Question
Find the equation of the straight line passing through the point (2, 1) and through the point of intersection of the lines x + 2y = 3 and 2x – 3y= 4.
Answer: Option A
:
A
1st method: - equation of any straight line passing through the intersection of the lines x + 2y = 3 and 2x – 3y= 4 is
λ(x + 2y – 3) + (2x – 3y – 4) = 0
Since it passes through the point (2, 1)
λ(2 + 2 – 3) + (4 – 3 – 4) = 0
λ - 3 = 0
λ = 3
Now substituting this value of λ in (i), we get
3(x + 2y – 3) + (2x – 3y – 4) = 0
5x + 3y – 13 = 0
2nd method: - The straight line passing through the point (2, 1), put x = 2 and y = 1only option (a) and (b) will satisfy. Now, intersection point of the lines x + 2y = 3 and 2x – 3y= 4 is (177,27), Put x = 177 and y = 27 only option(a) will satisfy. Option(a).
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:
A
1st method: - equation of any straight line passing through the intersection of the lines x + 2y = 3 and 2x – 3y= 4 is
λ(x + 2y – 3) + (2x – 3y – 4) = 0
Since it passes through the point (2, 1)
λ(2 + 2 – 3) + (4 – 3 – 4) = 0
λ - 3 = 0
λ = 3
Now substituting this value of λ in (i), we get
3(x + 2y – 3) + (2x – 3y – 4) = 0
5x + 3y – 13 = 0
2nd method: - The straight line passing through the point (2, 1), put x = 2 and y = 1only option (a) and (b) will satisfy. Now, intersection point of the lines x + 2y = 3 and 2x – 3y= 4 is (177,27), Put x = 177 and y = 27 only option(a) will satisfy. Option(a).
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