Tangent Lines - SSAT Upper Level Quantitative
Card 0 of 20
Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by 
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
Compare your answer with the correct one above
Find the equation of a tangent line at point 
if the function is 
.
Find the equation of a tangent line at point
To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function 
is already in the slope-intercept form, 
, and 
.
Substitute the slope and the given point 
into the slope-intercept equation.
Substitute the known slope and the y-intercept to the slope-intercept form.
To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function
Substitute the slope and the given point
Substitute the known slope and the y-intercept to the slope-intercept form.
Compare your answer with the correct one above
Find the slope of a tangent line at point 
if the equation of the function is 
.
Find the slope of a tangent line at point
To determine the slope of the function, 
, use the power rule to find the derivative function.
The slope at every point of the function 
has a slope of 
. To find the slope at the given point, substitute the x-value of the given point, 
, into the derivative function to find the slope.
To determine the slope of the function,
The slope at every point of the function
Compare your answer with the correct one above
There is a circle on a coordinate plain. Its perimeter passes through the point 
. At this point meets a tangent line, which also passes through the point 
. What is the slope of the line perpindicular to this tangent line?
There is a circle on a coordinate plain. Its perimeter passes through the point
In this kind of problem, it's important to keep track of information given about your line of interest. In this case, the coordinates given set up the stage for us to be able to get to our line of focus - the line perpendicular to the tangent line. In order to determine the perpendicular line's slope, the tangent line's slope must be calculated. Keeping in mind that:
where y2,x2 and y1,x1 are assigned arbitrarily as long as the order of assignment is maintained.
which is the slope of the tangent line.
To calculate the perpendicular line, we have to remember that the product of the tangent slope and the perpendicular slope will equal -1.
, the perpendicular slope can then be calculated as 
In this kind of problem, it's important to keep track of information given about your line of interest. In this case, the coordinates given set up the stage for us to be able to get to our line of focus - the line perpendicular to the tangent line. In order to determine the perpendicular line's slope, the tangent line's slope must be calculated. Keeping in mind that:
where y2,x2 and y1,x1 are assigned arbitrarily as long as the order of assignment is maintained.
To calculate the perpendicular line, we have to remember that the product of the tangent slope and the perpendicular slope will equal -1.
Compare your answer with the correct one above
Find the slope of a tangent line at point 
if the equation is 
.
Find the slope of a tangent line at point
Rewrite the linear equation in standard form to slope-intercept form, 
.
Since the slope of every point of this line is 
, the slope of the tangent line at the given point should also be 
.
Rewrite the linear equation in standard form to slope-intercept form,
Since the slope of every point of this line is
Compare your answer with the correct one above
Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
Compare your answer with the correct one above
Find the equation of a tangent line at point if the function is .
Find the equation of a tangent line at point
To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function is already in the slope-intercept form, , and .
Substitute the slope and the given point into the slope-intercept equation.
Substitute the known slope and the y-intercept to the slope-intercept form.
To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function
Substitute the slope and the given point
Substitute the known slope and the y-intercept to the slope-intercept form.
Compare your answer with the correct one above
Find the slope of a tangent line at point if the equation of the function is .
Find the slope of a tangent line at point
To determine the slope of the function, , use the power rule to find the derivative function.
The slope at every point of the function has a slope of . To find the slope at the given point, substitute the x-value of the given point, , into the derivative function to find the slope.
To determine the slope of the function,
The slope at every point of the function
Compare your answer with the correct one above
There is a circle on a coordinate plain. Its perimeter passes through the point . At this point meets a tangent line, which also passes through the point . What is the slope of the line perpindicular to this tangent line?
There is a circle on a coordinate plain. Its perimeter passes through the point
In this kind of problem, it's important to keep track of information given about your line of interest. In this case, the coordinates given set up the stage for us to be able to get to our line of focus - the line perpendicular to the tangent line. In order to determine the perpendicular line's slope, the tangent line's slope must be calculated. Keeping in mind that:
where y2,x2 and y1,x1 are assigned arbitrarily as long as the order of assignment is maintained.
which is the slope of the tangent line.
To calculate the perpendicular line, we have to remember that the product of the tangent slope and the perpendicular slope will equal -1.
, the perpendicular slope can then be calculated as
In this kind of problem, it's important to keep track of information given about your line of interest. In this case, the coordinates given set up the stage for us to be able to get to our line of focus - the line perpendicular to the tangent line. In order to determine the perpendicular line's slope, the tangent line's slope must be calculated. Keeping in mind that:
where y2,x2 and y1,x1 are assigned arbitrarily as long as the order of assignment is maintained.
To calculate the perpendicular line, we have to remember that the product of the tangent slope and the perpendicular slope will equal -1.
Compare your answer with the correct one above
Find the slope of a tangent line at point if the equation is .
Find the slope of a tangent line at point
Rewrite the linear equation in standard form to slope-intercept form, .
Since the slope of every point of this line is , the slope of the tangent line at the given point should also be .
Rewrite the linear equation in standard form to slope-intercept form,
Since the slope of every point of this line is
Compare your answer with the correct one above
Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
Compare your answer with the correct one above
Find the equation of a tangent line at point if the function is .
Find the equation of a tangent line at point
To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function is already in the slope-intercept form, , and .
Substitute the slope and the given point into the slope-intercept equation.
Substitute the known slope and the y-intercept to the slope-intercept form.
To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function
Substitute the slope and the given point
Substitute the known slope and the y-intercept to the slope-intercept form.
Compare your answer with the correct one above
Find the slope of a tangent line at point if the equation of the function is .
Find the slope of a tangent line at point
To determine the slope of the function, , use the power rule to find the derivative function.
The slope at every point of the function has a slope of . To find the slope at the given point, substitute the x-value of the given point, , into the derivative function to find the slope.
To determine the slope of the function,
The slope at every point of the function
Compare your answer with the correct one above
There is a circle on a coordinate plain. Its perimeter passes through the point . At this point meets a tangent line, which also passes through the point . What is the slope of the line perpindicular to this tangent line?
There is a circle on a coordinate plain. Its perimeter passes through the point
In this kind of problem, it's important to keep track of information given about your line of interest. In this case, the coordinates given set up the stage for us to be able to get to our line of focus - the line perpendicular to the tangent line. In order to determine the perpendicular line's slope, the tangent line's slope must be calculated. Keeping in mind that:
where y2,x2 and y1,x1 are assigned arbitrarily as long as the order of assignment is maintained.
which is the slope of the tangent line.
To calculate the perpendicular line, we have to remember that the product of the tangent slope and the perpendicular slope will equal -1.
, the perpendicular slope can then be calculated as
In this kind of problem, it's important to keep track of information given about your line of interest. In this case, the coordinates given set up the stage for us to be able to get to our line of focus - the line perpendicular to the tangent line. In order to determine the perpendicular line's slope, the tangent line's slope must be calculated. Keeping in mind that:
where y2,x2 and y1,x1 are assigned arbitrarily as long as the order of assignment is maintained.
To calculate the perpendicular line, we have to remember that the product of the tangent slope and the perpendicular slope will equal -1.
Compare your answer with the correct one above
Find the slope of a tangent line at point if the equation is .
Find the slope of a tangent line at point
Rewrite the linear equation in standard form to slope-intercept form, .
Since the slope of every point of this line is , the slope of the tangent line at the given point should also be .
Rewrite the linear equation in standard form to slope-intercept form,
Since the slope of every point of this line is
Compare your answer with the correct one above
Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?
The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by
The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.
The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.
The equation of the line is y – 4 = (3/4)(x – (–3))
Rearranging gives us: 3_x_ – 4_y_ = -25
Compare your answer with the correct one above
Find the equation of a tangent line at point if the function is .
Find the equation of a tangent line at point
To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function is already in the slope-intercept form, , and .
Substitute the slope and the given point into the slope-intercept equation.
Substitute the known slope and the y-intercept to the slope-intercept form.
To find the slope of the tangent line, it is necessary to determine the slope of the function.
The function
Substitute the slope and the given point
Substitute the known slope and the y-intercept to the slope-intercept form.
Compare your answer with the correct one above
Find the slope of a tangent line at point if the equation of the function is .
Find the slope of a tangent line at point
To determine the slope of the function, , use the power rule to find the derivative function.
The slope at every point of the function has a slope of . To find the slope at the given point, substitute the x-value of the given point, , into the derivative function to find the slope.
To determine the slope of the function,
The slope at every point of the function
Compare your answer with the correct one above
There is a circle on a coordinate plain. Its perimeter passes through the point . At this point meets a tangent line, which also passes through the point . What is the slope of the line perpindicular to this tangent line?
There is a circle on a coordinate plain. Its perimeter passes through the point
In this kind of problem, it's important to keep track of information given about your line of interest. In this case, the coordinates given set up the stage for us to be able to get to our line of focus - the line perpendicular to the tangent line. In order to determine the perpendicular line's slope, the tangent line's slope must be calculated. Keeping in mind that:
where y2,x2 and y1,x1 are assigned arbitrarily as long as the order of assignment is maintained.
which is the slope of the tangent line.
To calculate the perpendicular line, we have to remember that the product of the tangent slope and the perpendicular slope will equal -1.
, the perpendicular slope can then be calculated as
In this kind of problem, it's important to keep track of information given about your line of interest. In this case, the coordinates given set up the stage for us to be able to get to our line of focus - the line perpendicular to the tangent line. In order to determine the perpendicular line's slope, the tangent line's slope must be calculated. Keeping in mind that:
where y2,x2 and y1,x1 are assigned arbitrarily as long as the order of assignment is maintained.
To calculate the perpendicular line, we have to remember that the product of the tangent slope and the perpendicular slope will equal -1.
Compare your answer with the correct one above
Find the slope of a tangent line at point if the equation is .
Find the slope of a tangent line at point
Rewrite the linear equation in standard form to slope-intercept form, .
Since the slope of every point of this line is , the slope of the tangent line at the given point should also be .
Rewrite the linear equation in standard form to slope-intercept form,
Since the slope of every point of this line is
Compare your answer with the correct one above