Question

148. The distance of the point (1,2) from the linex+y+5=0
measured along the line parallel to 3x-y=7 is equal
to
(a) 4√10 (b) 40
(c) √40
(e) 2√20

(d) 10√2​

Answer 1

Answer:(c) √40

Explanation:

We have to find => distance of the point (1,2) from the line x+y + 5 = 0 measured along the line parallel to 3x-y = 7 *

∴Distance is measured along the line whose is equal to line 3x-y - 7 and passes through (1,2)

3x -y - k = 3* (1)- 2 - k = 1- k = k = 1So equation of the line is 3x-y -1 = 0

Now, we have two equation , x+y+5 =0_______(1)

and 3x-y - 1 = 0 __________(2)

substituting x = -(y+5) in (2) equation

∴ 3(-y-5) -y - 1 = 0

= -3y-y - 15 -1 -4y = 16y = -4 put y = -4 in (1) we get x = -1

now ,we have two points (1,2) and (-1,-4)

Using distance formula

we have , $\blue { \sqrt{(1 + 1) {}^{2} + (2 + 4) {}^{2} } \: = \sqrt{40} \pink \: {answer} }$

____________________________

Answer 2

Answer:

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