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# Find the equation of a line whose (i) slope = 3, y-intercept = – 5 , (ii) slope = -2/7, y-intercept = 3 , (iii) gradient = root 3, y-intercept = -4/3 , (iv) inclination = 30 degree , y-intercept = 2

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This is the Important question of class 10 Based on Equation of a Straight Line Chapter of M.L Aggarwal book for ICSE BOARD.
Here you have to find the equation of a line whose
Slope , intercept, gradient and inclination values are given  in this question.
This is the Question Number 06 Exercise 12.1 of M.L Aggarwal.

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1. Equation of a line whose slope and y-intercept is given by:

y = mx + c, where m is the slope and c is the y-intercept

(i) Given: slope = 3, y-intercept = – 5

⇒ y = 3x + (-5)

Hence, the equation of line is y = 3x – 5.

(ii) Given: slope = -2/7, y-intercept = 3

⇒ y = (-2/7)x + 3

y = (-2x + 21)/7

7y = -2x + 21

Hence, the equation of line is 2x + 7y – 21= 0.

(iii) Given: gradient = √3, y-intercept = -4/3

⇒ y = √3x + (-4/3)

y = (3√3x – 4)/3

3y = 3√3x – 4

Hence, the equation of line is 3√3x – 3y – 4 = 0.

(iv) Given: inclination = 30°, y-intercept = 2

Slope = tan 30o = 1/√3

⇒ y = (1/√3)x + 2

y = (x + 2√3)/ √3

√3y = x + 2√3

Hence, the equation of line is x – √3y + 2√3 = 0.

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