Question

A straight line parallel to the line y=√3x passes through Q(2,3) and cuts the line 2x+4y-27=0 find the length of PQ​

Answer 1

Given info : A straight line parallel to the line y = √3x and passes through Q(2,3).

To find : when the unknown line cuts the line 2x + 4y - 27 = 0 at P, find the length of PQ..

Solution : line parallel to y = √3x

so, slope of line is √3 = tan60°

Line passing through Q(2,3) and cuts 2x + 4y - 27 = 0 at P.

Let distance between P and Q = r

so parametric equation of line (2 + rcos60° , 3 + rsin60°) = (2 + r/2, 3 + √3r/2)

This point should satisfy the equation of line 2x + 4y - 27 = 0

⇒2(2 + r/2) + 4(3 + √3r/2) - 27 = 0

⇒4 + r + 12 + 2√3r - 27 = 0

⇒16 + r(1 + 2√3) - 27 = 0

⇒r = 11/(2√3 + 1) × (2√3 - 1)/(2√3 -1) [ just rationalising ]

⇒r = 2√3 - 1

Therefore the length of PQ is 2√3 - 1.

Answer 2

Answer:-

Given info :

A straight line parallel to the line y = √3x and passes through Q(2,3).

To find :

when the unknown line cuts the line 2x + 4y - 27 = 0 at P, find the length of PQ..

Solution :

line parallel to y = √3x

so, slope of line is √3 = tan60°

Line passing through Q(2,3) and cuts 2x + 4y - 27 = 0 at P.

Let distance between P and Q = r

so parametric equation of line (2 + rcos60° , 3 + rsin60°) = (2 + r/2, 3 + √3r/2)

This point should satisfy the equation of line 2x + 4y - 27 = 0

⇒2(2 + r/2) + 4(3 + √3r/2) - 27 = 0

⇒4 + r + 12 + 2√3r - 27 = 0

⇒16 + r(1 + 2√3) - 27 = 0

⇒r = 11/(2√3 + 1) × (2√3 - 1)/(2√3 -1) [ just rationalising ]

⇒r = 2√3 - 1

Therefore the length of PQ is 2√3 - 1.

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