Short Answer Questions 2 — Question 13

Given,

Equation : x + 2y = 4

⇒ 2y = -x + 4

⇒ y = $-\dfrac{1}{2}x + 4$

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

⇒ m = $-\dfrac{1}{2}$

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

We know that,

Product of slope of perpendicular lines = -1.

$\Rightarrow m_1 \times -\dfrac{1}{2} = -1 \\[1em] \Rightarrow m_1 = \dfrac{-1 \times 2}{-1} = 2.$

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 = $\dfrac{0}{5}$ = 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).