Question

Find a relation between $a$ and $b$ such that the point $(a,b)$ is equidistant from the points $(1,1)$ and $(7,9).$

• A. $4a+3b=32$
• B. $3a+4b=32$
• C. $a+b=10$
• D. $a-b=32$

Answer 1

The correct option is B $3a+4b=32$

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

The distance between the points $(1,1)$ and $(a,b)$
$=\sqrt{(a-1{)}^2+(b-1{)}^2}$

The distance between the points $(7,9)$ and $(a,b)$
$=\sqrt{(a-7{)}^2+(b-9{)}^2}$

Given, the point $(a,b)$ is equidistant from the points $(1,1)$ and $(7,9)$
$\Rightarrow \text{The distance between the points}(1,1)\text{and}(a,b)\text{= The distance between the points}(7,9)\text{and}(a,b)$
$\Rightarrow \sqrt{(a-1{)}^2+(b-1{)}^2}=\sqrt{(a-7{)}^2+(b-9{)}^2}$

Squaring both the sides
$(a-1{)}^2+(b-1{)}^2=(a-7{)}^2+(b-9{)}^2$

Using the formula $(x-y{)}^2=x^2+y^2-2xy$, we get

$\Rightarrow (a^2+1^2-2\times a\times 1)+(b^2+1^2-2\times b\times 1)=(a^2+7^2-2\times a\times 7)+(b^2+9^2-2\times b\times 9)$
$\Rightarrow (a^2+1-2a)+(b^2+1-2b)=(a^2+49-14a)+(b^2+81-18b)$

Cancel out $a^2+b^2$ from both the sides
$(1-2a)+(1-2b)=(49-14a)+(81-18b)$

On rearranging
$\Rightarrow 1+1-2a-2b=49+81-14a-18b$
$\Rightarrow 2-2a-2b=130-14a-18b$
$\Rightarrow 14a-2a=130-2-18b+2b$
$\Rightarrow 12a=128-16b\Rightarrow 12a+16b=128$

Dividing both the sides by 4
$\frac{12a}{4}+\frac{16b}{4}=\frac{128}{4}$
$\Rightarrow 3a+4b=32$

Hence, option(b.) is the correct choice.