Question

Find the equation of the line passing through the points (2, 3) and the point of intersection of the lines $4x-3y=7$ and $3x+4y+1=0$.

• A. 4x-y-2=0
• B. 4x+y-5=0
• C. 4x-y-5=0
• D. 4x-y+5=0

Answer 1

The correct option is C

4x-y-5=0

$4x-3y=7\dots \dots \left(i\right)3x+4y=-1\dots \dots \left(ii\right)$
On multiplying eq (i) by 4 and eq (ii) by 3, we get
$16x-12y=28\dots \left(iii\right)\underset{–}{9x+12y=-3}\dots \left(iv\right)adding25x=25\Rightarrow x=1$
Putting $x=1$, in equation $4x-3y=7$, we get
$\Rightarrow 4\times 1-3y=7\Rightarrow -3y=7-4\Rightarrow -3y=3\Rightarrow y=-1$
$\therefore$ Point of intersection = (1, -1)
Equation of line passing through (2, 3) and (1, -1) is
$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\Rightarrow y-3=\frac{-1-3}{1-2}(x-2)\Rightarrow y-3=4(x-2)\Rightarrow y-3=4x-8\Rightarrow 4x-y-5=0$