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.
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