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Mathematics

Find the equation of the straight line perpendicular to the line x + 2y = 4, which cuts an intercept of 2 units from the positive y-axis. Hence, find the intersection point of the two lines.

Straight Line Eq

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Answer

Given,

Equation : x + 2y = 4

⇒ 2y = -x + 4

⇒ y = −12x+4-\dfrac{1}{2}x + 4

Comparing above equation with y = mx + c we get :

⇒ m = −12-\dfrac{1}{2}

Let slope of line perpendicular to line x + 2y = 4 be m1.

We know that,

Product of slope of perpendicular lines = -1.

⇒m1×−12=−1⇒m1=−1×2−1=2.\Rightarrow m1 \times -\dfrac{1}{2} = -1 \\[1em] \Rightarrow m1 = \dfrac{-1 \times 2}{-1} = 2.

Substituting values we get :

⇒ y = mx + c

⇒ y = 2x + 2.

Simultaneously solving equation :

⇒ x + 2y = 4 ………(1)

⇒ y = 2x + 2 …….(2)

Substituting value of y from equation (2) in (1), we get :

⇒ x + 2(2x + 2) = 4

⇒ x + 4x + 4 = 4

⇒ 5x = 4 - 4

⇒ 5x = 0

⇒ x = 05\dfrac{0}{5} = 0.

Substituting value of x in equation (2), we get :

⇒ y = 2(0) + 2 = 2.

Hence, equation of required line is y = 2x + 2 and point of intersection = (0, 2).

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