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Find the equation of the line passing though the point (5,2) and prependicular to the line joining the point (2,3) & (3,1) is
Slope of the line passing through
(2,3) & (3,1) is
slope of required line
since, the equation of the line passing through (5,2) & having slope is
Find the points on the line \(x + y = 4\), which lie t a unit distance from the line \(4x + 3y = 10\)
Given , Let P(h,k) lie on this line
The distance of the point P(h,k) from the line 4x + 3y = 10 is
Taking positive sign,
Taking negative sign,
The point are (3,1) & (7,11).
The tangent of an angle between the lines \(\dfrac{x}{a} + \dfrac{y}{b} = 1\& \dfrac{x}{a} – \dfrac{y}{b} = 1\) is
Find the angle between the lines \(y = \left( {2 – \sqrt 3 } \right)\left( {x + 5} \right)\& y = \left( {2 + \sqrt 3 } \right)\left( {x – 7} \right)\)
Find the equation of the line passing through the points (7,9) and (5,7) is
Find the equation of the line passing through (1,2) and making angle \(45^\circ \) with yaxis in positive direction.
Given, angle with yaxis=
angle with xaxis=
slope (m)= tan
so, equation of line passing through (1,2) & having slope (m) 1, is y2=1(x1)
Slope of a line which cuts off intercepts of equal lengths on the axes is
Equation of line is
(Intercept from)
slope (m) = – 1
If P is the length of perpendicular from the origin on the line \(\dfrac{x}{a} + \dfrac{y}{b} = 1\) and \({a^2},{p^2}\& {b^2}\) are in A.P, then
Equation of the line is
perpendicular length from origin on the above line is given by p.
i.e. P=
Given that are in A.P
A line cutting off intercept 3 from the yaxis and the tangent at angle to the xaxis is \(\dfrac{3}{5}\), its equation is
Given that, c=3 & m =
Equation of the line is
The equation of the straight line passing through the point (3,2) & perpendicular to the line y=x is
Line passes through (3,2) & perpandicular to the line y=x
slope (m) = 1 [ line is perpendicular to line y=x]
Equation of the line is
y – 2 = 1 (x – 3)
For what values of a & b the intercepts cut off on the cordinate axes by the line \(ax + by + 8 = 0\) are equal length but opposite in signs to those cut off by the line \(2x – 3y + 6 = 0\) on the axes ?
Given,
so, the intercepts are
Another given equation of line is
, so the intercepts are 3, 2 According to the question.
If the intercept of a line between the coordinate axes is divided by the point (5,4) in the ratio 1:2, then find the equation of the line?
Let intercepts are h, k
The coordnates of A & B are (h,k) & (0,k) respectively
Equation of the line AB is
Find the equation of a line on which length of perpendicular from the orign is four units and the line makes an angle of \(120^\circ \) with the positive direction of xaxis.
Given, OC= P= 4 Units
Let
Thus, equation of line is
If the equation of the base of an equilateral triangle is x+y=2 & the vector is (2,1) then find the length of the side of the triangle.
Equation of the base is x+y=2.
In ,
Length of perpendicular from (2,1) to the
line x+y=2 is given by
From (i),
A variable line passes through a fixed point P. The algebraic sum of the perpendiculr drawn from the points (2,0), (0,2) & (1,1) on the line is zero. Find the coordinates of the point P.
Slope=m, P=(x,y)
Eqution of line is (Slopepoint form)
Given A(2,0), B(0,2) & C(1,1)
Perpendicular distance from A is
perpendicular distance from B is
Perpendicular distance from C is
Now,
Since, (1,1) lies on this line. P=(1,1)
A straight line moves so that sum of the reciprocals of its intercepts mde on axes is constant. So the line passes through a fixed point.
( intercept form)
Given Constant
(say)
So (k,k) lies on
Hence, the line passes through the
fixed point (k,k).
If the sum of distance of a moving point in a plane from the axes is 1, then find the locus of the point.
Let P be moving point of coordinates (x,y).
Given, sum of the distances of this point
in a plane from the xes is 1.
These are the equations of given locus. xy=1
The tangent of angle between the lines whose intercepts on the axes are a,b & b, a respectively, is
xintercept=a, yintercept=b
Equation of the line is
xintercept=b, yintercept=a
Eqution of the line is
The distance between two lines \(y = mx + {c_1}\)& \(y = mx + {c_2}\) is
Given,
These two lines are parallel lines
If the coordinates of a middle point of the portion of a line intercepted between the coordinate axes is (3,2), then the equation of the line will be
P=(3,2)
Thus, equation of line is
(intercept form)
Equation of diagonals of the square formed by the lines x=0, y=0, x=1 & y=1 are
Equation OB is
Equation of AC is
For specifying a straight line, how many geometrical parameter should be known?
Equation of straight line are
, parameter=2
, parameter=2
,parameter=2
, parameter=2
The equation of the line through the intersection of the lines 2x3y=0 & 4x5y=2 and through the point (2,1) is
Given, 2x3y=0 & 4x5y=2
Solving above equation we get x=3, y=2
(x,y)=(3,2) is intersection point
Equation of the line passes through(3,2) & (2,1) is
(Twopoint form)
The equation of the line through point (3,2) & perpendicular to the line x+2y+1=0
Point=(3,2), line is x+2y+1=0
Slope(m)=2
Thus, equation of the required line is y2=2(x3)
The equation of the line through the intersection of the lines 2x3y=0 & 4x5y=2 and inclined equally to the axes is
Given, 2x3y=0 & 4x5y=2
Solving above two equations we get x=3, y=2
Since, the line equally inclined to
coordinated axes, then
Taking negative value, then equation of line is
y2=1(x3)
x+y=5