y = -7x - 9, -3y = 21x + 7

without graphing, tell whether the graphs of each pair of equations are perpendicular.

Dear sharon,

the first equation
y = -7x - 9
slope= -7
y-intercept= -9

the second equation
-3y = 21x + 7
y = -7x -(7/3)
slope= -7
y-intercept= -(7/3)

To determine whether the two lines are perpendicular
slope of equation 1 X slope of equation 2 must be -1.
But in this case,
slope of equation 1 X slope of equation 2 = 49 instead of -1
so the two lines are not perpendicular.

(However, slope of equation 1 = slope of equation 2 and the y-intercept of the two equations are not the same, so we know that the two lines are parallel to each other.)

Hope it helps!
Wishing you the best of luck,
Barry

the slopes are the same, so they are either parallel or the same line.

If you multiply the 1st by -3, the constant term is different from 7, so the lines are parallel.

To determine whether the graphs of two equations are perpendicular, we need to compare the slopes of the two lines.

Let's rewrite the equations in slope-intercept form (y = mx + b), where m represents the slope.

For the first equation, y = -7x - 9, the slope is -7.

For the second equation, -3y = 21x + 7, we need to isolate y by dividing both sides by -3. This gives us y = -7x - (7/3), which means the slope is also -7.

Since both equations have the same slope (-7), the graphs are parallel, not perpendicular.

To determine if two lines are perpendicular, we need to examine their slopes. If the slopes of two lines are negative reciprocals of each other, then the lines are perpendicular.

Let's start by rearranging both equations into slope-intercept form, which is in the form y = mx + b, where m represents the slope.

For the first equation, y = -7x - 9, we can see that the slope is -7.

For the second equation, -3y = 21x + 7, let's divide both sides by -3 to isolate y: y = (-21/3)x - (7/3). This simplifies to y = -7x - (7/3). So, the slope of the second equation is -7 as well.

Since both equations have the same slope (-7), the two lines are parallel, not perpendicular.