Question

Distance of point p(6,-1) from the libe 6x-4y+9=0

Answer 1

Step-by-step explanation:

There’s a formula, d2=(ax0+by0+c)2a2+b2, boring.

There’s the perpendicular through the point, 4x+6y=4(6)+6(−1) or 2x+3y=9 that means 6x−4y+9=0 at a point, then we calculate the distance. Boring.

How about with vectors? The line passes through the point (−3/2,0). We read off the perpendicular direction vector to the line, (6,−4), or p=(3,−2) reduced.

The distance we seek is the magnitude of the vector from (6,−1) to (−3/2,0), h=(−15/2,1) projected on our perpendicular direction vector:

p⋅h=|p||h|cosθ

d=∣∣∣|h|cosθ∣∣∣=|p⋅h||p|=|3(−15/2)+(−2)(1)|32+(−2)2−−−−−−−−−√=|−49/2|13−−√

d=4913−−√26

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