Question

The number of lines which pass through point $(2, - 3)$ and are at a distance $8$ from point $( - 1,2)$ is

Answer 1

Hint : Here we will use the distance formula for the given points, it is given by
d $= \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$ then substitutes the value and simplify for the required resultant values. Then will compare the resultant value with the given term.

Complete step-by-step answer :
Given that:
Let us assume, $({x_1},{y_1}) = (2, - 3)$ and $({x_2},{y_2}) = ( - 1,2)$
Place the given terms in the standard distance formula,
d $= \sqrt {{{( - 1 - 2)}^2} + {{(2 - ( - 3))}^2}}$
Simplify the above expression, when you combine two negative terms together it becomes positive in terms it is multiplied. When you add two negative terms you have to add and give a negative sign to the resultant value.
d $= \sqrt {{{( - 3)}^2} + {{(2 + 3)}^2}}$
Square of any negative term always gives positive term –
$d = \sqrt {9 + {{(5)}^2}}$
Simplify the above expression finding the square of the term.
$d = \sqrt {9 + 25}$
Simplify the above expression finding the sum of the terms –
$d = \sqrt {34}$
Given that the distance between the points and the line is $8$ but by our solution the maximum distance of the line passing through the given points is $\sqrt {34}$ and hence there is no such line possible.
Therefore, zero lines can pass through it.
So, the correct answer is “Option B”.

Note : Always be careful about the sign convention while doing simplification. Remember that when there are two different signs then you have to do subtraction and give a sign of a bigger number to the resultant value and when there are two same signs do addition and give a common sign to the numbers. Remember the square of any positive or negative term always gives a positive term.