What is the equation of a line passing through the point (6,1) and parallel to the line whose equation 3x=2y+4

The equation of the line would be y = -2/3x + 5

Step-by-step explanation:

In order to solve this, start by finding the slope of the original line. You can do this by solving for y.

3x = 2y + 4

-2y = -3x + 4

y = 3/2x - 2

Now that we have a slope of 3/2, we know that the perpendicular slope is -2/3 (since perpendicular lines have opposite and reciprocal slopes). We can use this and the new point in point-slope form to find the equation.

y - y1 = m(x - x1)

y - 1 = -2/3(x - 6)

y - 1 = -2/3x + 4

y = -2/3x + 5

use this formula:

p(a\b) = p(a and b)/p(b).

take it from here.

step-by-step explanation:

assume that you have 3 different side lengths.   apply the pythagorean theorem here:   square each of the two shortest sides; call them a and b, so that their squares are a² and b².   also square the longest side, calling the result c².

now, if a² + b² = c² is true, the triangle is a right triangle; otherwise it is not a right triangle.