Question

Write the equation of the line passing through the midpoints of the line segment defined by the points $(2,8)$ and $(0,4)$ and perpendicular to the line whose equation is given by $-3x+6y=5$

Answer 1

The midpoint of the points $(2,8)$ and $(0,4)$ is given by:

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{2+0}{2},\frac{8+4}{2})=(\frac{2}{2},\frac{12}{2})=(1,6)$

We must must transform the standard form equation $-3x+6y=5$ into a slope-intercept form equation ($y=mx+b$) to find its slope.

$-3x+6y=5$ (Subtract $3x$ on both sides.)

$6y=3x+5$ (Divide both sides by $6$.)

$y=\frac{3}{6}x+\frac{5}{6}$

$y=\frac{1}{2}x+\frac{5}{6}$

The slope of our first line is equal to $\frac{1}{2}$ . Perpendicular lines have negative reciprocal slopes, so if the slope of one is $x$, the slope of the other is $\frac{1}{x}$ .

The negative reciprocal of $\frac{1}{2}$ is equal to $-2$, therefore, $-2$ is the slope of our line.

Since the equation of line passing through the midpoint $(1,6)$, therefore, substitute the given point in the equation $y=-2x+b$:

$6=(-2\times 1)+b$

$6=-2+b$

$b=6+2=8$

Substitute this value for $b$ in the equation $y=-2x+b$:

$y=-2x+8$

Hence, the equation of the line is $y=-2x+8$.