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Question

Through the point P(4,1) a line is drawn to meet the line 3xy=0 at Q where PQ=1122. Determine the equation of line.

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Solution

Any point on the 3xy=0 is of the form (a,3a).
Distance between (a,3a) and (4,1) is 1122.

(a4)2+(3a1)2=1122

(a28a+16+9a26a+1)=1218

10a214a+17=1218

10a214a+158=0

80a2112a+15=0

(4a5)(20a3)=0

a=54,320

Equation of line passing through (4,1) and (54,154) is

(yy1)=(y2y1x2x1)(xx1)

(y1)=⎜ ⎜ ⎜154154⎟ ⎟ ⎟(x4)

(y1)=114(x4)

4y4=11x44

11x4y=40

Equation of line passing through (4,1) and (320,920) is

(yy1)=(y2y1x2x1)(xx1)

(y1)=⎜ ⎜ ⎜92013204⎟ ⎟ ⎟(x4)

(y1)=1177(x4)

77y77=11x44

11x77y+33=0

x7y+3=0

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