Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Find the equation of the straight line passing through the point of intersection of 2x + y − 1 = 0 and x + 3y − 2 = 0 and making with the coordinate axes a triangle of area $\frac{3}{8}$ sq. units. - Mathematics

Find the equation of the straight line passing through the point of intersection of 2x + y − 1 = 0 and x + 3y − 2 = 0 and making with the coordinate axes a triangle of area $\frac{3}{8}$ sq. units.

#### Solution

The equation of the straight line passing through the point of intersection of 2x + y − 1 = 0 and x + 3y − 2 = 0 is given below:
2x + y − 1 + λ (x + 3y − 2) = 0

$\Rightarrow$ (2 + λ)x + (1 + 3λ)y − 1 − 2λ = 0

$\Rightarrow \frac{x}{\frac{1 + 2\lambda}{2 + \lambda}} + \frac{y}{\frac{1 + 2\lambda}{1 + 3\lambda}} = 1$

So, the points of intersection of this line with the coordinate axes are $\left( \frac{1 + 2\lambda}{2 + \lambda}, 0 \right) \text { and } \left( 0, \frac{1 + 2\lambda}{1 + 3\lambda} \right)$.

It is given that the required line makes an area of $\frac{3}{8}$ square units with the coordinate axes.

$\frac{1}{2}\left| \frac{1 + 2\lambda}{2 + \lambda} \times \frac{1 + 2\lambda}{1 + 3\lambda} \right| = \frac{3}{8}$

$\Rightarrow 3\left| 3 \lambda^2 + 7\lambda + 2 \right| = 4\left| 4 \lambda^2 + 4\lambda + 1 \right|$

$\Rightarrow 9 \lambda^2 + 21\lambda + 6 = 16 \lambda^2 + 16\lambda + 4$

$\Rightarrow 7 \lambda^2 - 5\lambda - 2 = 0$

$\Rightarrow \lambda = 1, - \frac{2}{7}$

Hence, the equations of the required lines are

$3x + 4y - 1 - 2 = 0 \text { and } \left( 2 - \frac{2}{7} \right)x + \left( 1 - \frac{6}{7} \right)y - 1 + \frac{4}{7} = 0$

$\Rightarrow 3x + 4y - 3 = 0 \text { and } 12x + y - 3 = 0$

Concept: Straight Lines - Equation of Family of Lines Passing Through the Point of Intersection of Two Lines
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 23 The straight lines
Exercise 23.19 | Q 8 | Page 131