Along with the material on parabolas in the workshop on Functions,
Equations and Polynomials, some useful formulae include:
If the line is
reflected in the line , find
the equation of the reflected line.
Solution
Recall that after reflection in a line, the distance from the image
of any point to the line of reflection is the same as the distance from
the original point to the line. Thus, the line segment joining any point
to its image is perpendicular to the line of reflection and the midpoint
of this line segment is on the line of reflection.
The line has
intercepts of and . Since the image of a line after
reflection is another line, we reflect these two points and then find
the equation of the line through their image points. The image of the
point upon reflection in the
line is . This result follows since the
line segment from to has slope 1, which makes it
perpendicular to , and its
midpoint is , which is on the line . Similarly, the image of the point
reflected in the line is ,. Since the line passes through and , the slope of the line is and the equation of the line
is or .
If and are fixed points, find the
point(s) on the -axis such that the area of the triangle
equals .
Solution 1
The length of is . The slope of the line through and is . Therefore, the equation of the line is or . Now if we think of as the base of the triangle, then the
distance from , to the line must be the height of the triangle and
therefore, this distance is . Thus,
The points are and .
Solution 2
Let . Then using the
formula for the area of a triangle given its vertices we obtain Therefore, and so or .
The points are and .
Given two circles, the line joining their points of intersection
is called their common chord. It can be shown that the common
chord is perpendicular to the line connecting the centres of the
circles. (Can you prove this?) Given the circles and , find the length of their
common chord.
Solution
The first circle has centre and radius . By completing the square, the equation
for the second circle can be written as and therefore, it has
centre and radius . Since the line joining the
centres is horizontal, the common chord is vertical. We can find the
intersection points of the two circles by solving
Substituting this value back into the equation for either circle
gives intersection points . Thus, the length of the common chord is .
A line has slope and is a
distance of units from the
origin. What is the area of the triangle formed by this line and the
axes?
Solution
Let the -intercept of the line
be . Since the line has slope
, the -intercept is , and the equation of the line is . Therefore, the formula for the
distance from a line to a point shows that the distance from this line
to is .
Since this distance is , we have
that . The base of the
triangle is the distance from the origin to the -intercept. This distance is . The height of the triangle
is the distance from the origin to the -intercept. This distance is . Therefore, the area of
the required triangle is .
A vertical line divides the triangle with vertices , , and into two regions of equal area.
Find the equation of the line.
Find all value(s) of such
that the line is tangent to
the circle .
Find all value(s) of so
that the circle with equation will intersect the circle
with equation
in exactly one point.
A circle intersects the axes at ,, , and ,. A line through , cuts the circle in half. What is the
intercept of the line?
If triangle has vertices
, , and , determine the equation of the
bisector of .
What are the length(s) and the slope(s) of the tangent(s) from
the origin to the point of tangency with the circle ?
Find the equation of the set of points equidistant from and .
In quadrilateral , is at the origin, is on the positive -axis and and are in the first quadrant. The
midpoints of and are and , respectively. Given that , prove that is parallel to .
The point is on the line
and the point is on the line . If the midpoint of is , find the co-ordinates of and .