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  • Find the points on the x–axis which are at a distance of 2 sqrot 5 from the point(7, –4). How many such points are there

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

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Solution.   Let point on x-axis is (x, 0) {\because y is zero on x-axis}
Given point (7, –4)
Distance = 2\sqrt{5}
2\sqrt{5}= \sqrt{\left ( 7-x \right )^{2}+\left ( -4-0 \right )^{2}}
Squaring both sides
\left ( 2\sqrt{5} \right )^{2}= \left ( 7-x \right )^{2}+\left ( -4 \right )^{2}
20= \left ( 7 \right )^{2}+\left ( x \right )^{2}-2\times 7\times x+16\left \{ \because \left ( a-b \right )^{2}= a^{2}+b^{2}-2ab \right \}
20=49+x^{2}-14x+16
x^{2}-14x+65-20= 0
x^{2}-14x+45= 0
x^{2}-9x-5x+45= 0
x\left ( x-9 \right )-5\left ( x-9 \right )= 0
\left ( x-9 \right )\left ( x-5 \right )= 0
x = 9, x =5
Point are (9, 0) and (5, 0)
Hence, two points are there.

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