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. $\left(\frac{1}{7},\frac{8}{7}\right)$
• D. $\left(0,\frac{5}{6}\right)$

Answer 1

The correct option is B

$(15,–10)$

Compute the required point:

Let $\left(x,y\right)$ be the point on the line.

Given: Point is equidistant from $(1,2)$ and $(3,4)$.

Using the distance formula,

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

$\Rightarrow$ ${\left(x-1\right)}^2+{(y-2)}^2={(x-3)}^2+{(y-4)}^2$

$\Rightarrow$$x^2+1-2x+y^2+4-4y=x^2+9-6x+y^2+16-8y$

$\Rightarrow$ $4x+4y=20$……………$\left(1\right)$

The equation of the line is $3x+4y=5$……………..$\left(2\right)$

subtract equation $\left(2\right)$ from equation $\left(1\right)$,

$4x+4y-3x-4y=20-5$

$\Rightarrow$ $x=15$

Substitute the value of $x$ in equation $\left(1\right)$,

$4\left(15\right)+4y=20$

$\Rightarrow$ $4y=-40$

$\Rightarrow$ $y=-10$

Therefore the point is $\left(15,-10\right)$.

Hence, option $\left(B\right)$ is the correct option.