English

Find the Equations of the Bisectors of the Angles Between the Coordinate Axes.

Advertisements
Advertisements

Question

Find the equations of the bisectors of the angles between the coordinate axes.

Answer in Brief
Advertisements

Solution

There are two bisectors of the coordinate axes.
Their inclinations with the positive x-axis are

\[{45}^\circ \text { and } {135}^\circ\]

So, the slope of the bisector is \[m = \tan {45}^\circ \text { or } m = \tan {135}^\circ , \text { i . e . m = 1 or } m = - 1\] and c = 0.

Substituting the values of m and c in y = mx + c, we get,
y = x + 0

\[\Rightarrow\] x \[-\] y = 0 or y = - x + 0

\[\Rightarrow\] x + y = 0

Hence, the equation of the bisector is \[x \pm y = 0\].

shaalaa.com
  Is there an error in this question or solution?
Chapter 23: The straight lines - Exercise 23.3 [Page 21]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 23 The straight lines
Exercise 23.3 | Q 3 | Page 21

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Draw a quadrilateral in the Cartesian plane, whose vertices are (–4, 5), (0, 7), (5, –5) and (–4, –2). Also, find its area.


Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).


Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, –4) and B (8, 0).


The slope of a line is double of the slope of another line. If tangent of the angle between them is `1/3`, find the slopes of the lines.


A line passes through (x1, y1) and (h, k). If slope of the line is m, show that k – y1 = m (h – x1).


Find the equation of a line drawn perpendicular to the line `x/4 + y/6 = 1`through the point, where it meets the y-axis.


Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.


Find the slope of a line passing through the following point:

\[(a t_1^2 , 2 a t_1 ) \text { and } (a t_2^2 , 2 a t_2 )\]


Find the slope of a line passing through the following point:

(3, −5), and (1, 2)


Find the value of x for which the points (x, −1), (2, 1) and (4, 5) are collinear.


By using the concept of slope, show that the points (−2, −1), (4, 0), (3, 3) and (−3, 2) are the vertices of a parallelogram.


Find the equation of a line which is perpendicular to the line joining (4, 2) and (3, 5) and cuts off an intercept of length 3 on y-axis.


Show that the perpendicular bisectors of the sides of a triangle are concurrent.


Find the equations of the altitudes of a ∆ ABC whose vertices are A (1, 4), B (−3, 2) and C (−5, −3).


Find the equation of the right bisector of the line segment joining the points (3, 4) and (−1, 2).


The line through (h, 3) and (4, 1) intersects the line 7x − 9y − 19 = 0 at right angle. Find the value of h.


Find the angles between the following pair of straight lines:

3x + y + 12 = 0 and x + 2y − 1 = 0


Find the angles between the following pair of straight lines:

3x − y + 5 = 0 and x − 3y + 1 = 0


Show that the line a2x + ay + 1 = 0 is perpendicular to the line x − ay = 1 for all non-zero real values of a.


If two opposite vertices of a square are (1, 2) and (5, 8), find the coordinates of its other two vertices and the equations of its sides.


The reflection of the point (4, −13) about the line 5x + y + 6 = 0 is  


If the slopes of the lines given by the equation ax2 + 2hxy + by2 = 0 are in the ratio 5 : 3, then the ratio h2 : ab = ______.


If x + y = k is normal to y2 = 12x, then k is ______.


The two lines ax + by = c and a′x + b′y = c′ are perpendicular if ______.


The tangent of angle between the lines whose intercepts on the axes are a, – b and b, – a, respectively, is ______.


The coordinates of the foot of perpendiculars from the point (2, 3) on the line y = 3x + 4 is given by ______.


The points (3, 4) and (2, – 6) are situated on the ______ of the line 3x – 4y – 8 = 0.


If the vertices of a triangle have integral coordinates, then the triangle can not be equilateral.


The line `x/a + y/b` = 1 moves in such a way that `1/a^2 + 1/b^2 = 1/c^2`, where c is a constant. The locus of the foot of the perpendicular from the origin on the given line is x2 + y2 = c2.


The equation of the line through the intersection of the lines 2x – 3y = 0 and 4x – 5y = 2 and

Column C1 Column C2
(a) Through the point (2, 1) is (i) 2x – y = 4
(b) Perpendicular to the line (ii) x + y – 5
= 0 x + 2y + 1 = 0 is
(ii) x + y – 5 = 0
(c) Parallel to the line (iii) x – y –1 = 0
3x – 4y + 5 = 0 is
(iii) x – y –1 = 0
(d) Equally inclined to the axes is (iv) 3x – 4y – 1 = 0

The line which passes through the origin and intersect the two lines `(x - 1)/2 = (y + 3)/4 = (z - 5)/3, (x - 4)/2 = (y + 3)/3 = (z - 14)/4`, is ______.


A ray of light coming from the point (1, 2) is reflected at a point A on the x-axis and then passes through the point (5, 3). The co-ordinates of the point A is ______.


The lines whose vector equations are `r = 2hati - 3hatj + 7hatk + lambda (2hati + phatj + 5hatk) and r = hati - 2hatj + 3hatk + µ(3hati + phatj + phatk)` are perpendicular for all values of λ and µ if p =


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×