Math  /  Algebra

QuestionWhat is the slope of a line that is perpendicular to the line whose equation is Ax+By+C=0,A0A x+B y+C=0, A \neq 0 and B0B \neq 0 ?
The slope of the line perpendicular to the line Ax+By+C=0A x+B y+C=0 is \square

Studdy Solution

STEP 1

What is this asking? We're looking for the slope of a line that's perpendicular to another line given in a somewhat complicated form. Watch out! Don't mix up the slope of the original line with the slope of the perpendicular line!
Also, remember that AA, BB, and CC are just constants – they're just numbers we don't know yet!

STEP 2

1. Rewrite the equation
2. Find the slope
3. Perpendicular slope

STEP 3

We want to rewrite the equation Ax+By+C=0Ax + By + C = 0 in slope-intercept form, which looks like y=mx+by = mx + b, where mm is the slope and bb is the y-intercept.
To do this, let's **isolate** the ByBy term.
We can move the AxAx and CC to the other side of the equation by adding Ax-Ax and C-C to both sides.
This gives us By=AxCBy = -Ax - C.
We did this because we want to get the yy by itself!

STEP 4

Now, we want to **solve for** yy.
Since BB is multiplied by yy, we can divide both sides of the equation by BB (remember, we're told B0B \neq 0, so we're allowed to divide by it).
This gives us y=ABxCBy = \frac{-A}{B}x - \frac{C}{B}.
Awesome!

STEP 5

Now our equation is in slope-intercept form!
Remember, slope-intercept form is y=mx+by = mx + b, where mm is the slope.
Looking at our equation, y=ABxCBy = \frac{-A}{B}x - \frac{C}{B}, we can see that the **slope** of the original line is m=ABm = \frac{-A}{B}.
Boom!

STEP 6

To find the slope of a line *perpendicular* to our original line, we need to take the **negative reciprocal** of the original slope.
This means we flip the fraction and change the sign.

STEP 7

The negative reciprocal of AB\frac{-A}{B} is BA\frac{B}{A}.
Notice how the AA and BB switched places, and the negative sign disappeared!
So, the slope of the perpendicular line is BA\frac{B}{A}.
Fantastic!

STEP 8

The slope of the line perpendicular to Ax+By+C=0Ax + By + C = 0 is BA\frac{B}{A}.

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