Question

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

Answer 1

Answer:

Let the equation of the line parallel to x−2y=1 is x−2y+λ=0

Since, it passes through (3,5)

⇒3−10+λ=0

⇒λ=7

Therefore, the line is x−2y+7=0.

The point of intersection of x−2y+7=0 and 2x+3y−14=0 is (1,4).

The distance between (3,5) and (1,4)

=

(3−1)

2

+(5−4)

2

=

4+1

= 5

Step-by-step explanation:

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