Find the coordinates of the point in which the line `2y-3x+ 7=0`
meets the line joining the two points (6, - 2) and ( - 8, 7). Find also the angle between them.

Answer

To solve the problem step by step, we need to find the intersection point of the line given by the equation \(2y - 3x + 7 = 0\) and the line joining the points (6, -2) and (-8, 7). We will also calculate the angle between the two lines.

### Step 1: Find the equation of the line joining the points (6, -2) and (-8, 7).

Using the two-point form of the equation of a line:

\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]

Where \((x_1, y_1) = (6, -2)\) and \((x_2, y_2) = (-8, 7)\).

Calculating the slope:

\[ \text{slope} = \frac{7 - (-2)}{-8 - 6} = \frac{9}{-14} = -\frac{9}{14} \]

Now substituting into the line equation:

\[ y + 2 = -\frac{9}{14}(x - 6) \]

Multiplying through by 14 to eliminate the fraction:

\[ 14(y + 2) = -9(x - 6) \]

Expanding:

\[ 14y + 28 = -9x + 54 \]

Rearranging gives:

\[ 9x + 14y - 26 = 0 \]

### Step 2: Find the intersection point of the two lines.

We have the two equations:

1. \(2y - 3x + 7 = 0\) (Equation 1)
2. \(9x + 14y - 26 = 0\) (Equation 2)

From Equation 1, we can express \(y\) in terms of \(x\):

\[ 2y = 3x - 7 \implies y = \frac{3x - 7}{2} \]

Now substitute this expression for \(y\) into Equation 2:

\[ 9x + 14\left(\frac{3x - 7}{2}\right) - 26 = 0 \]

Multiplying through by 2 to eliminate the fraction:

\[ 18x + 14(3x - 7) - 52 = 0 \]

Expanding:

\[ 18x + 42x - 98 - 52 = 0 \]

Combining like terms:

\[ 60x - 150 = 0 \implies 60x = 150 \implies x = \frac{150}{60} = \frac{5}{2} \]

Now substitute \(x = \frac{5}{2}\) back into the equation for \(y\):

\[ y = \frac{3\left(\frac{5}{2}\right) - 7}{2} = \frac{\frac{15}{2} - 7}{2} = \frac{\frac{15}{2} - \frac{14}{2}}{2} = \frac{\frac{1}{2}}{2} = \frac{1}{4} \]

Thus, the intersection point is:

\[ \left(\frac{5}{2}, \frac{1}{4}\right) \]

### Step 3: Find the angle between the two lines.

The slopes of the lines are:

- For line 1: \(m_1 = \frac{3}{2}\) (from \(2y - 3x + 7 = 0\))
- For line 2: \(m_2 = -\frac{9}{14}\)

The formula for the angle \(\theta\) between two lines is given by:

\[ \tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \]

Substituting the values:

\[ \tan \theta = \left|\frac{\frac{3}{2} - \left(-\frac{9}{14}\right)}{1 + \frac{3}{2} \cdot \left(-\frac{9}{14}\right)}\right| \]

Calculating \(m_1 - m_2\):

\[ \frac{3}{2} + \frac{9}{14} = \frac{21}{14} + \frac{9}{14} = \frac{30}{14} = \frac{15}{7} \]

Calculating \(1 + m_1 m_2\):

\[ 1 + \frac{3}{2} \cdot \left(-\frac{9}{14}\right) = 1 - \frac{27}{28} = \frac{28}{28} - \frac{27}{28} = \frac{1}{28} \]

Thus,

\[ \tan \theta = \left|\frac{\frac{15}{7}}{\frac{1}{28}}\right| = \frac{15}{7} \cdot 28 = \frac{420}{7} = 60 \]

Finally, we can find \(\theta\) using \(\tan^{-1}(60)\).

### Summary of Results:
- The coordinates of the intersection point are \(\left(\frac{5}{2}, \frac{1}{4}\right)\).
- The angle between the two lines is \(\tan^{-1}(60)\).