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Mathematics

Find the equation of the line, whose :

(i) x-intercept = 5 and y-intercept = 3

(ii) x-intercept = -4 and y-intercept = 6

(iii) x-intercept = -8 and y-intercept = -4

Straight Line Eq

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Answer

(i) x-intercept = 5 and y-intercept = 3

When x-intercept = 5; corresponding point on the x-axis = (5, 0)

When y-intercept = 3; corresponding point on the y-axis = (0, 3).

Slope =3005=35.\text{Slope }= \dfrac{3 - 0}{0 - 5} = -\dfrac{3}{5}.

By point-slope form,

⇒ y - y1 = m(x - x1)

⇒ y - 0 = 35-\dfrac{3}{5}(x - 5)

⇒ 5y = -3(x - 5)

⇒ 5y = -3x + 15

⇒ 3x + 5y = 15.

Hence, equation of line is 3x + 5y = 15.

(ii) x-intercept = -4 and y-intercept = 6

When x-intercept = -4; corresponding point on the x-axis = (-4, 0)

When y-intercept = 6; corresponding point on the y-axis = (0, 6).

Slope =600(4)=64.\text{Slope }= \dfrac{6 - 0}{0 - (-4)} = \dfrac{6}{4}.

By point-slope form,

⇒ y - y1 = m(x - x1)

⇒ y - 0 = 64\dfrac{6}{4}[x - (-4)]

⇒ y = 32\dfrac{3}{2}(x + 4)

⇒ 2y = 3x + 12

⇒ 2y - 3x = 12

⇒ 2y = 3x + 12

Hence, equation of line is 2y = 3x + 12.

(iii) x-intercept = -8 and y-intercept = -4

When x-intercept = -8; corresponding point on the x-axis = (-8, 0)

When y-intercept = -4; corresponding point on the y-axis = (0, -4).

Slope =400(8)=48=12.\text{Slope }= \dfrac{-4 - 0}{0 - (-8)} = \dfrac{-4}{8} = -\dfrac{1}{2}.

By point-slope form,

⇒ y - y1 = m(x - x1)

⇒ y - 0 = 12-\dfrac{1}{2}[x - (-8)]

⇒ y = 12-\dfrac{1}{2}(x + 8)

⇒ 2y = -x - 8

⇒ 2y + x + 8 = 0.

Hence, equation of line is x + 2y + 8 = 0.

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