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Intermediate Geometry : Tangent Lines

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Example Questions

Example Question #1 : Tangent Lines

Given the function y=3x+10 , find the equation of the tangent line passing through x=2 .

Possible Answers:

y=3x+6

y=3x

y=3x+10

y=3x+16

$\textup{The equation does not exist.}$

Correct answer:

y=3x+10

Explanation:

Find the slope of y=3x+10 . The slope is 3.

Substitute x=2 to determine the y-value.

y=3x+10=3(2)+10=16

The point is (2,16) .

Use the slope-intercept formula to find the y-intercept, given the point and slope.

y=mx+b

Substitute the point and the slope.

16=3(2)+b

b=10

Substitute the y-intercept and slope back to the slope-intercept formula.

The correct answer is: y=3x+10

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Example Question #1 : How To Find The Equation Of A Tangent Line

Find the equation of the tangent line at the point (2,4) if the given function is 2y=3x+4 .

Possible Answers:

y=3x+1

y=3x+2

y=3x-2

$y=\frac{3}{2}x+4$

$y=\frac{3}{2}x+1$

Correct answer:

$y=\frac{3}{2}x+1$

Explanation:

Write 2y=3x+4 in slope-intercept form y=mx+b and determine the slope.

Rearranging our equation to be in slope-intercept form we get:

$y=\frac{3}{2}x+2$ .

The slope is our value which is $\frac{3}{2}$ .

Substitute the slope and the point to the slope-intercept form.

y=mx+b

$4=\frac{3}{2}(2)+b$

b=1

Substitute the slope and y-intercept to find our final equation.

$y=\frac{3}{2}x+1$

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Example Question #1 : Tangent Lines

What is the equation of the tangent line at x=0 to the equation 5y-3x+2=0 ?

Possible Answers:

$y=\frac{3}{5}x-1$

$y=\frac{3}{5}x-\frac{2}{5}$

y=-3x-1

y=-3x+1

y=-3x-2

Correct answer:

$y=\frac{3}{5}x-\frac{2}{5}$

Explanation:

Rewrite 5y-3x+2=0 in slope-intercept form to determine the slope. Remember slope-intercept form is y=mx+b .

Therefore, the equation becomes,

5y=3x-2

$y=\frac{3}{5}x-\frac{2}{5}$ .

The slope is the value of the function thus, it is $\frac{3}{5}$ .

Substitute x=0 back to the original equation to find the value of .

5y-3(0)+2=0

5y+2=0

$y=-\frac{2}{5}$

Substitute the point $\left(0,-\frac{2}{5}\right)$ and the slope of the line into the slope-intercept equation.

y=mx+b

$-\frac{2}{5}=\left(\frac{3}{5}\right)(0)+b$

$b=-\frac{2}{5}$

Substitute the point and the slope back in to the slope-intercept formula.

$y=\frac{3}{5}x-\frac{2}{5}$

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Example Question #4 : Tangent Lines

Write the equation for the tangent line of the circle x^2 + (y-3)^2 = 41 through the point (4, -2) .

Possible Answers:

$y = \frac{4}{5}x - \frac{26}{5 }$

y = 4x + 3

$y =- \frac{1}{4} x - 1$

$y = \frac{4}{5} x + 3$

y = 4x - 18

Correct answer:

$y = \frac{4}{5}x - \frac{26}{5 }$

Explanation:

The circle's center is (0,3) . The tangent line will be perpendicular to the line going through the points (0,3) and (4, -2) , so it will be helpful to know the slope of this line:

$\frac{ \Delta y }{ \Delta x } = \frac{ 3 - - 2 }{ 0 - 4 } = \frac{ 5 }{ -4 }$

Since the tangent line is perpendicular, its slope is $\frac{4}{5 }$

To write the equation in the form y = mx + b , we need to solve for "b," the y-intercept. We can plug in the slope for "m" and the coordinates of the point (4, -2) for x and y:

$-2 = \frac{4}{5}(4) + b$

$-2 = \frac{16} { 5 } + b$

$-\frac{10}{5} = \frac{16} {5} + b$

$-\frac{26}{5} = b$

The equation is $y = \frac{4}{5} x - \frac{26} {5 }$

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Example Question #5 : Tangent Lines

Find the equation for the tangent line at (-3,8) for the circle (x+4)^2 + (y - 2 )^2 = 37 .

Possible Answers:

y = 6x + 10

$y = -\frac{7}{6}x + 4 \frac{1}{2}$

$y = -\frac{1}{6} x + 7\frac{1}{2 }$

$y = \frac{7}{6} x + 11 \frac{1}{2}$

$y = -\frac{1}{6} x + 2\frac{2}{3}$

Correct answer:

$y = -\frac{1}{6} x + 2\frac{2}{3}$

Explanation:

The circle's center is (-4, 2) . The tangent line will be perpendicular to the line going through the points (-4,2) and (-3, 8) , so it will be helpful to know the slope of this line:

$\frac{ \Delta y }{ \Delta x } = \frac{ 8-2 }{ -3+4 } = \frac{ 6 }{ 1 }$

Since the tangent line is perpendicular, its slope is $-\frac{1}{6 }$

To write the equation in the form y = mx + b , we need to solve for "b," the y-intercept. We can plug in the slope for "m" and the coordinates of the point (-3, 8 ) for x and y:

$8= -\frac{1}{6}(-3) + b$

$8 = \frac{1} { 2 } + b$

$7 \frac{1}{2} = b$

The equation is $y = -\frac{1}{6} x+7 \frac{1}{2}$

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Example Question #6 : Tangent Lines

Find the equation for the tangent line of the circle (y+1)^2 + (x- 4 ) ^ 2 = 8 at the point (-2, 3 ) .

Possible Answers:

$y = \frac{3}{2} x + 6$

$y = -\frac{3}{2}x$

$y = \frac{3}{2 } x - 2\frac{1}{2}$

$y = -\frac{2}{3} x - \frac{5}{3}$

$y = -\frac{2}{3} x + \frac{13}{3 }$

Correct answer:

$y = \frac{3}{2} x + 6$

Explanation:

The circle's center is (4, -1) . The tangent line will be perpendicular to the line going through the points (4, -1 ) and (-2,3) , so it will be helpful to know the slope of this line:

$\frac{ \Delta y }{ \Delta x } = \frac{-1 - 3 }{ 4 - - 2 } = \frac{-4}{6 } = -\frac{2}{3}$

Since the tangent line is perpendicular, its slope is $\frac{3}{2 }$ .

To write the equation in the form y = mx + b , we need to solve for "b," the y-intercept. We can plug in the slope for "m" and the coordinates of the point (-2,3) for x and y:

$3= \frac{3}{2}(-2) + b$

3=-3 + b

6= b

The equation is $y = \frac{3}{2} x +6$

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Example Question #7 : Tangent Lines

Find the equation for the tangent line to the circle (y+ 2 )^2 + (x -3)^2 = 169 at the point (15, 3 ) .

Possible Answers:

y = -18x + 273

y = -12x + 183

y = -18x + 69

$y = -\frac{12}{5} x + 39$

$y = -\frac{12} {5 } x + \frac{26}{5 }$

Correct answer:

$y = -\frac{12}{5} x + 39$

Explanation:

The circle's center is (3, -2) . The tangent line will be perpendicular to the line going through the points (3, -2) and (15,3) , so it will be helpful to know the slope of this line:

$\frac{ \Delta y }{ \Delta x } = \frac{ 3 --2 }{ 15 -3 } = \frac{ 5 }{ 12 }$

Since the tangent line is perpendicular, its slope is $-\frac{12}{5 }$

To write the equation in the form y = mx + b , we need to solve for "b," the y-intercept. We can plug in the slope for "m" and the coordinates of the point (15,3 ) for x and y:

$3 = -\frac{12}{5}(15) + b$

3 = -36+ b

39 = b

The equation is $y = - \frac{12}{5} x+39$

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Example Question #211 : Lines

Circle

Refer to the above diagram,

Give the equation of the line tangent to the circle at the point shown.

Possible Answers:

None of the other choices gives the correct response.

$y = \frac{4}{9} x+ \frac{65}{9}$

$y = \frac{9}{4}x$

$y = -\frac{9}{4}x+ 13$

$y = -\frac{4}{9} x+ \frac{97}{9}$

Correct answer:

$y = -\frac{4}{9} x+ \frac{97}{9}$

Explanation:

The tangent to a circle at a given point is perpendicular to the radius that has the center and the given point as its endpoints.

The circle has its center at origin (0,0) ; since this and (4,9) are the endpoints of the radius - and the line that includes this radius includes both points - its slope can be found by setting $x_{1} = y_{1} = 0 , x_{2}= 4, y_{2} = 9$ in the following slope formula:

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m = \frac{9-0}{4-0} = \frac{9 }{4 }$

The tangent line, being perpendicular to this radius, has as its slope the opposite of the reciprocal of this, which is $m = -\frac{4}{9}$ . Since the tangent line includes point (4, 9) , set $m = -\frac{4}{9}, x_{1} = 4, y_{1} = 9$ in the point-slope formula and simplify:

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

$y - 9= -\frac{4}{9} (x-4)$

$y - 9= -\frac{4}{9} \cdot x- \left (-\frac{4}{9} \right ) \cdot 4$

$y - 9= -\frac{4}{9} x+ \frac{16}{9}$

$y - 9+ 9 = -\frac{4}{9} x+ \frac{16}{9} + 9$

$y = -\frac{4}{9} x+ \frac{16}{9} + \frac{81}{9}$

$y = -\frac{4}{9} x+ \frac{97}{9}$

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Example Question #9 : Tangent Lines

A line is tangent to the circle $(x-4)^{2} + (y-2)^{2} = 100$ at the point (10,-6)

What is the equation of this line?

Possible Answers:

$y= \frac{4}{3}x-\frac{58}{3}$

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

None of the other answers are correct.

$y= \frac{3}{4}x-\frac{27}{2}$

$y= -\frac{4}{3}x+\frac{22}{3}$

Correct answer:

$y= \frac{3}{4}x-\frac{27}{2}$

Explanation:

The center of this circle is (4,2)

Therefore, the radius with endpoint (10,-6) has slope

$\frac{2-(-6)}{4-10}= \frac{8}{-6}= -\frac{4}{3}$

The tangent line at (10,-6) is perpendicular to this radius; therefore, its slope is the opposite of the reciprocal of $-\frac{4}{3}$ , or $\frac{3}{4}$ .

Now use the point-slope formula with this slope and the point of tangency:

$y-(-6)= \frac{3}{4} \left ( x-10\right )$

$y+6= \frac{3}{4}x-\frac{15}{2}$

$y= \frac{3}{4}x-\frac{27}{2}$

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Example Question #1 : Tangent Lines

What is the slope of the tangent line to the graph of y=x^2-6 when x=3 ?

Possible Answers:

Correct answer:

Explanation:

To find the slope of the tangent line, first we must take the derivative of x^2-6 , giving us . Next we simply plug in our given x-value, which in this case is x=3 . This leaves us with a slope of .

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Intermediate Geometry : Tangent Lines

Study concepts, example questions & explanations for Intermediate Geometry

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Example Questions

Example Question #1 : Tangent Lines

Example Question #1 : How To Find The Equation Of A Tangent Line

Example Question #1 : Tangent Lines

Example Question #4 : Tangent Lines

Example Question #5 : Tangent Lines

Example Question #6 : Tangent Lines

Example Question #7 : Tangent Lines

Example Question #211 : Lines

Example Question #9 : Tangent Lines

Example Question #1 : Tangent Lines

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