Question

What is the equation of the line that passes through the point (6,8) and is perpendicular to a line with the equation y= 3/2x + 5?

Answer 1

Answer:

$y=-\frac{2}{3}x+12$

Step-by-step explanation:

step 1

Find the slope of the line

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

The slope of the given line is $m=\frac{3}{2}$

so

The slope of the line perpendicular to the given line is $m=-\frac{2}{3}$

step 2

Find the equation in point slope form

$y-y1=m(x-x1)$

we have

$m=-\frac{2}{3}$

$point\ (6,8)$

substitute

$y-8=-\frac{2}{3}(x-6)$

step 3

Convert to slope intercept form

isolate the variable y

$y-8=-\frac{2}{3}(x-6)\\\\y-8=-\frac{2}{3}x+4\\\\y=-\frac{2}{3}x+4+8\\\\y=-\frac{2}{3}x+12$

Answer 2

The equation of the line that passes through the points is mathematically given as

y=1/1/5x+5.6

What is the equation of the line that passes through the point?

Question Parameters:

the equation of the line that passes through the point (6,8)

perpendicular to a line with the equation y= 3/2x + 5

Generally, the equation for the  slope is mathematically given as

slope=  -a/b

Therefore

y-3/2x - 5=

slope=-(-3/2)/1

Since, perpendicular lines  slopes are  multiplied to get -1

-(-3/2)/1=1

slope=1/2.5

gives

y=1/2.5x+b

In conclusion at point  (6,8)

(8)=1/2.5(6)+b

b=5.6

With an intercept as 5.6

The equation is

y=1/1/5x+5.6

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