Verified answer

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°