Question

The point on the line $3x+4y=5$ which is equidistant from $(1,2)$ and $(3,4)$ is:

• A. $(7,-4)$
• B. $(15,-10)$
• C. $(\frac{1}{7},\frac{8}{7})$
• D. $(0,\frac{5}{4})$

Answer 1

Answer: (B)

$(15,-10)$

Let's take a general point on the line $3x+4y-5=0$, which is $(x,\frac{5-3x}{4})$

According to given condition, this point is equidistant from points $(1,2)$ and $(3,4)$

Using distance formula, by equalizing their distances we get,

$\Rightarrow \sqrt{{(2-\frac{5-3x}{4})}^2+{(1-x)}^2}=\sqrt{{(4-\frac{5-3x}{4})}^2+{(3-x)}^2}$

$\Rightarrow (2{)}^2+(\frac{5-3x}{4}{)}^2-(5-3x)+1^2+x^2-2x=(4{)}^2+(\frac{5-3x}{4}{)}^2-2(5-3x)+3^2+x^2-6x$

$\Rightarrow x=15$

$\Rightarrow$ Hecne $y=\frac{5-3\left(15\right)}{4}=-10$

So the point is $(15,-10)$

The correct option is $B$.