Find the equation of the line perpendicular to the
given line through the point (1,9). y=1/3x+5

Slope-intercept form:  y = mx + b

(m is the slope, b is the y-intercept or the y value when x = 0 --> (0, y) or the point where the line crosses through the y-axis)

For lines to be perpendicular, their slopes have to be the negative reciprocal of each other. (Basically flip the sign +/- and the fraction[switch the numerator and the denominator])

For example:

Slope =

Perpendicular line's slope =  or  -3

Slope = -2 or

Perpendicular line's slope =

   The given line's slope is , so the perpendicular line's slope is -3. Now that you know the slope, substitute/plug it into the equation:

y = mx + b

y = -3x + b    To find b, plug in the point (1, 9) and isolate/get the variable "b" by itself in the equation

9 = -3(1) + b      Add 3 on both sides to get "b" by itself

9 + 3 = -3 + 3 + b

12 = b

y = -3x + 12

sss (side, side, side)

sas (side, angle, side)

asa (angle, side, angle)

aas (angle, angle, side)

hl (hypotenuse, leg)

the fist one is not equalvalent but i think th second one isand the third isnt definetly or the last one is i think. hope this !

step-by-step explanation: