If the equation of the straight line passing through the point of intersection of x+2y-19=0 ,x-2y-3=0 and which is at a perpendicular distance of 5 units from the point (-2 ,4 ) is 5x+ by+c=0, then a…

Question

If the equation of the straight line passing through the point of intersection of x+2y-19=0 ,x-2y-3=0 and which is at a perpendicular distance of 5 units from the point (-2 ,4 ) is 5x+ by+c=0, then a possible value of 5x+ by+c is

Accepted Answer

Correct answer: D. (d) Equation of line passing through the point of intersection of lines x+2y-19=0 and x-2y-3=0 is By family of lines (x+2y-19 )+ (x-2y-3 )=0 ⇒ (1+ )x+ (2-2 )y- (19+3 )=0, is at a perpendicular distance of 5 units from point (-2 ,4 ) So ,5= |-2-2 +8-8 -19-3 | (1+ )^2+4 (1- )^2 ⇒ 25(5 ^2-6 +5)= (13 +13 )^2 ⇒125 ^2-150 +125=169 ^2+338 +169 ⇒44 ^2+488 +44=0 ⇒11 ^2+122 +11=0 ⇒ (11 +1 ) ( +11 )=0 ⇒ =-11 or- 111 ∴ The possible equation are -10x+24y+14=0 ⇒5x-12y-7=0 or10x+24y-206=0 ⇒5x+12y-103=0 Now, on comparing with 5x+by+c=0, we get The possible value of 5+b+c=-14 or -86

$If the equation of the straight line passing through the point of intersection of x + 2 y - 19 = 0, x - 2 y - 3 = 0 and which is at a perpendicular distance of 5 units from the point (- 2, 4) is 5 x + by + c = 0, then a possible value of 5 x + by + c is$

Solution

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