Answer :
To find the equation of the line that passes through the midpoint of the line segment joining the points \((4, 3)\) and \((2, 5)\) and is perpendicular to the line \(2x + 5y + 3 = 0\), follow these steps:
1. Find the Midpoint of the Line Segment:
The midpoint \((x_m, y_m)\) of the line segment joining the points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
[tex]\[ x_m = \frac{x_1 + x_2}{2}, \quad y_m = \frac{y_1 + y_2}{2} \][/tex]
For the points \((4, 3)\) and \((2, 5)\):
[tex]\[ x_m = \frac{4 + 2}{2} = 3, \quad y_m = \frac{3 + 5}{2} = 4 \][/tex]
So, the midpoint is \((3, 4)\).
2. Find the Slope of the Given Line:
The equation of the given line is \(2x + 5y + 3 = 0\). The general form of a linear equation is \(Ax + By + C = 0\), and the slope \(m\) of the line can be found using:
[tex]\[ m = -\frac{A}{B} \][/tex]
Here, \(A = 2\) and \(B = 5\), so the slope of the given line is:
[tex]\[ m = -\frac{2}{5} \][/tex]
3. Find the Slope of the Perpendicular Line:
The slope of a line that is perpendicular to another line is the negative reciprocal of the slope of the original line. So, if the slope of the original line is \(-\frac{2}{5}\), the slope \(m'\) of the perpendicular line is:
[tex]\[ m' = -\left(-\frac{5}{2}\right) = \frac{5}{2} \][/tex]
4. Write the Equation of the Perpendicular Line:
The equation of a line in slope-intercept form is given by:
[tex]\[ y = mx + c \][/tex]
We need to find the value of \(c\), the y-intercept. For a line passing through the point \((x_m, y_m) = (3, 4)\) with slope \(m' = \frac{5}{2}\), we have:
[tex]\[ 4 = \left(\frac{5}{2}\right) \cdot 3 + c \][/tex]
Now, solve for \(c\):
[tex]\[ 4 = \frac{15}{2} + c \\ 4 - \frac{15}{2} = c \\ c = \frac{8}{2} - \frac{15}{2} = -\frac{7}{2} = -3.5 \][/tex]
Therefore, the y-intercept \(c\) is \(-3.5\).
5. Combine the Slope and Y-Intercept into the Final Equation:
So, the equation of the line passing through \((3, 4)\) and perpendicular to \(2x + 5y + 3 = 0\) is:
[tex]\[ y = \frac{5}{2}x - 3.5 \][/tex]
Summarizing, the equation of the line is:
[tex]\[ y = \frac{5}{2}x - 3.5 \][/tex]
1. Find the Midpoint of the Line Segment:
The midpoint \((x_m, y_m)\) of the line segment joining the points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
[tex]\[ x_m = \frac{x_1 + x_2}{2}, \quad y_m = \frac{y_1 + y_2}{2} \][/tex]
For the points \((4, 3)\) and \((2, 5)\):
[tex]\[ x_m = \frac{4 + 2}{2} = 3, \quad y_m = \frac{3 + 5}{2} = 4 \][/tex]
So, the midpoint is \((3, 4)\).
2. Find the Slope of the Given Line:
The equation of the given line is \(2x + 5y + 3 = 0\). The general form of a linear equation is \(Ax + By + C = 0\), and the slope \(m\) of the line can be found using:
[tex]\[ m = -\frac{A}{B} \][/tex]
Here, \(A = 2\) and \(B = 5\), so the slope of the given line is:
[tex]\[ m = -\frac{2}{5} \][/tex]
3. Find the Slope of the Perpendicular Line:
The slope of a line that is perpendicular to another line is the negative reciprocal of the slope of the original line. So, if the slope of the original line is \(-\frac{2}{5}\), the slope \(m'\) of the perpendicular line is:
[tex]\[ m' = -\left(-\frac{5}{2}\right) = \frac{5}{2} \][/tex]
4. Write the Equation of the Perpendicular Line:
The equation of a line in slope-intercept form is given by:
[tex]\[ y = mx + c \][/tex]
We need to find the value of \(c\), the y-intercept. For a line passing through the point \((x_m, y_m) = (3, 4)\) with slope \(m' = \frac{5}{2}\), we have:
[tex]\[ 4 = \left(\frac{5}{2}\right) \cdot 3 + c \][/tex]
Now, solve for \(c\):
[tex]\[ 4 = \frac{15}{2} + c \\ 4 - \frac{15}{2} = c \\ c = \frac{8}{2} - \frac{15}{2} = -\frac{7}{2} = -3.5 \][/tex]
Therefore, the y-intercept \(c\) is \(-3.5\).
5. Combine the Slope and Y-Intercept into the Final Equation:
So, the equation of the line passing through \((3, 4)\) and perpendicular to \(2x + 5y + 3 = 0\) is:
[tex]\[ y = \frac{5}{2}x - 3.5 \][/tex]
Summarizing, the equation of the line is:
[tex]\[ y = \frac{5}{2}x - 3.5 \][/tex]