LINE IN GEOMETRY

LINE

A line is collection of points extending in both directions.

line has length but has zero thickness.

or

A line is formed by joining two or more than two points.

If the line formed by joining three points on straight line then these point are called collinear points.

Below given line is formed by joining points A and B

Equations of line in different form:

1. Slope intercept form 

If the slope of the line with the axis and intercept on the y axis are given then the equation of line is given by

y=mx+c

where m is the slope of line and c is the intercept on the y axis.

2. Point slope form

If the slope of the line and point through which the line passes are given then the equation of line is given by

y-y1=m(x-x1)

where m is slope of line and (x1 and y1) are points through which the line passes.

3. Two point form

If the two points through which line passes are given then the equation of line id given by

 


Find the equation of line passing through (2,-3) and making an angle 135 with the positive direction of X axis?

This equation of line will be in point slope form.

now,

slope of line(m)=tanθ=tan135=-1

and point (2,-3)

general form y-y1=m(x-x1)

or

y-(-3)=-1(x-2)

y+3=-1(x-2)

y+3=-x+2

x+y+1=0

What is equation of straight line having slope 1/2 and passing through the mid point joining the point (-4,7) and (-2,3)?
Mid point of given point is (-4-2)/2, (7+3)/2

i.e. (-3,5)

slope=1/2

so the equation of line in one point form

is

y-y1=m(x-x1)

where m=1/2

and (x1,y1)=(-3,5)

y-5=1/2((x-(-3))

y-5=1/2(x+3)

2y-10=x+3

x-2y+13=0

What is the area of the triangle formed by the graph of x + y + 6 = 0 with coordinate axis?
The line cut the x and y axis at (-6,-6)

hence the distance from origin is 6

and area of triangle=1/2*(6*6)=36/2=18 squre unit

What is the gradient of a line given by y = 3x + 5?

y=3X+5

y=mx+c

compare both

m=slope=3

and

diff w.r.t x

dy/dx=3

What is the equation of line passing through the point of intersection of the lines X+3y-1=0,X-2y+4=0, and perpendicular to the line 3X+2y=0?
Point of intersection of lines x+3y=1 and x-2y=-4

is (-14,5) and slope of line perpendicular to given line 3x+2y=0

is 2/3

then one point form

(y-y1)=m(x-x1)

y-5=2/3(x-(-14))

3(y-5)=2(x+14)

3y-15=2x+28

2x-3y+28+15=0

2x-3y+33=0

At what point does the line 3x + 2y = 12 cuts the Y-axis?
3x+2y=12
if thwe line cut Y axis then x coordinate of line will be zero i.e. x=0
then put x=0 in the above line
3x + 2y = 12
3*0+2y=12
2y=12
y=6
so the coordinate is (0,6)

What is the equation of the perpendicular bisector of the line segment with points (2,3) and (5,6)?

Mid point of (2,3) and (5,6) is (7/2,9/2)

and slope of line joining the given point is (y2-y1)/(x2-x1)  

i.e. (6-3)/(5-2)=3/3=1 lets say m1=1

and slope of perpendicular bisector is m1*m2=-1

1*m2=-1

m2=-1

so the equation of perpendicular bisector is y-y3=m2(x-x3)

where (x3,y3)=(7/2,9/2)

y-9/2=-1(x-9/2)

or y-9/2=-x+9/2

x+y=9/2+9/2

x+y=9