Question

Transform the equation x/a+y/b=1 into the normal form. If the perpendicular distance of straight line from the origin hp, distance from 1/p²=1/a²=1/b²​

Answer 1

EXPLANATION.

Transform the equation ⇒ x/a + y/b = 1.

If perpendicular distance of straight line from the origin.

distance from 1/p² = 1/a² + 1/b².

As we know that,

Equation of line whose intercept on the axis a and b is,

x/a + y/b = 1.

Length of perpendicular,

From (x₁ , y₁) to the straight line ax + by + c = 0, then

P = | ax₁ + by₁ + c |/√a² + b².

Compare to the straight line ax + by + c = 0.

A = 1/a , B = 1/b , c = -1.

Distance from origin (0,0).

P = | (0).(1/a) + (0).(1/b) - 1 |/√1/a² + 1/b².

P = |-1|/√1/a² + 1/b².

P = 1/√1/a² + 1/b².

1/P = √1/a² + 1/b².

Squaring on both sides, we get.

(1/P)² = (√1/a² + 1/b²)².

we get,

1/p² = 1/a² + 1/b².

Answer 2

Required solution :

Transform the equation x/a + y/b = 1 into the normal form.

If the perpendicular distance of straight line from the origin hp, distance from 1/p²=1/a²=1/b².

Now,

We know that,

Equation of line which intersects on x axis and g axis at any point :

→ x/a + y/b = 1

So, Length of perpendicular from axis (x1, Y1) to straight line is ax + by + c = 0.

Then,

→ P = | ax1 + by1 + c |/ √a² + b².

Now,

Comparing the above equation by straight line.

We get,

ax + by + c = 0

• A = 1/a
• B = 1/b
• C = -1

Distance from origin (0,0) will be :-

→ P = | 0 × (1/a) + 0 × (1/b) - 1 |/ √1/a² + 1/b²

→ P = | -1 |/ √1/a² + 1/b²

→ 1/P = √1/a² + 1/b².

Now,

Squaring on the both sides.

We get,

1/P² = 1/a² + 1/b².

Hence :

Therefore, The equation x/a + y/b = 1 in normal form is 1/p² = 1/a² + 1/b².