# Help me solve the math?

Find the equation of the line passing through the points A(4;3) B (-3;-3)

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• Line L passes through A(4,3) & B(-3,-3). Slope of a line

joining 2 points P1(x1,y1) & P2(x2,y2) = (y2-y1)/(x2-x1).

Here (x1.y1) = (4,3) & (x2,y2) = (-3,-3). Then slope L =

(-3-3)/(-3-4) = (6/7). We shall now determine the equation

of L in slope, y-intercept form. We have y = mx +b as the equation of L where m=slope L and b is the y-intercept of

L. We have already determined m =(6/7) so y = (6/7)x +b. Now B(-3,-3) is on L. Then -3=(6/7)(-3)+b & b = 3[(6/7)-1] = 3(6-7)/7 = -(3/7). So equation of L is y = (6/7)x - (3/7).

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• The equation of the line passing through the points

A (4, 3) and B (-3, -3):

The equation of a line is y = mx where m is the slope.

That line passes through the origin.

If you want it to pass through some point (a, b) you subtract the coordinates like this: y - b = m (x - a).

That is the point-slope form, and you can rewrite it in other forms if it is convenient. You may do this with any point on the line.

y - 3 = m(x - 4)

y + 3 = m(x + 3)

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• The equation of the line passing through the points A(4, 3) and B (-3, -3):

Line equation

y=0.8571428571428571x−0.4285714285714288

Slope

0.86

Intercept

-0.43

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• solve for its slope

.........y₂ - y₁

m =-------------

.........x₂ - x₁

.........- 3 - 3

m =------------- = 6/7

..........- 3 - 4

apply slope intercept form

y = mx + b

3 = 6/7(4) + b

b = -3/7

The equation of the line is y = 6/7x - 3/7 ...//

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• Two point equation of a line:

y-y₀ = [(y₁-y₀)/(x₁-x₀)](x-x₀)

Given:

(x₀, y₀) = (4, 3)

(x₁, y₁) = (-3, -3)

Substituting the given points into the formula:

y-3 = [(-3-3)/(-3-4)](x-4)

You asked of "the equation", but there are multiple forms of the equation of a line. I'll convert the two point equation into the standard form of the equation. If you want any of the other forms, you can convert the standard form into whatever form you wish.

y - 3 = (-6/-7)(x-4)

-7y + 21 = -6x + 24

Ans:

Standard form: 6x - 7y = 3

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You can verify this equation by substituting the given points into the equation to show they are points on the line.

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