Equation of a Straight Line — Analytical & Application Based Questions

Square ABCD lies in the third quadrant of a XY plane such that its vertex A is at (-3, -1) and the diagonal DB produced is equally inclined to both the axes. The diagonals AC and BD meet at P(-2, -2). Find the:

(a) slope of BD

(b) equation of AC

Answer

(a) Given,

BD is equally inclined to both axes.

∴ Slope of BD = tan 45° = 1.

Hence, slope of BD = 1.

(b) Since, diagonals of a square are perpendicular and product of slope of perpendicular lines equals to -1.

∴ Slope of BD × Slope of AC = -1

⇒ 1 × Slope of AC = -1

⇒ Slope of AC = -1.

Since, diagonals meet at point (-2, -2).

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-2) = -1[x - (-2)]

⇒ y + 2 = -1[x + 2]

⇒ y + 2 = -x - 2

⇒ x + y + 2 + 2 = 0

⇒ x + y + 4 = 0.

Hence, equation of AC is x + y + 4 = 0.

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.

Answer

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).

Given the equations of two straight lines, L₁ and L₂ are x - y = 1 and x + y = 5 respectively. If L₁ and L₂ intersect at point Q(3, 2), find:

(a) the equation of line L₃ which is parallel to L₁ and has y-intercept 3.

(b) the value of k, if the line L₃ meets the line L₂ at a point P(k, 4).

(c) the coordinates of R and the ratio PQ : QR, if line L₂ meets the x-axis at point R.

Answer

(a) (a) L₁ : x - y = 1

⇒ y = x - 1

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

⇒ m = 1.

We know that,

Slope of parallel lines are equal.

∴ Slope of L₃ = 1.

Given,

L₃ has y-intercept = 3.

∴ y = mx + c

⇒ y = 1.x + 3

⇒ y = x + 3.

Hence, equation of line L₃ : y = x + 3.

(b) L₂ : x + y = 5 and L₃ : y = x + 3

⇒ x + y = 5 .........(1)

⇒ y = x + 3 .........(2)

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

⇒ x + (x + 3) = 5

⇒ 2x + 3 = 5

⇒ 2x = 5 - 3

⇒ 2x = 2

⇒ x = $\dfrac{2}{2}$

⇒ x = 1.

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

⇒ y = 1 + 3 = 4.

⇒ (x, y) = (1, 4)

∴ P(k, 4) = (1, 4)

Hence, value of k = 1.

(c) Given,

L₂ meets x-axis at point R.

At point on x-axis, y-coordinate = 0.

L₂ : x + y = 5

⇒ x + 0 = 5

⇒ x = 5

⇒ R = (x, y) = (5, 0).

P = (1, 4), Q = (3, 2) and R = (5, 0)

Let PQ : QR = k : 1

By section formula,

$\Rightarrow (x, y) = \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] \Rightarrow (3, 2) = \Big(\dfrac{k \times 5 + 1 \times 1}{k + 1}, \dfrac{k \times 0 + 1 \times 4}{k + 1}\Big) \\[1em] \Rightarrow (3, 2) = \Big(\dfrac{5k + 1}{k + 1}, \dfrac{4}{k + 1}\Big) \\[1em] \Rightarrow \dfrac{5k + 1}{k + 1} = 3 \text{ and } 2 = \dfrac{4}{k + 1} \\[1em] \Rightarrow 5k + 1 = 3(k + 1) \text{ and } 4 = 2(k + 1) \\[1em] \Rightarrow 5k + 1 = 3k + 3 \text{ and } 4 = 2k + 2 \\[1em] \Rightarrow 5k - 3k = 3 - 1 \text{ and } 2k = 4 - 2 \\[1em] \Rightarrow 2k = 2 \text{ and } 2k = 2 \\[1em] \Rightarrow k = \dfrac{2}{2} = 1.$

∴ PQ : QR = 1 : 1.

Hence, R = (5, 0) and PQ : QR = 1 : 1.