Solved Examples on Two Point Form
Now that we have learned various things about the Two Point Form. Let’s get some practice on the concepts through some examples.
Example 1: Find the general equation of a line passing through the points (-1, 1) and (3, -7).
Solution:The two points on the straight line are (-1, 1) and (3, -7).
Equation of a line in two-point form:
\(\frac{(y – y_1)}{(y_2 – y_1)} = \frac{(x – x_1)}{(x_2 – x_1)}\)
Substitute \((x_1, y_1) = (-1, 1)\) and \((x_2, y_2) = (3, -7)\).
\(\frac{(y – 1)}{(-7 – 1)} = \frac{(x + 1)}{(3 + 1)}\)
\(\frac{(y – 1)}{(-8)} =\frac{(x + 1)}{(4)}\)
\(4(y – 1) = -8(x + 1)\)
Distribute.
\(4y – 4 = -8x – 8\)
Simplify.
\(8x + 4y + 4 = 0\)
Divide each side by 2.
\(8x + 4y + 4 = 0\)
Example 2: Find the equation of the line joining the points (3, 6) and (2, -5).
Solution:\(x_1 = 3, y_1 = 6, x_2 = 2, y_2 = -5\)
Equation of line in two-point form:
\(\frac{(y – y_1)}{(y_2 – y_1)} = \frac{(x – x_1)}{(x_2 – x_1)}\)
Substitute \((x_1, y_1) = (3, 6)\) and \((x_2, y_2) = (6, -5)\).
\(\frac{(y – 6)}{-5 – 6)} = \frac{(x – 3)}{(2 – 3)}\)
\(\frac{(y – 6)}{(-11)} =\frac{(x – 3)}{(-1)}\)
\(-1(y – 6) = -11(x – 3)\)
\(1(y – 6) = 11(x – 3)\)
\(y – 6 = 11x – 33\)
\(11x – 33 – y + 6 = 0\)
Equation of the Line: \(11x – y – 27 = 0\)
Example 3: Find the equation of a line in a slope-intercept form which passing through the points (-2, 5) and (3, 6).
Solution:The two points on the straight line are (-2, 5) and (3, 6).
Equation of line in two-point form : \(\frac{(y – y_1)}{(y_2 – y_1)} = \frac{(x – x_1)}{(x_2 – x_1)}\)
Substitute \((x_1 , y_1) = (-2, 5)\) and \((x_2, y_2) = (3, 6)\).
\(\frac{(y – 5)}{(6 – 5)} = \frac{(x + 2)}{(3 + 2)}\)
\(\frac{(y – 1)}{1} = \frac{(x + 2)}{5}\)
\(5(y – 1) = x + 2\)
\(5y – 5 = x + 2\)
Add 5 to each side.
\(5y = x + 7\)
Divide each side by 5.
\(y = \frac{x}{5} + \frac{7}{5}\)
\(y = (\frac{1}{5})x + \frac{7}{5}\)
Hope this article on the Two Point Form was informative. Get some practice of the same on our free Testbook App. Download Now!
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