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Exercise 6.4 · Q14

Q.Show that the equation 9x2−24xy+16y2−12x+16y−12=09x^2 - 24xy + 16y^2 - 12x + 16y - 12 = 0 represents a pair of parallel lines. Find the distance between them.

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h2−ab=0h^2-ab=0 confirms a parallel pair; the quadratic part is a perfect square (3x−4y)2(3x-4y)^2, so substitute u=3x−4yu=3x-4y to split into two parallel lines, then use the parallel-line distance formula.

Step 1. Identify a,h,ba,h,b. From 9x2−24xy+16y2−12x+16y−12=09x^2-24xy+16y^2-12x+16y-12=0: a=9, 2h=−24⇒h=−12, b=16a=9,\ 2h=-24\Rightarrow h=-12,\ b=16.

Step 2. Test the parallel condition.

h2−ab=(−12)2−(9)(16)=144−144=0,h^2-ab=(-12)^2-(9)(16)=144-144=0,

confirming equal slopes (parallel pair), pending confirmation there are two distinct lines.

Step 3. Factor the quadratic part as a perfect square.

9x2−24xy+16y2=(3x−4y)2.9x^2-24xy+16y^2=(3x-4y)^2.

Step 4. Reduce using u=3x−4yu=3x-4y. The linear part is −12x+16y=−4(3x−4y)=−4u-12x+16y=-4(3x-4y)=-4u. So the equation becomes

u2−4u−12=0.u^2-4u-12=0.

Step 5. Solve and factor.

u2−4u−12=(u−6)(u+2)=0 ⟹ u=6 or u=−2,u^2-4u-12=(u-6)(u+2)=0\ \Longrightarrow\ u=6\ \text{or}\ u=-2,

giving the two lines

3x−4y−6=0and3x−4y+2=0.3x-4y-6=0\qquad\text{and}\qquad 3x-4y+2=0. …

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