The equation of line passing through the point of intersection of the lines \(3 \mathrm{x}-2 \mathrm{y}=0\) and \(5 \mathrm{x}+\mathrm{y}-2=0\) and making the angle of measure \(\tan ^{-1}(-5)\) with the positive direction of \(\mathrm{x}\) -axis is \(\ldots \ldots\) (a) \(3 x-2 y=0\) (b) \(5 x+y-2=0\) (c) \(5 \mathrm{x}+\mathrm{y}=0\) (d) \(3 \mathrm{x}+2 \mathrm{y}+1=0\)

Understanding the slope-intercept form is essential for anyone studying linear equations. It is an expression of a straight line in the format:
\[ y = mx + b \]
where \( m \) represents the slope of the line and \( b \) denotes the y-intercept, which is the point at which the line crosses the y-axis.

The slope \( m \) measures the steepness and direction of the line: a positive slope means the line inclines upwards, while a negative slope indicates it goes downwards as you move from left to right. The y-intercept \( b \) provides a starting point to graph the line or understand where the line 'starts' on the graph.

When working with the slope-intercept form, distinct values of \( m \) and \( b \) characterize different lines, unless they are parallel. This is why it's a powerful tool for graphing lines and solving for various line equations quickly. You can also convert other forms of linear equations into the slope-intercept form.

The point where two lines cross each other is known as the 'point of intersection.' To find this point, you solve the two equations of the lines simultaneously. Let’s consider the lines given in the exercise:
\[ 3x-2y=0 \] (Eq1)
\[ 5x+y-2=0 \] (Eq2)

Finding their point of intersection involves setting up a system of equations and solving for the variables \( x \) and \( y \). In our exercise, the solution done in steps showed how to substitute and eliminate variables to find the precise point where these two lines intersect. The values obtained for \( x \) and \( y \) then define a single point on the 2D plane that lies on both lines, which is significant as it serves as the starting point for the equation we are trying to find.

As an exercise improvement advice, always check your final intersection point by plugging the calculated values of \( x \) and \( y \) back into the original line equations to confirm they satisfy both equations. This ensures accuracy in finding your point of intersection.

The tangent of an angle in trigonometry, often shortened to \( \tan \), is a ratio that represents the slope of the angle's line when drawn within a circle of radius one, known as the unit circle. It's particularly useful in right-angle triangles, where the tangent of an angle is the proportion of the opposite side to the adjacent side.

For any angle \( \theta \), the tangent can be found by:\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
In the context of the given exercise, the \( \tan^{-1}(-5) \) notation is used to indicate the angle whose tangent is -5, effectively 'reversing' the tangent operation to find the angle. This shows us the slope of a line that forms this angle with the positive x-axis.

When dealing with linear equations, knowing the tangent of the angle can directly give us the slope of the line, as seen in step 3 of the solution. This correlation between trigonometry and algebra is invaluable for solving problems involving angles and slopes. For improved comprehension, always remember that the tangent provides a measure of how a line inclines relative to the x-axis and is a concept that appears frequently in calculus as well, often in the study of derivatives.