Last updated at Dec. 8, 2016 by Teachoo

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Misc 17 The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle. Let ∆ ABC be a right angle triangle where AC is the hypotenuse & ∠B = 90° Given that the hypotenuse has its ends at points (1, 3) & (–4, 1) ∴ A = (1, 3) & C = (–4, 1) We need to calculate equations of the legs of triangle i.e. we need to find equation of line AB & BC Let Slope of line AB = m We know if two lines are perpendicular, then product of their slope is –1 Here, AB & BC are perpendicular ∴ Slope of AB × Slope of BC = –1 m × Slope of BC = –1 Slope of BC = ( − 1)/𝑚 Now, We know that equation of line having slope m and passing through point (x1,y1) is y – y1 = m(x – x1) Equation of line AB passing through point A(1, 3) & having slope m is (y – 3) = m(x – 1) Similarly, Equation of line BC passing through point C(–4, 1) & having slope ( − 1)/𝑚 is (y – 1) = ( − 1)/𝑚 (x – (-4)) (y – 1) = ( − 1)/𝑚 (x + 4) We can have any value of m Finding equation of legs of triangle (AB & BC) by taking specific value of m For m = 0 Equation of AB (y – 3) = m(x – 1) y – 3 = 0(x – 1) y – 3 = 0 y = 3 Equation BC y – 1 = ( − 1)/𝑚 (x + 4) y – 1 = ( − 1)/0 (x + 4) 0(y – 1) = –1(x + 4) 0 = – x – 4 x = – 4

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.