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Byju's Answer

Standard XII

Mathematics

Foot of Perpendicular from a Point on a Line

If 2y-x=8; ...

Question

If $2 y - x = 8; 5 y - x = 14, y - 2 x = 1$ intersect at $(a_{1}, b_{1}), (a_{2}, b_{2}), (a_{3}, b_{3})$
Find $(a_{1} + a_{2} + a_{3} + b_{1} + b_{2} + b_{3})$

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Solution

Given,

$2 y - x = 8$ .... (1)

$5 y - x = 14$ .....(2)

$y - 2 x = 1$ .....(3)

Solving (1) and (2) we get, $x = - 4, y = 2$

Solving (2) and (3) we get, $x = 1, y = 3$

Solving (1) and (3) we get, $x = 2, y = 5$

i.e. when these lines intersect with each other, they form a triangle, with vertices:

$a (- 4, 2), b (1, 3), c (2, 5)$

$∴ a_{1} + a_{2} + a_{3} + b_{1} + b_{2} + b_{3}$

$= - 4 + 1 + 2 + 2 + 3 + 5 = 9$

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If 2y−x=8;5y−x=14,y−2x=1 intersect at (a1,b1),(a2,b2),(a3,b3)Find (a1+a2+a3+b1+b2+b3)

If $2 y - x = 8; 5 y - x = 14, y - 2 x = 1$ intersect at $(a_{1}, b_{1}), (a_{2}, b_{2}), (a_{3}, b_{3})$
Find $(a_{1} + a_{2} + a_{3} + b_{1} + b_{2} + b_{3})$