Question

1 point
In a triangle made by the points of the intersection of the two axes and the line
b
x
+
a
y
=
a
b
,
d
is the height of the perpendicular from the origin to the
opposite side. Then
1
a
2
+
1
b
2
is equal to

√
2
d
2

√
2
d
2

1
d
2

d
2​

Answer 1

Answer:

1/d^2 is the answer...

Step-by-step explanation:

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Answer 2

Answer:

Answer is 1/d^2.

Step-by-step explanation:

1. We are given with perpendicular distance from origin to a straight line and we are asked to find $\frac{1}{a^2} + \frac{1}{b^2}$.
2. We know the formula for perpendicular distance from origin to a straight line ax+by+c = 0 is,
3. distance = $\frac{|c|}{\sqrt{a^2 + b^2}}$. For our case c = -ab, a = a, b = b.
4. $d = \frac{ab}{\sqrt{a^2+b^2}} = \frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\\\sqrt{\frac{1}{a^2}+\frac{1}{b^2}} = \frac{1}{d}\\\\\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{d^2}$
5. Hence we arrived at our answer.

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