Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form?
1 answer:
You can start with the form
∆y(x -x1) -∆x(y -y1) = 0
Here, we have
∆y = 11-(-3) = 14
∆x = -3-1 = -4
and we can choose (x1, y1) = (1, -3). This gives
14(x -1) -(-4)(y -(-3)) = 0
14x +4y -2 = 0
All these terms have a common factor of 2 that we can remove. Adding 1 to the result puts it in standard form:
7x +2y = 1
∆y(x -x1) -∆x(y -y1) = 0
Here, we have
∆y = 11-(-3) = 14
∆x = -3-1 = -4
and we can choose (x1, y1) = (1, -3). This gives
14(x -1) -(-4)(y -(-3)) = 0
14x +4y -2 = 0
All these terms have a common factor of 2 that we can remove. Adding 1 to the result puts it in standard form:
7x +2y = 1
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