The answer to your question the equation of the line BC is C [tex]-7x - 5y = -48[/tex]
Since AB and BC form a right angle at point B, it means that AB is perpendicular to BC.
To find the equation of BC, we first need to find the gradient of BC.
Let m be the gradient of AB and m' be the gradient of BC, since AB is perpendicular to BC, mm' = -1. Thus m' = -1/m
So, we need to find the gradient of AB and thus find the gradient of BC.
To find the gradient of AB, we use the equation for the gradient of a line in slope-point form. Having point A = (x₁, y₁) = (-3, -1) and point B = (x₂, y₂) = (4, 4).
[tex]m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}[/tex]
Substituting the values of the variables into the equation, we have
[tex]m = \frac{4 - (-1) }{4 - (-3)}[/tex]
[tex]m = \frac{4 + 1 }{4 + 3}[/tex]
[tex]m = \frac{5}{7}[/tex]
Since m' = -1/m
m' = -1 ÷ 5/7
m' = -7/5
Since we know the gradient of the line BC and the line BC passes through the point B = (4, 4) = (x₂, y₂) we find the equation of the line BC using the equation of a line in slope-point form.
[tex]m' = \frac{y - y_{2} }{x - x_{2}}[/tex]
[tex]-\frac{7}{5} = \frac{y - 4 }{x - 4}[/tex]
cross-multiplying, we have
[tex]-7(x - 4) = 5(y - 4)[/tex]
Expanding the brackets, we have
[tex]-7x + 28 = 5y - 20[/tex]
Subtracting 28 from both sides, we have
[tex]-7x = 5y - 48[/tex]
Subtracting 5y from both sides, we have
[tex]-7x - 5y = -48[/tex]
So, the equation of the line BC is [tex]-7x - 5y = -48[/tex]
The answer to your question the equation of the line BC is C [tex]-7x - 5y = -48[/tex]
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