directed distance from a line

Question

Find the distance from the line 5x= 12y+26 to the point P1(3,-5) P2(-4,1) & P3(9,0)

Answer 1

The formula for calculating the magnitude of the perpendicular distance from a point P(X,Y) to a line Ax+By+C=0 is:

|AX+BY+C|/√(A²+B²). See later for derivation of this formula.

The directed distance depends on the sign of C and is given by (AX+BY+C)/√(A²+B²). C determines whether the positive or negative square root in the denominator applies. If C≥0, the sign is negative, otherwise positive.  The sign of AX+BY+C is also important when deciding the direction. In this question, P1 is on the opposite side of the line to the origin so the direction is positive; P2 is on the same side of the line as the origin so the direction is negative; P3 is on the opposite side of the line to the origin so the direction is positive.

We need to apply the formula where A=5, B=-12, C=-26, X={3,-4,9} and Y={-5,1,0} depending respectively on {P1,P2,P3}.

d_P1=|5×3-12×(-5)-26|/√(5²+12²)=|15+60-26|/13=49/13; since C<0 the directed distance is +49/13.

d_P2=|5×(-4)-12×1-26|/√(5²+12²)=|-20-12-26|/13=|-58|/13=58/13; since C<0 the directed distance is -58/13.

d_P3=|5×9+0-26|/√(5²+12²)=|45-26|/13=|19|/13; since C<0 the directed distance is +19/13.

DERIVATION OF FORMULA

Line: Ax+By+C=0. Line has slope -A/B so perpendicular has negative inverse slope=B/A.

Perpendicular: y-Y=(B/A)(x-X), y=(B/A)(x-X)+Y, which can be written Bx-Ay+AY-BX=0.

As a system of simultaneous equations, we can find Q where the perpendicular meets the line:

Ax+By=-C

Bx-Ay=BX-AY

A²x+ABy=-AC

B²x-ABy=B²X-ABY

(A²+B²)x=B²X-ABY-AC, x=(B²X-ABY-AC)/(A²+B²);

  ABx+B²y=-BC

-ABx+A²y=-ABX+A²Y

y(A²+B²)=A²Y-ABX-BC, y=(A²Y-ABX-BC)/(A²+B²).

Q(x,y)=((B²X-ABY-AC)/(A²+B²),(A²Y-ABX-BC)/(A²+B²)).

d=√((P_x-Q_x)²+(P_y-Q_y)²) where the subscripts represent the x and y coordinates of the points P and Q.

d=√(X-(B²X-ABY-AC)/(A²+B²))²+(Y-(A²Y-ABX-BC)/(A²+B²))²),

d=√(X(A²+B²)-(B²X-ABY-AC))²+(Y(A²+B²)-(A²Y-ABX-BC))²)/(A²+B²).

P_x-Q_x=X(A²+B²)-(B²X-ABY-AC)=A²X+ABY+AC=A(AX+BY+C);

P_y-Q_y=Y(A²+B²)-(A²Y-ABX-BC)=B²Y+ABX+BC=B(AX+BY+C).

Let p=AX+BY+C, then:

d=√((Ap)²+(Bp)²)=p√(A²+B²)/(A²+B²) rationalised form of:

p/√(A²+B²)=(AX+BY+C)/√(A²+B²). Note that p can be negative, depending on the coordinates of P and the value of C.

So the formula for the (absolute) distance d=|AX+BY+C)|/√(A²+B²).