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# Find the equation for the perpendicular line. watch

1. In each part find the equation of the line throgh the given point which is perpendicular to the iven line. Write your final answer so that it doesn't contain fractions.

2L)

(-1, -2) ax + by = c

________________________________ _

ax + by = c
by = -ax + c
y = -a/b x + c

so perpendicular line equation is:

y = b/a x + c

and when x = -1 and y = -2 then:

-2 = b/a x -1 + c

-2 = -b/a + c

therefore c = (b-2/a) is this where i'm making a mistake??

so the equation can be written as y = b/a x + (b-2/a)

However the question asks for no fractions so multiply all terms by a so

ay = bx + a(b-2/a)
so
ay = bx + b - 2
rearranged to

bx - ay = 2 - b

however the correct answer in the text books is: bx - ay = 2a - b

Any suggestions? Thanks..
2. (Original post by Bleak Lemming)
In each part find the equation of the line throgh the given point which is perpendicular to the iven line. Write your final answer so that it doesn't contain fractions.

2L)

(-1, -2) ax + by = c

________________________________ _

ax + by = c
by = -ax + c
y = -a/b x + c
by = -ax + c
y = -a/b x + c/b

(Original post by Bleak Lemming)

CORRECT

so perpendicular line equation is:

y = b/a x + c
y - y1 = m(x - x1)

can be used to create the new equation, are you familiar with it?
3. (Original post by Bleak Lemming)
-2 = -b/a + c

therefore c = (b-2/a) is this where i'm making a mistake??
That's one mistake, yes. c = (b/a) -2

Then if you put the RHS over a common denominator you will have c= (b-2a)/a

Gdunne42's first point is another error, although it doesn't effect the final outcome.

@gdunne42, OP is using the form "y=mx+c", which works fine.
4. (Original post by ghostwalker)
That's one mistake, yes. c = (b/a) -2

Then if you put the RHS over a common denominator you will have c= (b-2a)/a

Gdunne42's first point is another error, although it doesn't effect the final outcome.

@gdunne42, OP is using the form &quot;y=mx+c&quot;, which works fine.

(Original post by gdunne42)
by = -ax + c
y = -a/b x + c/b

y - y1 = m(x - x1)

can be used to create the new equation, are you familiar with it?
Thx both,

I've just started teaching myself AS/A2 maths after being out of school for a few years and I've forgotten the basics like the mistake i made in this question.

Thx for the y1 - y2 = m(x1 - x2) thought, twas one that never sprung to mind.

Sean

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