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Question
Find the equation of a line passing through (6, −2) and perpendicular to the line joining the points (6, 7) and (2, −3)
Solution
Let the vertices A(6, 7), B(2, −3), D(6, −2)
Slope of a line = `(y_2 - y_1)/(x_2 - x_1)`
Slope of AB = `(-3 - 7)/(2 - 6) = (-10)/(-4) = 5/2`
lope of its perpendicular (CD) = `-2/5`
Equation of the line CD is y – y1 = m(x – x1)
y + 2 = `-2/5(x - 6)`
5(y + 2) = – 2(x – 6)
5y + 10 = – 2x + 12
2x + 5y + 10 – 12 = 0
2x + 5y – 2 = 0
The equation of the line is 2x + 5y – 2 = 0
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