Question

The Possible value(s) of $a$ such that, the distance between the points $(2,3)$ and $(a,4)$ is equal to $\sqrt{17}$ units is/are:

• A. $-6$
• B. $2$
• C. $-2$
• D. $6$

Answer 1

Answer: (C), (D)

$6$

We know that, the distance between the points $(x_1,y_1)$ and $(x_2,y_2)$
$=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

Given, distance between the points $(2,3)$ and $(a,4)=\sqrt{17}$
$\Rightarrow \sqrt{(a-2{)}^2+(4-3{)}^2}=\sqrt{17}$

Squaring both the sides, we get
$(a-2{)}^2+(4-3{)}^2=(\sqrt{17}{)}^2$
$(\because$ The square root and the square cancel out each other)
$\Rightarrow (a-2{)}^2+1^2=17$

Using the formula $(x-y{)}^2=x^2+y^2-2xy$
$\Rightarrow (a^2+2^2-2\times a\times 2)+1=17$
$\Rightarrow a^2+4-4a+1=17\Rightarrow a^2+(4+1)-4a=17$
$\Rightarrow a^2+5-4a=17$

Subtract 5 from both the sides
$a^2+5-4a-5=17-5$
$\Rightarrow a^2-4a=12$

Subtract 12 from both the sides
$a^2-4a-12=12-12$
$\Rightarrow a^2-4a-12=0$
$\Rightarrow a^2-4a-2a+2a-12=0\text{(adding and subtracting}2a\text{)}$
$\Rightarrow a^2-6a+2a-12=0$

Taking $a$ as a common term from the first two terms and also $2$ as a common term from the next two terms
$\Rightarrow a(a-6)+2(a-6)=0$

$\because \text{Here (a - 6) is a common term, so taking out the term (a-6) we get}$
$\Rightarrow (a-6)(a+2)=0$

Now, each term can be equated to 0
$\Rightarrow a-6=0\to a=6\Rightarrow a+2=0\to a=2$


Hence, options (c.) and (d.) are correct choices.