Find an equation of the line having slope and containing the given point.
M=4/5, (3, -7)
The equation of the line is Y=
I have tried to work this problem and could not get the answer. Please help me.
y = (4/5) x + b
-7 = (4/5)(3) + b
b = -7 -(12/5)
b = -35/5 - 12/5
b = - 47/5
so
y = (4/5) x - 47/5
or
5 y = 4 x - 47
m=-4/5, T(1/3)=7/4
To find the equation of a line with a given slope and containing a specific point, we can use the point-slope form of a linear equation. The equation is given by:
y - y₁ = m(x - x₁)
Where (x₁, y₁) is the given point and m is the slope.
In this case, the given point is (3, -7) and the slope is 4/5. Plugging these values into the equation:
y - (-7) = (4/5)(x - 3)
Simplifying,
y + 7 = (4/5)(x - 3)
Expanding the right side,
y + 7 = (4/5)x - 12/5
Subtracting 7 from both sides,
y = (4/5)x - 12/5 - 7
Simplifying further,
y = (4/5)x - 12/5 - 35/5
Combining like terms,
y = (4/5)x - 47/5
Therefore, the equation of the line with slope 4/5 and passing through the point (3, -7) is y = (4/5)x - 47/5.
To find the equation of a line with a given slope and containing a given point, you can use the point-slope form of a linear equation.
The point-slope form of a linear equation is:
y - y1 = m(x - x1)
where:
- (x1, y1) represents the coordinates of the given point
- m represents the slope of the line
In your case, the given point is (3, -7) and the slope is 4/5. Substituting these values into the point-slope form, we have:
y - (-7) = (4/5)(x - 3)
Simplifying, we get:
y + 7 = (4/5)(x - 3)
Now, we re-arrange the equation to be in slope-intercept form (y = mx + b), where b represents the y-intercept:
y + 7 = (4/5)(x - 3)
y + 7 = (4/5)x - (12/5)
y = (4/5)x - (12/5) - 7
y = (4/5)x - (12/5) - (35/5)
y = (4/5)x - (47/5)
Therefore, the equation of the line with a slope of 4/5 and passing through the point (3, -7) is:
y = (4/5)x - (47/5)