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Exercise 6.3 · Q12

Q.Find the distance between the parallel lines

(i) 12x+5y=712x + 5y = 7 and 12x+5y+7=012x + 5y + 7 = 0
(ii) 3x−4y+5=03x - 4y + 5 = 0 and 6x−8y−15=06x - 8y - 15 = 0.
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For parallel lines a1x+b1y+c1=0a_1x+b_1y+c_1=0 and a1x+b1y+c2=0a_1x+b_1y+c_2=0 (same a1,b1a_1,b_1 after scaling), the distance between them is D=∣c2−c1∣a12+b12D=\dfrac{|c_2-c_1|}{\sqrt{a_1^2+b_1^2}}.

Both parts give two parallel lines directly (or after scaling to the same a,ba,b); write each in ax+by+c=0ax+by+c=0 form and apply the parallel-line distance formula.

Step 1 (i). Write both lines with matching coefficients.

12x+5y=7 ⇒ 12x+5y−7=012x+5y=7\ \Rightarrow\ 12x+5y-7=0, so c1=−7c_1=-7.

12x+5y+7=012x+5y+7=0, so c2=7c_2=7. Here a=12, b=5a=12,\ b=5.

Step 2 (i). Apply the distance formula.

D=∣c2−c1∣a2+b2=∣7−(−7)∣122+52=14144+25=14169=1413D=\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}=\frac{|7-(-7)|}{\sqrt{12^2+5^2}}=\frac{14}{\sqrt{144+25}}=\frac{14}{\sqrt{169}}=\frac{14}{13}

Step 3 (ii). Scale the second line so its x,yx,y coefficients match the first.

3x−4y+5=03x-4y+5=0 has a=3, b=−4, c1=5a=3,\ b=-4,\ c_1=5. …

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