70. Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.
70. Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.
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1 Answer
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70. The given equation of the line is
l1: x + y = 4
Let P (x0, y0) be the point of intersect of l1 and the line to be drawn.
Then, x0 + y0 = 4 ⇒ y0 = 4? x
Given, distance between P (x0, y0) and Q (? 1, 2) is 3
ie,
⇒ (x0 + 1)2 + (y? 2)2= 9
⇒x20+1+ 2x0 + (4? x? 2)2 = 9
⇒ x20+ 2x0 + 1 + (2? x0 )2 = 9
⇒x20+ 2x0 + 1 + 4 + x20 ? 4x0 ?9 = 0
⇒ 2 x20 ?2x0 ? 4 = 0
x20 ? x0 ? 2 = 0
x20 + x0 ? 2x0 ? 2 = 0
x0 (x +1)? 2 (x0 +1) = 0
(x0 +1) (x0 ? 2) = 0
x0 = 2 and x0 =? 1
When, x0 = 2, y0 = 4 ?2 = 2.
and when x0 =? 1, y0 = y? (?1) =5.
The points of interaction of line l1which are
...more
Similar Questions for you
Eqn : y – 0 = tan45° (x – 9) Þ y = (x – 9)
Option (B) is correct
|r1 – r2| < c1c2 < r1 + r2
->
Now,
(y – 2) = m (x – 8)
⇒ x-intercept
⇒
⇒ y-intercept
⇒ (–8m + 2)
⇒ OA + OB =
->
->
->
->Minimum = 18
Kindly consider the following figure
According to question,
Equation of required line is
Obviously B (2, 2) satisfying condition (i)
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