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VITEEE 2012 Paper

Section: Mathematics

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Question : 103 of 120

Marks: +1, -0

The equation of straight line through the intersection of the lines x - 2y = 1 and x + 3y = 2 and parallel 3x + 4y = 0 is

3x + 4y + 5 = 0
3x + 4y - 10 = 0
3x + 4y - 5 = 0
3x + 4y + 6 = 0

Solution:

The intersection point of lines x - 2y = 1

and x + 3y = 2 is

\left(\frac{7}{5},\frac{1}{5}\right)

Since, required line is parallel to 3x+4y = 0.

Therefore, the slope of required line is

-\frac{3}{4}

.

∴ Equation of required line which passes through

\left(\frac{7}{5},\frac{1}{5}\right)

is given by

y -

\frac{1}{5}

= -

\frac{3}{4} \left(x-\frac{7}{5}\right)

⇒

\frac{3x}{4}

+ y =

\frac{21}{20} + \frac{1}{5}

⇒ 3x + 4y - 5 = 0

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