91Ó°ÊÓ

To find the slope of the given line, we need to rewrite the linear equation \(2x-y+3=0\) in the slope-intercept form, i.e., \(y=mx+b\). In this form, \(m\) signifies the slope of the line. Rearrange the equation \[y = 2x + 3\] Now we can see that the slope of the given line is \(m = 2\).

We know that if two lines are perpendicular, the product of their slopes is equal to \(-1\). Therefore, let the slope of the line perpendicular to the given line be \(m_p\). We have: \[m \cdot m_p = -1\] Substituting the slope of the given line (m = 2) in the formula, we get: \[2 \cdot m_p = -1\] \[m_p = -\frac{1}{2}\]

We have the slope of the perpendicular line \(m_p = -\frac{1}{2}\), and we know that it passes through the point \((5,3)\). Using the point-slope form of a linear equation, \(y-y_1 = m_p(x-x_1)\), we can find the equation of the perpendicular line. Here, \((x_1,y_1) = (5,3)\). Thus, we get \[y - 3 = -\frac{1}{2} (x - 5)\]

To find the intersection point \((x_i, y_i)\) of the given line and the perpendicular line, we need to solve the linear equations simultaneously. The given line: \[y = 2x + 3\] The perpendicular line: \[y - 3 = -\frac{1}{2} (x - 5)\] We can solve these equations using substitution. Substitute the perpendicular line equation into the given line equation: \[2x + 3 = 3 -\frac{1}{2} (x - 5)\] Solve for \(x_i\): \[2x +3 = 3 - \frac{1}{2}x + \frac{5}{2}\] \[\frac{5}{2}x = \frac{2}{2}\] \[x_i = \frac{2}{5}\] Now substitute the value of \(x_i\) back into either of the equations. We can use the given line: \[y_i = 2\left(\frac{2}{5}\right) + 3\] \[y_i = \frac{4}{5} + \frac{15}{5}\] \[y_i = \frac{19}{5}\] So, the point of intersection is \(\left(\frac{2}{5}, \frac{19}{5}\right)\).

Lastly, we use the distance formula to find the distance between the given point \((5,3)\) and the point of intersection \(\left(\frac{2}{5}, \frac{19}{5}\right)\): \[d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\] Substitute the points into the distance formula: \[d = \sqrt{\left(\frac{2}{5} - 5\right)^2 + \left(\frac{19}{5} - 3\right)^2}\] \[d = \sqrt{\left(\frac{-23}{5}\right)^2 + \left(\frac{4}{5}\right)^2}\] \[d = \sqrt{\frac{529 + 16}{25}}\] \[d = \sqrt{\frac{545}{25}}\] \[d = \frac{\sqrt{545}}{5}\] Thus, the distance from the point \((5,3)\) to the line \(2x-y+3=0\) is \(\frac{\sqrt{545}}{5}\).