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Find the equation of the bisector of two lines 3x – 4y = 9 & 4x + 3y = 7
Equation of bisector
(Taking negative signs)
The line passes through (2,2) & is perpendicular to the line 3x+y=3, its y-intercept is
Given, y=3-3x, slope= -3
slope of the required line=
equation of the required line is
for y-intercept put x=0
The ratio in which the line 3x+4y+2=0 divides the distance between the lines 3x+4y+5=0 & 3x+4y-5=0 is
Perpendicular distance from A to 3x+4y+2=0
Perpendicular distance from B to 3x+4y+2=0
Required ratio =
=7:3
The co-ordinates of the foot of perpendiculars from the point (2,3) on the line y=3x+4 is
Given, y=3x+4, slope
slope of OP
Equation of OP is
using the value of y from given
x+3(3x+4)-11=0
If a line is at a distance K and parallel to x-axis, then the equation of the line is
Equation of the line is y=k
If a line is parallel to y-axis and at a distance C, then the equation of line is
L || Y-axis, AB=C
Equation of L is
x=-c
In \(\triangle ABC\) vertices are A(3,6), B(7,9) & C(3,9) & a,b,c are the length of \(\overline {BC}, \overline {AC} \& \overline {AB} \) respectively, find the co-ordinate of excentre to A.
is the excentre to A.
Then
=(6,12)
In \(\triangle ABC\) A(3,6), B(7,9) & C(3,9) are vertices & a,b,c represents the length of \(\overline {BC,} \overline {AC} \& \overline {AB} \) respectively, find the co-ordinate of excentre to B.
Given A(3,6), B(7,9) & C(3,9)
are the vertices of
Here a=4, b=3, c=5
co-ordinate of excentre OB
=(1,7)
In \(\triangle ABC\) A(3,6), B(7,9) & C(3,9) are vertices & a,b,c represents the length of \(\overline {BC,} \overline {AC} \& \overline {AB} \) respectively, find the co-ordinate of excentre to C.
Here a=4, b=3, c=5
co-ordinate of excentre to C (I_3)
A(3,6), B(7,9) & C(3,9) are vertices in \(\triangle ABC\) And a,b,c are length of \(\overline {BC} ,\overline {AC} \& \overline {AB} \) respectively. Find the co-ordinates of incentre of \(\triangle ABC\):
Let I be the incentre in
Here a=4, b=3, c=5
The points (2,5) & (5,1) are two opposite vertices of a rectangle. If other two vertices are points on the Straight line y=2x+k, then the value of k is
C bisects AB
since C line on the line y=2k+k
k= -4
If a straight line perpendicular to the line 2x+y=3 is passing through (1,1). Its y-intercept is
Straight line perpendicular to 2x+y=3 is 2y-x+c=0
Since, it passes through (1,1)
2(1)-1+c=0
of the line is 2y-x-1=0
so y-intrcept is
The ratio by which the line 2x+5y=7 divided the straight line joining the points (-4,7) & (6,-5) is
2x+5y-7=0-(i) Given points A(-4,7) & B(6,-5)
(two-point from)
5y – 35 = -6x -24
6x+3y=11 -(ii)
solving (i) & (ii) we get
x=1, & y=1 P(x,y)=(1,1)
since P(1,1) Is midpoint of A(-4,7) & B(6,-5)
then the ratio is 1:1.
The straight lines x+y=0, 5x+y=4 & x+5y=4 from
x+y=0 -(i)
5x+y=4 -(ii)
x+5y=4 -(iii)
solving above equation, vertices are
Let A(2,-3) & B(-2,1) be vertices of a \(\triangle ABC\). If the centroid of the triangle moves on the line 2x+3y=1, then the locus of the verter C is the line
Let
G satisfies the equation 2x+3y=1
Hence the locus of C is
2x+3y=9
Joint equation of pair of lines through (3,-2) & parallel to \({x^2} – 4xy + 3{y^2} = 0\) is
Solving,
The above lines pases through (3,-2)
Required equation of pair of lines
(3x-x+9) (y-x+5)=0
If the lines x+3y-9=0, 4x+by-2=0 & 2-y-4=0 are concurrent, then b is equal to
Given, x+3y-9=0, 4x+by-2=0, 2x-y-4=0 are concurrent if
(-4b-2)-3(-16+4)-9(-4-2b)=0
b= -5
If non-zero numbers, a,b,c are in HP, then the straight line \(\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{1}{c} = 0\) always passes through a fixed point, is
so, the line always passes through the fixed point (1,-2)
The values of \(\lambda \), for which the equation \({x^2} – {y^2} – x + \lambda y – 2 = 0\) represents a pair of straight lines is
Given
Here a=1, b=-1, c=-2, h=0
represents pair of straight lines.
The equation of pair of lines joinning origin to the points of intersection of \({x^2} + {y^2} = 9\& x + y = 3\) is
From above two equation, we make a homogeneous equation.
2xy=0
Find the area of the triangle with vertices A(5,0), B(5,5) & C(2,1)
Area
Find the co-ordinate of the centroid of the triangle ABC with vertices A(4,7), B(2,-1) & C(3,9)
Find the distance from P(2,-5) to the equation of line 4x-3y+2=0
Find the angle between the lines 2x+3y-5=0 & 3x-2y+11=0
: 2X+3Y-5=0
:3X-2Y+11=0
slope of
slope of