[ -\frac{84}{25}x^2 - \frac{576}{25}xy + \frac{84}{25}y^2 - 7 = 0 \]

### Final Step: Multiply to Clear Denominators
Multiply through by \(-25\) to clear the denominators:
\[ 84x^2 + 576xy - 84y^2 + 175 = 0 \]

This is the transformed equation in rectangular axes.