Question

Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x – 2y - 4 = 0 to the point of intersection of 7x – 3y = -12 and 2y = x + 3.

• A. 13x + 5y + 3 = 0
• B. 31x - 15y - 30 = 0
• C. 31x + 15y + 30 = 0
• D. 19x - 25y + 8 = 0

Answer 1

The correct option is C 31x + 15y + 30 = 0

The given equations of the first two lines are:
$3x+y+2=0\Rightarrow 3x+y=-2\dots ..\left(1\right)$
$x–2y-4=0\Rightarrow x–2y=4\dots ......\left(2\right)$
Now, $\left(1\right)\times 2\Rightarrow 6x+2y=-4\dots .....\left(3\right)$
$\left(2\right)\times 1\Rightarrow x-2y=4\dots ........\left(4\right)$
$\left(3\right)+\left(4\right)\Rightarrow 7x=0\Rightarrow x=0$
Substitute the value of $x=0$ in (1)
$3\left(0\right)+y=-2\Rightarrow y=-2$
The point of intersection is $(0,-2)$.
The given equations of the other two lines are:
$7x–3y=-12\dots ..\left(5\right)$
$2y=x+3\Rightarrow -x+2y=3\dots ..\left(6\right)$
Now, $\left(5\right)\times 1\Rightarrow 7x–3y=-12\dots ..\left(7\right)$
$\left(6\right)\times 7\Rightarrow -7x+14y=21\dots ..\left(8\right)$
$\left(7\right)+\left(8\right)\Rightarrow 11y=9\Rightarrow y=\frac{9}{11}$
Substitute the value of $y=\frac{9}{11}$ in (6)
$–x+2\left(\frac{9}{11}\right)=3\Rightarrow -x+\frac{18}{11}=3$
$\Rightarrow -x=3–\frac{18}{11}=\frac{33-18}{11}=\frac{15}{11}$
$x=-\frac{15}{11}$
The point of intersection is $(-\frac{15}{11},\frac{9}{11})$
Equation of the line joining the points $(0,-2)$ and $(-\frac{15}{11},\frac{9}{11})$ is:
$\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$
$\frac{y+2}{\frac{9}{11}+2}=\frac{x-0}{-\frac{15}{11}-0}$
$\frac{y+2}{\frac{31}{11}}=-\frac{11x}{15}$
$\Rightarrow -31x=15y+30$
$\Rightarrow 31x+15y+30=0$
The required equation is $31x+15y+30=0$