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Exercise 6.4 · Q15

Q.Show that the equation 4x2+4xy+y2−6x−3y−4=04x^2 + 4xy + y^2 - 6x - 3y - 4 = 0 represents a pair of parallel lines. Find the distance between them.

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Same equation as Q.2: factoring gives (2x+y−4)(2x+y+1)=0(2x+y-4)(2x+y+1)=0, two parallel lines of slope −2-2; apply the distance-between-parallel-lines formula.

Note

This is the identical equation to Q.2 (4x2+4xy+y2−6x−3y−4=04x^2+4xy+y^2-6x-3y-4=0); Q.2 only asked to show the lines are parallel, while this question additionally asks for the distance between them, so the full derivation is repeated here for completeness before the new final step.

Step 1. Identify a,h,ba,h,b. a=4, 2h=4⇒h=2, b=1a=4,\ 2h=4\Rightarrow h=2,\ b=1.

Step 2. Test the parallel condition.

h2−ab=22−(4)(1)=4−4=0,h^2-ab=2^2-(4)(1)=4-4=0,

confirming equal slopes.

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

4x2+4xy+y2=(2x+y)2.4x^2+4xy+y^2=(2x+y)^2.

Step 4. Reduce using u=2x+yu=2x+y. The linear part is −6x−3y=−3(2x+y)=−3u-6x-3y=-3(2x+y)=-3u, so the equation becomes

u2−3u−4=0 ⟹ (u−4)(u+1)=0 ⟹ u=4 or u=−1.u^2-3u-4=0\ \Longrightarrow\ (u-4)(u+1)=0\ \Longrightarrow\ u=4\ \text{or}\ u=-1.

Step 5. Write the two lines. …

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