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Mathematics

Given the equations of two straight lines, L1 and L2 are x - y = 1 and x + y = 5 respectively. If L1 and L2 intersects 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 coordinate of R and the ratio PQ : QR, if line L2 meets x-axis at point R.

Straight Line Eq

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Answer

(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 = 22\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,

(x,y)=(m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)(3,2)=(k×5+1×1k+1,k×0+1×4k+1)(3,2)=(5k+1k+1,4k+1)5k+1k+1=3 and 2=4k+15k+1=3(k+1) and 4=2(k+1)5k+1=3k+3 and 4=2k+25k3k=31 and 2k=422k=2 and 2k=2k=22=1.\Rightarrow (x, y) = \Big(\dfrac{m1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\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.

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