Q.6 - Precalculus Enhanced with Graphing Utilities

Question

Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

If $\left(x,y\right)$ is a point on the unit circle in quadrant $\mathrm{IV}$ and if $x=\frac{\sqrt{3}}{2}$ what is $y$?

Step-by-Step Solution

Answer

When $x=\frac{\sqrt{3}}{2}$and the point $\left(x,y\right)$is a point on the unit circle in quadrant $\mathrm{IV}$, the value of $y$ is $y=-\frac{1}{2}$.

Given information

Consider $x=\frac{\sqrt{3}}{2}$ and the point $\left(x,y\right)$ lies on the unit circle in quadrant $\mathrm{IV}$.

The main objective is to find the value of $y$.

Definition

The general form of the circle equation with center $\left(h,k\right)$and radius $r$ is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$.

Here, given the circle is a unit circle.

This means the radius is $r=1$ and the center is at the origin, $\left(h,k\right)=\left(0,0\right)$.

Then the equation of the unit circle is,

$\begin{array}{rcl}{\left(x-0\right)}^2+{\left(y-0\right)}^2& =& 1^2\\ x^2+y^2& =& 1\end{array}$

Evaluate

Substitute $x=\frac{\sqrt{3}}{2}$in $\begin{array}{rcl}{x}^2+y^2& =& 1\end{array}$, to find $y$.

$\begin{array}{rcl}{x}^2+y^2& =& 1\\ {\left(\frac{\sqrt{3}}{2}\right)}^2+y^2& =& 1\\ \frac{3}{4}+y^2& =& 1\\ y^2& =& 1-\frac{3}{4}\\ y^2& =& \frac{1}{4}\\ y& =& \pm \frac{1}{2}\end{array}$

So, $y=\frac{1}{2},y=-\frac{1}{2}.$

But, the point is lies in 4th quadrant, so, the value of $y$ must be negative.

Then, $y=-\frac{1}{2}.$