Write an equation of the line that passes through the given point and is perpendicular to the given line: (6,-4); y=-3x+4 (-4,-2); 4x+2y=-6 explain how to do !

Case 1:  (6,-4); y=-3x+4

The slope of a line perpendicular to y= -3x+4 is the negative reciprocal of -3 and is m = +(1/3).

Let's use the slope-intercept form of the eq'n of a line:  y = mx + b.

Subbing 6 for x, -4 for y and (1/3) for m, we get:

-4 = (1/3)(6) + b, or -4 = 2 + b.  Then b = -6, and the desired equation is:

y = (1/3)m - 6.

Case 2:   (-4,-2); 4x+2y=-6.

This equation, when div. by 2, results in 2x + y = -3, or y = -2x - 3.

The slope of a line perpendicular to y = -2x - 3 is (+1/2).

Using the point-slope equation for a straight line, y-k = m(x-h), and subbing -2 for k, -4 for x and (1/2) for m, we get:

y + 2 = (1/2)(x + 4).  This could be put into other forms as well:

y = (1/2)x + 2 - 2, or y = (1/2)x.

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this