Geometry help..please?

Question

write an equation for the line perpendicular to line XY that contains point Z....... line XY: 3x+2y= -6,Z(3,2)... how do i do this?? please show me steps.. if there is any

Answer 1

Note that I am using properties of standard form of an equation of a line (Ax+By=C) below.

slope of a line from an equation in standard form is -(A/B);

so the slope of your original line is -3/2. Slope of a line perpendicular is the opposite reciprocal of the slope, so 2/3.

Using -(A/B), the left side of the equation will be:

2x-3y; to find the right side, plug the point in:

2(3)-2(2) = 6-4 = 2

So, an equation is 2x - 3y = 2

Answer 2

Solve the equation for y. The new equation is (-3/2)X - 6 = y. Perpendicular lines always have inverse slopes. Also, you must change the sign. For example, the slope perpendicular to y= -2x is y=1/2x. In your problem, the new slope would be 2/3. Then, put the coordinates of the point into your equation. The x value into the x and the y into the y. It would be 2= 3(2/3) + b. Solve for b.
2= 2 + b. B=0. The new equation is y = 2/3 X.

Answer 3

Solve for y. Then determine the slope (y=mx+b m is the slope). You want the negative reciprocal for the slope in the new equation. So you'll have y=mx+b, with something for m. Then you substitute (3,2) for the x and y. Then solve for b. The resulting answer you use in another equation for b. You already have the slope, so put it in for m. So the result will be y=mx+b, with numbers for b and m.