Задание 2. Дано уравнение. Известно, что точка A принадлежит графику этого уравнения. Найдите коэффициент a.

Решение задания 2: 1) Дано уравнение $$ax + 2y - 10 = 0$$. a) $$A(2; 3)$$: $$a(2) + 2(3) - 10 = 0 => 2a + 6 - 10 = 0 => 2a - 4 = 0 => 2a = 4 => a = 2$$ b) $$A(4; 1)$$: $$a(4) + 2(1) - 10 = 0 => 4a + 2 - 10 = 0 => 4a - 8 = 0 => 4a = 8 => a = 2$$ c) $$A(1; 5)$$: $$a(1) + 2(5) - 10 = 0 => a + 10 - 10 = 0 => a = 0$$ d) $$A(-2; 7)$$: $$a(-2) + 2(7) - 10 = 0 => -2a + 14 - 10 = 0 => -2a + 4 = 0 => -2a = -4 => a = 2$$ 2) Дано уравнение $$3x + ay - 12 = 0$$. a) $$A(1; 3)$$: $$3(1) + a(3) - 12 = 0 => 3 + 3a - 12 = 0 => 3a - 9 = 0 => 3a = 9 => a = 3$$ b) $$A(2; 2)$$: $$3(2) + a(2) - 12 = 0 => 6 + 2a - 12 = 0 => 2a - 6 = 0 => 2a = 6 => a = 3$$ c) $$A(4; 1)$$: $$3(4) + a(1) - 12 = 0 => 12 + a - 12 = 0 => a = 0$$ d) $$A(-1; 5)$$: $$3(-1) + a(5) - 12 = 0 => -3 + 5a - 12 = 0 => 5a - 15 = 0 => 5a = 15 => a = 3$$ 3) Дано уравнение $$-2x + 8y - a = 0$$. a) $$A(4; 2)$$: $$-2(4) + 8(2) - a = 0 => -8 + 16 - a = 0 => 8 - a = 0 => a = 8$$ b) $$A(1; 4)$$: $$-2(1) + 8(4) - a = 0 => -2 + 32 - a = 0 => 30 - a = 0 => a = 30$$ c) $$A(-2; 6)$$: $$-2(-2) + 8(6) - a = 0 => 4 + 48 - a = 0 => 52 - a = 0 => a = 52$$ d) $$A(6; -1)$$: $$-2(6) + 8(-1) - a = 0 => -12 - 8 - a = 0 => -20 - a = 0 => a = -20$$