SAT II Math II : Graphing Linear Functions

Study concepts, example questions & explanations for SAT II Math II

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Example Questions

Example Question #1 : Graphing Linear Functions

Circle

Note: Figure NOT drawn to scale.

Refer to the above figure. The circle has its center at the origin; the line is tangent to the circle at the point indicated. What is the equation of the line in slope-intercept form?

Possible Answers:

Insufficient information is given to determine the equation of the line.

\displaystyle y = \frac{3}{4} x+ \frac{7}{2}

\displaystyle y = \frac{4}{3} x

\displaystyle y = - \frac{3}{4} x+ \frac{25}{2}

\displaystyle y = - \frac{4}{3} x+ 16

Correct answer:

\displaystyle y = - \frac{3}{4} x+ \frac{25}{2}

Explanation:

A line tangent to a circle at a given point is perpendicular to the radius from the center to that point. That radius, which has endpoints \displaystyle (0,0),(6,8),  has slope

\displaystyle m = \frac{8-0}{6-0} = \frac{8}{6} = \frac{4}{3}.

The line, being perpendicular to this radius, will have slope equal to the opposite of the reciprocal of that of the radius. This slope will be \displaystyle -\frac{3}{4}. Since it includes point \displaystyle (6,8), we can use the point-slope form of the line to find its equation:

\displaystyle y - y_{1} = m(x-x_{1})

\displaystyle y - 8= - \frac{3}{}4\left ( x-6 \right )

\displaystyle y - 8= - \frac{3}{4} x+ \frac{9}{2}

\displaystyle y - 8+ 8 = - \frac{3}{4} x+ \frac{9}{2} + 8

\displaystyle y = - \frac{3}{4} x+ \frac{25}{2}

 

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