Question

Given this unit circle what is the value of x

Answer 1

Remember that the unit circle has a radius of 1. In other words, every point on the unit circle is a distance of exactly 1 unit away from the origin, (0,0). Knowing that, you can use the distance formula to find your missing coordinate:

`\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2_{}}\text{ = Distance = 1}`

We get,

`\sqrt[]{(0-x)^2+(0-(3)/(4))^2_{}}\text{ = 1}`
`\sqrt[]{(x)^2+((3)/(4))^2_{}}\text{ = 1}`
`\sqrt[]{x^2+(9)/(16)^{}_{}}\text{ = 1}`
`(\sqrt[]{x^2+(9)/(16)^{}_{}})^2=(1)^2`
`x^2^{}+(9)/(16)^{}_{}=1`
`x^2=1\text{ - }(9)/(16)^{}_{}`
`x^2=(7)/(16)^{}_{}`
`√(x^2)=\sqrt{(7)/(16)^{}_{}}`
`\text{ x = }\pm\text{ }\frac{\sqrt[]{7}}{4}`

Since the x, 3/4 falls on the second quadrant, x is a negative value.

Therefore,

`\text{ x = -}\frac{\sqrt[]{7}}{4}`