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Find the equation of the line that passes through the point (−8,13)(−8,13), with gradient 1221​ . Give your answer in the form: 𝑎𝑥+𝑏𝑦+𝑐ax+by+c==00

Question

Find the equation of the line that passes through the point (−8,13)(−8,13), with gradient 1221​ . Give your answer in the form: 𝑎𝑥+𝑏𝑦+𝑐ax+by+c==00

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Solution

The equation of a line in slope-intercept form is given by y = mx + c, where m is the slope and c is the y-intercept.

Given that the line passes through the point (-8,13) and has a gradient of 12/21, we can substitute these values into the equation to find the y-intercept.

13 = (12/21)*(-8) + c 13 = -96/21 + c c = 13 + 96/21 c = 273/21 + 96/21 c = 369/21 c = 17.57 (approx)

So, the equation of the line in slope-intercept form is y = 12/21*x + 17.57.

To convert this into the form ax + by + c = 0, we can subtract y from both sides and then multiply through by 21 to get rid of the fraction:

21y = 12x + 369 -12x + 21y - 369 = 0

So, the equation of the line in the form ax + by + c = 0 is -12x + 21y - 369 = 0.

This problem has been solved

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