Question

The distance between $A(1,3)$ and $B(x,7)$ is $5$. Then the positive value of $x$ satisfying the given condition is:

• A. $4$
• B. $2$
• C. $1$
• D. $3$

Answer 1

The correct option is A $4$

Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ can be calculated using the formula $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}.$
Distance between the points $(1,3)$ and $(x,7)$ is equal to $5.$ So,

$\begin{array}{cc}\sqrt{{\left(x-1\right)}^2+{\left(7-3\right)}^2}& =5\\ \sqrt{{(x-1)}^2+16}& =5\\ (x-1{)}^2+16& =25\\ (x-1{)}^2& =9\\ x-1& =3\\ x& =4\end{array}$