Question

find the equation of a line whose gradient is 7/3 and which passes through the point (6,0)​

Answer 1

The equation is

y = 7/3 (x-6)

3y = 7x - 42

Answer 2

Answer:

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Step-by-step explanation:

Let equation of the straight line is Y = mx + c. According to problem,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: m = \frac{7}{3}$

∴ Equation of the straight line

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y = \frac{7}{x} + c$

$⇒ \: \: \: \: \: \: \: \: 3y = 7x + 3c$

The line passes through point (6, 0) so it coordinates will satisfy the straight line.

$∴ \: \: \: \: \: \: \: \: \: \: \: \: 3 \times 0 = 7 \times 6 + 3c$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: 0 = 42 + 3c$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: \: 3c = - 42$

$∴ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: c = - \frac{42}{3} = - 14$

Hence, equation of the required straight line is,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3y = 7x + 3( - 14)$

$⇒ \: \: \: 7x - 3y - 42 = 0.$