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Exercise 6.5 · Q23

Q.If one of the lines given by 6x2−xy+4cy2=06x^2-xy+4cy^2=0 is 3x+4y=03x+4y=0, then cc equals

(1) −3-3
(2) −1-1
(3) 33
(4) 11
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Since 3x+4y=03x+4y=0 is one factor of 6x2−xy+4cy2=06x^2-xy+4cy^2=0, write it as (3x+4y)(2x+ky)(3x+4y)(2x+ky) and match coefficients: c=−3c=-3.

If 3x+4y=03x+4y=0 is one of the two lines represented by 6x2−xy+4cy2=06x^2-xy+4cy^2=0, then (3x+4y)(3x+4y) must be a factor of the quadratic, and the other factor must be linear of the form (2x+ky)(2x+ky) so that the x2x^2 coefficient works out to 66 (since 3×2=63\times2=6).

Step 1. Set up the factorisation.

6x2−xy+4cy2=(3x+4y)(2x+ky)6x^2-xy+4cy^2=(3x+4y)(2x+ky)

Step 2. Expand the right side.

(3x+4y)(2x+ky)=6x2+3kxy+8xy+4ky2=6x2+(3k+8)xy+4ky2(3x+4y)(2x+ky)=6x^2+3kxy+8xy+4ky^2=6x^2+(3k+8)xy+4ky^2

Step 3. Match the coefficient of xyxy. Comparing with −xy-xy (i.e. coefficient −1-1):

3k+8=−1  ⟹  3k=−9  ⟹  k=−33k+8=-1 \implies 3k=-9 \implies k=-3

Step 4. Match the coefficient of y2y^2. Comparing 4ky24ky^2 with 4cy24cy^2:

4k=4c  ⟹  c=k=−34k=4c \implies c=k=-3 …

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