Question

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

• A. (3,0)
• B. (5,0)
• C. (9,0)
• D. (4,0)

Answer 1

Answer: (B), (C)

The correct options are
B

(5,0)

C

(9,0)

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