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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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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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.