Write the equation of the line perpendicular to

9x−5y=4
that passes through the point (1,−6)

in slope-intercept form and in standard form

Solving for y, we add 5y to both sides and subtract 4, getting 9x-4=5y. Dividing both sides by 5, we get 9x/5-4/5=y. Since the slope is 9/5 (since 9/5*x=9x/5), we multiply it by -1 and find the reciprocal of it to get -5/9 as the perpendicular slope, so -5x/9+b=y. Plugging 1 in for x and -6 in for y, we get -5*1/9+b=-6 and by adding 5/9 to both sides we get -5-4/9=b , and since in y=mx+b y and x are variables, we end up with y=-5x/9+(-5-4/9) for slope intercept form.

To get it into standard form, we need it in ay+cx=b with a, b, and c being constants. Adding 5x/9 to both sides, we end up with y+5x/9=(-5-4/9) for standard form

Arectangle is a quadrilateral and a parallelogram.here is all that apply to a rectangle!