Let d1 and d2 be the lengths of the perpendiculars drawn from any point of the line 7x - 9y + 10 = 0 upon the lines 3x + 4y = 5 and 12x + 5y = 7 respectively. Then

Question

Let d1 and d2 be the lengths of the perpendiculars drawn from any point of the line 7x - 9y + 10 = 0 upon the lines 3x + 4y = 5 and 12x + 5y = 7 respectively. Then (A) d1 > d2 (B) d1 = d2 (C) d1 < d2 (D) d1 = 2d2

Tardigrade Admin · Accepted Answer

Let (h, k) be any point on the line 7 x-9 y+10=0, then 7 h-9 k+10=0 ⇒ 7 h =9 k-10 ⇒ h =(9 k-10/7) ... (i) Now, perpendicular distance from point (h, k) to the line 3 x+4 y=5 is d1 d1=(3 h+4 k-5/√32+42) ⇒ d1=(3 h+4 k-5/5) ⇒ d1=(3 h+4 k-5/5) ... (ii) and perpendicular distance from (h, k) to the line 12 x+5 y=7 is d2 ∴ d2=(12 h+5 k-7/√122+52) ⇒ d2=(12 h+5 k-7/13) ... (iii) Now, d1-d2=(3 h+4 k-5/5)-(12 h+5 k-7/13) ⇒ d1-d2 =(13(3 h+4 k-5)-5(12 h+5 k-7)/65) =(39 h+52 k-65-60 h-25 k+35/65) =(-21 h+27 k-30/65) =(-21((9 k-10/7))+27 k-30/65) =(-27 k+30+27 k-30/65)=0 [fromEq.(i)] ⇒ d1-d2=0 ⇒ d1=d2

Q. Let $d_{1}$ and $d_{2}$ be the lengths of the perpendiculars drawn from any point of the line $7 x - 9 y + 10 = 0$ upon the lines $3 x + 4 y = 5$ and $12 x + 5 y = 7$ respectively. Then

$d_{1} > d_{2}$

$d_{1} = d_{2}$

$d_{1} < d_{2}$

$d_{1} = 2 d_{2}$

Q. Let d1​ and d2​ be the lengths of the perpendiculars drawn from any point of the line 7x−9y+10=0 upon the lines 3x+4y=5 and 12x+5y=7 respectively. Then

d1​>d2​

d1​=d2​

d1​<d2​

d1​=2d2​

Q. Let $d_{1}$ and $d_{2}$ be the lengths of the perpendiculars drawn from any point of the line $7 x - 9 y + 10 = 0$ upon the lines $3 x + 4 y = 5$ and $12 x + 5 y = 7$ respectively. Then

$d_{1} > d_{2}$

$d_{1} = d_{2}$

$d_{1} < d_{2}$

$d_{1} = 2 d_{2}$