What is an equation of the line that passes through the point (- 1, - 3) and is perpendicular to the line x - 2y = 14 ?

**Perpendicular lines have slopes which are negative reciprocals.**

To find the slope of the line x - 2y = 14, get it in slope-intercept form.

**Slope-intercept form is y = mx + b**

m is the slope

b is the y-intercept

m is the slope

b is the y-intercept

x - 2y = 14

-2y = -x + 14 {subtracted x from each side}

y = (1/2)x - 7 {divided each side by -2}

slope = 1/2

The perpendicular slope would be -2.

Using the point (-1,-3) and the slope -2, get the equation of the line.

y = mx + b {slope-intercept form}

-3 = (-2)(-1) + b {substituted -3 for y, -2 for m, and -1 for x}

-3 = 2 + b {multiplied -2 by -1}

b = -5 {subtracted 2 from each side}

On the new line, the slope is -2 and the y-intercept is -5

y = mx + b {slope-intercept form, once again}

y = -2x - 5 {substituted -2 for m and -5 for b}

The equation of the line passing through (-1,-3) and perpendicular to x - 2y = 14 is:

**y = -2x - 5**

**Ask the House**