Question

If $x-y=2$, then point $(x,y)$ is equidistant from $(7,1)$ and ____.

• A. $(3,5)$
• B. $(5,3)$
• C. $(6,2)$
• D. $(2,6)$

Answer 1

The correct option is A $(3,5)$

Line perpendicular to the line $x-y=2$ and passing through point $(7,1)$ will be $y=mx+c$
Two lines with slope $1$ and $m$ are perpendicular to each other.
So, $m_1{m}_2=-1$ where $m_1=1$ and $m_2=m$......Condition for perpendicular lines
$\therefore m\times 1=-1$
$\therefore m=-1$
Put $m=-1$ in the straight line $y=mx+c$, we get $y=-x+c$.
Now, line $y=-x+c$ also passes through the point $(7,1)$ and hence, it should satisfy the line equation.
Put $(7,1)$ in the equation $y=-x+c$, we get
$1=-7+c$
$\therefore c=8$
Therefore, final equation comes out to be $y=-x+8$.
Now, find the intersection of both the lines which comes out to be $(5,3)$ which will be a midpoint of the point $(7,1)$ and $(a,b)$ since it is equidistant from both the points.
So, from midpoint formula, we get
$\frac{a+7}{2}=5$ and $\frac{b+1}{2}=3$
$\therefore a=3,b=5$
Hence, option $A$ is the correct answer.