Question

If P(2,4) is equidistant from Q(7,0) and R(x,9) find the value of x, x is not equal to -3

Answer 1

Answer:

x = -2 or 6

Step-by-step explanation:

As  P(2,4) is equidistant from Q(7,0) and R(x,9)

so magnitude of PQ & PR will be equal

| PQ | = | PR |

| PQ | =  

$\sqrt{(2-7)^{2} + (4-0)^{2} } \\\\=\sqrt{(-5)^{2} + 4^{2} } \\\\=\sqrt{25 + 16} \\\\=\sqrt{41}$

| PR | =

$\sqrt{(2-x)^{2} + (4-9)^{2} } \\\\=\sqrt{(4 +x^{2} -4x) + (-5)^{2} } \\\\=\sqrt{(4 +x^{2} -4x) +25} \\\\=\sqrt{x^{2} -4x +29}$

| PQ | = | PR |

PQ² = PR²

=> x² - 4x + 29 = 41

=> x² - 4x - 12 = 0

=> x² -6x + 2x - 12 = 0

=> x (x-6) + 2(x-6) = 0

=> (x + 2) ( x-6) = 0

=> x = -2 or 6

Answer 2

hope this helps you!!!