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Equation of Circle Whose Extremities of a Diameter Given

Let d 1 and d...

Question

Let $d_{1}$ and $d_{2}$ be the lengths of the perpendiculars drawn from any point of the line $7 x - 9 y + 10 = 0$ upon the lines $3 x + 4 y = 5$ and $12 x + 5 y = 7$ respectively. Then

d_{1} > d_{2}

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d_{1} = d_{2}

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d_{1} < d_{2}

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d_{1} = 2 d_{2}

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Solution

The correct option is B $d_{1} = d_{2}$
Let $(α, β)$ be any point on $7 x - 9 y + 10 = 0,$
so, $\Rightarrow 7 α - 9 β + 10 = 0 \Rightarrow β = \frac{7 α + 10}{9}$
Perpendicular distances $d_{1} = \frac{3 α + 4 β - 5}{\sqrt{9 + 16}} = \frac{3 α + 4 (\frac{7 α + 10}{9}) - 5}{\sqrt{9 + 16}} d_{1} = \frac{11 α - 1}{9}$
$d_{2} = \frac{12 α + 5 β - 7}{\sqrt{144 + 25}} = \frac{12 α + 5 (\frac{7 α + 10}{9}) - 7}{\sqrt{169}} d_{2} = \frac{11 α - 1}{9}$
$∴ d_{1} = d_{2}$

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Similar questions

Q. Let

d_{1}

and

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be the lengths of the perpendiculars drawn from any point of the line

7 x - 9 y + 10 = 0

upon the lines

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and

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respectively. Then

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A line meets the co-ordinate axes in A & B, A circle is circumscribed about the triangle OAB. If $d_{1} a n d d_{2}$ are the distances of the tangent to the circle at the origin O from the points A and B respectively the diameter of the circle is:

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6 x + 8 y + 20 = 0

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and

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Let d1 and d2 be the lengths of the perpendiculars drawn from any point of the line 7x−9y+10=0 upon the lines 3x+4y=5 and 12x+5y=7 respectively. Then

The correct option is B d1=d2Let (α,β) be any point on 7x−9y+10=0, so, ⇒7α−9β+10=0⇒β=7α+109 Perpendicular distances d1=3α+4β−5√9+16=3α+4(7α+109)−5√9+16d1=11α−19 d2=12α+5β−7√144+25=12α+5(7α+109)−7√169d2=11α−19 ∴d1=d2

Let $d_{1}$ and $d_{2}$ be the lengths of the perpendiculars drawn from any point of the line $7 x - 9 y + 10 = 0$ upon the lines $3 x + 4 y = 5$ and $12 x + 5 y = 7$ respectively. Then