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let the equation bel :
y- y1 = m(x-x1)
Given the slope m = 2/3
==> y-y1 = (2/3)(x-x1)
Let (x1,y1) be any point of the line.
We know that the y-intercept = -8
Then the point (0,-8) is on the line:
==> y-(-8) = (2/3)(x-0)
==> y+ 8 = (2/3)x
==> y= (2/3)x - 8
The equation of the line with slope m and y intercept c is y = mx+c. The line is inclined to x axis with a slope of arctanm and intersects y axis at c units from the origin.
Given m = 2/3 and c = -8.
So substituting the values of m and c in y = mx+c, we get:
y = (2/3)x+(-8) .....(1)
y = (2/3)x - 8. is the required equation in slope intercept form. We can write this in ax+by+k = 0 form also.
3y = 2x-8*3
3y = 2x-24.
Subtract 3y and rearrrange:
2x-3y-24 = 0 which is in the standard ax+by+k = 0 form.
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