Question

The straight lines $3x+4y=5$ and $4x-3y=15$ intersect at the point $A$. On these lines points $B$ and $C$ are chosen so that $AB=AC$.Possible equation(s) of the line $BC$ passing through $(1,2)$ is/are

• A. $x+7y=9$
• B. $y+7x=9$
• C. $7x-y=13$
• D. $7y-x=13$

Answer 1

Answer: (B), (D)

The correct options are
B $y+7x=9$
D $7y-x=13$

The lines $3x+4y=5$ and $4x-3y=15$ respectively.
Hence the two given straight lines are at right angles.

So, according to the condition $AB=AC$ there are $2$ such lines are possible which make an angle of $45°$ with both lines
Now slope of the line which makes an angle of $45°$ with $4x-3y+5=0$is
$m=\frac{4/3\pm 1}{1\mp 4/3}=-7,\frac{1}{7}$
Hence the required equations are
$y-2=-7(x-1)$ and $y-2=\frac{1}{7}(x-1)$
$\Rightarrow y+7x=9$ and $7y-x=13$