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Q.If the line (x−y+2)+k(2x+3y+5)=0(x-y+2)+k(2x+3y+5)=0 is parallel to the line 3x+y=03x+y=0, then the value of kk will be

(a) k=1k=1
(b) k=47k=\dfrac{4}{7}
(c) k=74k=\dfrac{7}{4}
(d) None of these
Jharkhand JacJAC Intermediate Board (1st Year) 2024MCQ· 1mImportance★★★★★
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Expand the family of lines to standard form Ax+By+C=0Ax+By+C=0, write its slope −A/B-A/B, and set it equal to the slope of the given parallel line.

Expand (x−y+2)+k(2x+3y+5)=0(x-y+2)+k(2x+3y+5)=0:

x−y+2+2kx+3ky+5k=0x - y + 2 + 2kx + 3ky + 5k = 0

(1+2k)x+(3k−1)y+(2+5k)=0(1+2k)x + (3k-1)y + (2+5k) = 0

The slope of a line Ax+By+C=0Ax+By+C=0 is −A/B-A/B, so this line's slope is:

m1=−1+2k3k−1m_1 = -\frac{1+2k}{3k-1}

The line 3x+y=03x+y=0 has slope m2=−3m_2 = -3.

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