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Simplify (6x^2 − 3 + 5x^3) − (4x^3 − 2x^2 − 16).
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The equation of line in standard form is 5x - 4y = -38
<em><u>Solution:</u></em>
Given that we have to find the equation of line perpendicular to 4x + 5y = 40 that includes the point (-10, -3)
<em><u>The equation of line in slope intercept form is given as:</u></em>
y = mx + c ------ eqn 1
Where "m" is the slope of line and "c" is the y - intercept
Given equation of line is:
4x + 5y = 40
Rearranging to slope intercept form, we get
5y = -4x + 40
On comparing the above equation with eqn 1,
We know that product of slope of line and slope of line perpendicular to given line is equal to -1
Therefore,
Now find the equation of line with slope 5/4 and passing through (-10, -3)
Substitute and (x, y) = (-10, -3) in eqn 1
Substitute and
in eqn 1
The standard form of an equation is Ax + By = C
Therefore,
Thus the equation of line in standard form is found