Question

Question 2
Find the points on the X-axis which are at a distance of $2\sqrt{5}$ from the point (7, -4). How many such points are there?

Answer 1

We know that, every point on the X-axis in the form (x,0).
Let the point on the X-axis be P(x,0) .
Given: P(x,0) is at a distance of $2\sqrt{5}$ units from the point Q(7,-4).
By given condition,$PQ=2\sqrt{5}$
$[\because \text{Distance formula}\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}]$
$\Rightarrow (PQ{)}^2=4\times 5\Rightarrow (x-7{)}^2+(0+4{)}^2=20\Rightarrow x^2+49-14x+16=20\Rightarrow x^2-14x+65-20=0\Rightarrow x^2-14x+45=0\Rightarrow x^2-9x-5x+45=0\text{[by factorisation method]}\Rightarrow x(x-9)-5(x-9)=0\Rightarrow (x-9\left)\right(x-5)=0\therefore x=5and9$
Hence, there are two such points i.e., point P can be (5,0) or (9,0).
[$\because$ Both the points are at a distance of $2\sqrt{5}$ from (7, -4) ]