The intercepts made by a line on the axes are equal in magnitude but opposite in signs . If the line passes through the point (3,2) then find the equation of the line . Express the equation of the line . ¡) in gradient from , ¡¡) in normal form .
Answers
Given :-
The intercepts made by a line on the axes are equal in magnitude but opposite in signs . If the line passes through the point (3,2) then find the equation of the line .
To find :-
The equation of the line
¡) in gradient from ,
¡¡) in normal form .
Solution :-
We know that
The intercept form of the equation of a line is
( x/a ) + ( y/b ) = 1 --------(1)
Given that
The intercepts made by a line on the axes are equal in magnitude but opposite in signs
Let they be a and -a
The above equation becomes
=> (x/a)+(y/-a) = 1
=> (x/a)-(y/a) = 1
=> (x-y)/a = 1
=> x-y = 1×a
=> x-y = a ---------------(2)
The point (3,2) passes throught the above lene then (2) becomes
=> 3-2 = a
=> 1 = a
Therefore, a = 1
If a = 1 then (2) becomes x-y = 1
The required equation of the line is x-y = 1
i)
The line is x-y = 1
It can be written as x = 1+y
=> y = x-1
=> y = 1x+(-1)
It is in the form of y = mx+c ,
where m = 1 and c = -1
The slope or gradient of the line = 1
ii)
We have, x - y = 1
We know that
The normal form of a line is x cos θ + y sin θ = p
where p is the perpendicular distance from the origin and θ is the angle between perpendicular and positive x-axis .
now
x-y = 1
On dividing the line by √2
since, √(a²+b²) = √[1²+(-1)²] = √(1+1) = √2 then
(x/√2) - (y/√2) = (1/√2)
=> x (1/√2) + y(-1/√2) = (1/√2)
On comparing this with x cos θ + y sin θ = p
cos θ = 1/√2 and sin θ = -1/√2 and p = 1/√2
We know that
sin 45° = cos 45° = 1/√2
As sin is negative and cos is positive then θ must be in 4th quadrant .
Therefore,the required angle = 360°-45° = 315°
The normal form is x cos 315°+y sin 315° = 1/√2
Answer :-
i) The gradient form of the line and is y = x-1
Slope or gradient = 1
ii) Normal form of the line is
x cos 315°+y sin 315° = 1/√2
p = 1/√2
The angle = 315°