Question

Find the points on the line 3x – 4y - 1= 0
which are at a distance of 5 units from the
point (3,2).
​

Answer 1

Step-by-step explanation:

Let P (h,k) is a point on given line 3x-4y-1=0 .

3h-4k-1=0 , or h=(4k+1)/3……………….(1)

Distance between a point A(2,3) and P(h,k) is

5 units.

PA = 5

or PA^2 = 25

(h-2)^2+(k-3)^2=25 , on putting h=(4k+1)/3 from eq. (1).

{(4k+1)/3–2}^2 +(k-3)^2 = 25

(1/9).(4k-5)^2 +(k-3)^2 = 25

(4k-5)^2 +9.(k-3)^2 = 225

16k^2–40k+25 +9k^2–54k+81 = 225

25k^2 -94 k + 106–225 =0

25k^2 -94k - 119 = 0

25k^2 -119k +25 k - 119 = 0

k(25k -119) +1 ( 25k -119 ) = 0

(25k -119) (k +1 )=0

k = 119/25 , -1 . On putting the value of k in

eq.(1).

h = (4k+1)/3……………..(1)

h =(476/25+1)/3 =501/(25×3) =167/25

or h=(-4+1)/3 = -1 .

Required points are (167/25,119/25) and (-1,-1).

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