Question

Find the equation of the line through point (8,−9) and perpendicular to 3+8=4.

Answer 1

The equation of the line through the point (8,−9) and perpendicular to 3x+8y=4 is given by 8x - 3y = 91.

We are aware that any straight line's equation can be expressed as

y = mx + c,

where m denotes the slope and c denotes a constant.

Also, two perpendicular lines' slopes are the negative reciprocals of one another.

Here, the equation of the given straight line is

3x+8y=4

i.e. 8y = 4 -3x

i.e. y = (4/8) - (3/8)x

Now the negative reciprocal of - 3/8 is 8/3.

Then we can write the equation of the perpendicular line is

y = (8/3)x + c ...(1)

Since (1) passes through the point (8, -9), so we can put x = 8 and y = -9 in (1) to get the value of c.

So, -9 = (8/3)*8 + c

i.e. -9 = 64/3 + c

i.e. c = -9 -64/3 = - (27 + 64)/3 = - 91/3

(1) can be written as

y = (8/3)x - (91/3)

i.e. 3y = 8x - 91

i.e. 8x - 3y = 91

Therefore the equation of the line through the point (8,−9) and perpendicular to 3x+8y=4 is given by 8x - 3y = 91.

Learn more about perpendicular lines here -

brainly.com/question/12209021

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