Note: If is a line, then the equation perpendicular to this line is given by
where
.
If is a line, then the equation parallel to this line is given by
where
.
Question 1: Find the equation of a line passing through the point and parallel to the line
.
Answer:
Therefore the line parallel to the given line is:
… … … … … i)
This line passes through . Therefore substituting this in i) we get
Therefore the equation of the required line is
Question 2: Find the equation of a line passing through and perpendicular to the line
Answer:
Therefore the line perpendicular to the given line is:
… … … … … i)
This line passes through . Therefore substituting this in i) we get
Therefore the equation of the required line is
Question 3: Find the equation of the perpendicular bisector of the line joining the points .
Answer:
Therefore the slope of perpendicular bisector
Hence the equation of the perpendicular bisector:
Question 4: Find the equation of the altitude of a whose vertices are
.
Answer:
Please refer to the adjoining figure.
Therefore the slope of AD
Hence the equation of AD:
Therefore the slope of BE
Hence the equation of BE:
Therefore the slope of CF
Hence the equation of CF:
Question 5: Find the equation of a line which is perpendicular to the line and which cuts off an intercept of
units with the negative direction of y-axis.
Answer:
Therefore the line perpendicular to the given line is:
… … … … … i)
This line passes through . Therefore substituting this in i) we get
Therefore the equation of the required line is
Question 6: If the image of the point with respect to a line mirror is
find the equation of the mirror.
Answer:
.
Hence the equation of mirror:
Question 7: Find the equation of the straight line through the point and perpendicular to the line
.
Answer:
Therefore the line perpendicular to the given line is:
… … … … … i)
This line passes through . Therefore substituting this in i) we get
Therefore the equation of the required line is
Question 8: Find the equation of the straight line perpendicular to and cutting off an intercept
on the positive direction of the x-axis.
Answer:
Therefore the line perpendicular to the given line is:
… … … … … i)
This line passes through . Therefore substituting this in i) we get
Therefore the equation of the required line is
Question 9: Find the equation of the straight line perpendicular to and which passes through the mid-point of the line segment joining
.
Answer:
Therefore the line perpendicular to the given line is:
… … … … … i)
Given points
This line passes through . Therefore substituting this in i) we get
Therefore the equation of the required line is
Question 10: Find the equation of the straight line line which has y-intercept equal to and is perpendicular to
.
Answer:
Therefore the line perpendicular to the given line is:
… … … … … i)
This line passes through . Therefore substituting this in i) we get
Therefore the equation of the required line is
Question 11: Find the equation of the right bisector of the line segment joining the points .
Answer:
.
Therefore the slope of right bisector
Hence the right bisector:
Question 12: Find the image of the point with respect to the line mirror
.
Answer:
.
Mid point C lies on , therefore
… … … … … i)
Slope of mirror
Hence the image of the point on mirror
Question 13: If the image of the point with respect to the line mirror be
find the equation of the mirror.
Answer:
.
Hence the equation of mirror:
Question 14: Find the equation to the straight line parallel to and passing through the middle point of the join of points
.
Answer:
Therefore the line parallel to the given line is:
… … … … … i)
.
This line passes through . Therefore substituting this in i) we get
Therefore the equation of the required line is
Question 15: Prove that the lines form a parallelogram
Answer:
Given lines:
Line 1:
Line 2:
Line 3:
Line 4:
Therefore slope of Line 1 and Line 3 is equal. Therefore the lines are parallel to each other.
Similarly, slope of Line 2 and Line 4 is equal. Therefore the lines are parallel to each other.
Hence the four lines will form a parallogram.
Question 16: Find the equation of a line drawn perpendicular to the line through the point where it meets the y-axis.
Answer:
The point of intersection with y axis means
Therefore for
Hence the point of intersection is
Therefore the line perpendicular to the given line is:
… … … … … i)
This line passes through . Therefore substituting this in i) we get
Therefore the equation of the required line is
Question 17: The perpendicular from the origin to the line meets it at the point
). Find the values of
.
Answer:
… … … … … i)
Therefore the slope of the line
Hence the slope of the line perpendicular to the given line
Therefore the equation of the required line:
… … … … … ii)
The point of intersection of line i) and ii) is given as .
Substituting in equation ii) we get
Therefore from i)
Hence,
Question 18: Find the equation of the right bisector of the given line: segment, joining the points .
Answer:
.
Therefore the slope of right bisector
Hence the equation of right bisector:
Question 19: The line through intersects the line
at right angle. Find the value of
.
Answer:
Question 20: Find the image of the point with respect to the line
assuming the line to be a plane mirror.
Answer:
… … … … … i)
Slope of mirror
Therefore slope of a line perpendicular to mirror
Let the image be
Equation of the perpendicular line:
… … … … … ii)
satisfies both line i) and ii), therefore
… … … … … iii)
Also,
… … … … … iv)
Solving iii) and iv) we get
Therefore the image of the point with respect to line
.
Question 21: Find the coordinates of the foot of the perpendicular from the point to the line
.
Answer:
… … … … … i)
Slope of the line
Therefore slope of a line perpendicular
Equation of the perpendicular line:
… … … ii)
Solving i) and ii) we get the point of intersection which is which is also the foot of the perpendicular from
Question 22: Find the projection of the point on the line joining the points
.
Answer:
.
Therefore the slope of perpendicular to
Equation of mirror:
… … … … … i)
Equation of perpendicular:
… … … … … ii)
Solving i) and ii) we get the point of intersection which is
Question 23: Find the equation of a line perpendicular to the line and at a distance of
, units from the origin.
Answer:
Therefore the line perpendicular to the given line is:
… … … … … i)
Given that line i) is 3 units
Therefore ,
Therefore the equation of the required line:
Question 24: The line meets the x-axis at
and y-axis at
. The line through
perpendicular to
meets the x-axis and the line
at
respectively. If
is the origin of coordinates, find the area of figure
.
Answer:
… … … … … i)
The x and y intercept are respectively.
Line perpendicular to i) is … … … … … ii)
Line ii) passes through , substituting we get
Therefore equation of the perpendicular is … … … … … iii)
Line iii) meets x-axis at
Point of intersection of line iii) and i) is
Area of Area
Area
Therefore Area of
Area of
Therefore the area of sq. units.
Question 25: Find the equation of the straight line which cuts off intercepts on x-axis twice that on y-axis and is at a unit distance from the origin.
Answer:
intercepts be
Therefore the equation of line is:
Changing this into normal form
( this is the length of perpendicular from origin)
sign is because the line could be on both sides of origin.
Therefore the equation of the required line is .
Question 26: The equations of perpendicular bisectors of the sides of a
are
respectively. If the point
find the equation of the line.
Answer:
Given lines: … … … … … i)
… … … … … ii)
Please refer to the adjoining figure.
Therefore equation of :
… … … … … iii)
Let the coordinate of be
Now is the mid point of
, therefore
lies on equation i)
… … … … … iv)
is also on line iii). Therefore,
… … … … … v)
Solving iv) and v) we get
Similarly, we calculate
Therefore the slope of equation :
… … … … … vi)
be the mid point of
. Therefore,
lies on line ii). Therefore,
… … … … … vii)
Also lies on vi)
… … … … … viii)
Solving vii) and viii) we get , and
Hence, equation of :