Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form?
2 answers:
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<h2>Answer:</h2>
y = x + 3
As shown in the graph, the line is a straight line. Therefore, the general equation of a straight line can be employed to derive the equation of the line.
The general equation of a straight line is given by:
y = mx + c <em>or </em>-------------(i)
y - y₁ = m(x - x₁) -----------------(ii)
Where;
y₁ is the value of a point on the y-axis
x₁ is the value of the same point on the x-axis
m is the slope of the line
c is the y-intercept of the line.
Equation (i) is the slope-intercept form of a line
Steps:
(i) Pick any two points (x₁, y₁) and (x₂, y₂) on the line.
In this case, let;
(x₁, y₁) = (0, 3)
(x₂, y₂) = (4, -2)
(ii) With the chosen points, calculate the slope <em>m</em> given by;
m =
m =
m =
(iii) Substitute the first point (x₁, y₁) = (0, 3) and m = into equation (ii) as follows;
y - 3 = (x - 0)
(iv) Solve for y from (iii)
y - 3 = x
y = x + 3 [This is the slope intercept form of the line]
Where the slope is and the intercept is 3