Write an equation of the line containing the given point and perpendicular to the given line. express your answer in the fory y=mx+b.

(8,9); 7x+y=2
the equation of the line is y=?

given line 7x+y=2 or

y = -7x + 2
the slope of the given line is -7 so the slope of the new line is +1/7

then y = (1/7)x + b
plug in the point
9 = (1/7)(8) + b
63 = 8 + 7b
b = 55/7

so y = (1/7)x + 55/7

check:
Ls = 9
Rs = (1/7)(8) + 55/7 = 63/7 = 9 , Yeah!

can you help on this im lost do to the negative x-3

y+4 = 8/7 (x-3)
now rearrange to y=mx+b form

(8,9);x+6y=5

To find the equation of a line that is perpendicular to the given line, first, we need to determine the slope of the given line. The slope-intercept form of a line equation is y = mx + b, where m represents the slope.

Let's rearrange the equation 7x + y = 2 in slope-intercept form:
y = -7x + 2

The given line has a slope of -7.

Since a line perpendicular to another line has a slope that is the negative reciprocal of the original slope, we can find the slope of the line perpendicular to it. The negative reciprocal of -7 is 1/7.

Now we have the slope of the perpendicular line, and we also have a point the line passes through, which is (8, 9).

Using the point-slope form, we can write the equation for the perpendicular line:

y - y1 = m(x - x1)

Substitute the values of the point (8, 9) and the slope 1/7:

y - 9 = (1/7)(x - 8)

Rearrange the equation to put it in slope-intercept form:

y - 9 = (1/7)x - (1/7)(8)

y - 9 = (1/7)x - 8/7

Adding 9 to both sides:

y = (1/7)x - 8/7 + 9

y = (1/7)x - 8/7 + 63/7

y = (1/7)x + 55/7

So, the equation of the line passing through the point (8, 9) and perpendicular to the line 7x + y = 2 is y = (1/7)x + 55/7.