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Find the equation of a line which passes through the point of intersection of lines x + 2y = 5 and x – 3y = 7 and also passes through the point (0, –1) ml js

a

3x – 29y + 1 = 0

b

3x – 29y – 29 = 0

c

3x + 4y – 6 = 0

d

–3x + 29y + 7 = 0

Answer : Option B
Explanation :

Let the equation of line be

(x + 2y – 5) + O(x – 3y – 7) = 0

As it passes through (0, –1)

025+λ(0+37)=0\therefore 0-2-5+\lambda(0+3-7)=0

74λ=0-7-4 \lambda=0

λ=74\lambda=\frac{-7}{4}

\therefore Equation of line is

(x+2y5)74(x3y7)=0(x+2 y-5) \frac{-7}{4}(x-3 y-7)=0

4x+8y207x+21y+49=0\Rightarrow 4 x+8 y-20-7 x+21 y+49=0

3x+29y+29=0-3 x+29 y+29=0

3x29y29=03 x-29 y-29=0

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