A line passes through point P(4, 9) and Q(3, 2) is perpendicular to the line 3y + ax + 3 = 0. Find the value of a and write down the equation of line PQ.

Hey I — ready when you are. Here is the solution to question 34 from the image you sent: 34. A line passes through point P(4, 9) and Q(3, 2) is perpendicular to the line 3y + ax + 3 = 0. Find the value of a and write down the equation of line PQ. Step 1: Find the gradient of line PQ. The coordinates of P are (x_1, y_1) = (4, 9) and Q are (x_2, y_2) = (3, 2). The gradient m_PQ is given by: m_PQ = (y_2 - y_1)/(x_2 - x_1) = (2 - 9)/(3 - 4) = (-7)/(-1) = 7 Step 2: Find the gradient of the line 3y + ax + 3 = 0. Rearrange the equation into the form y = mx + c: 3y = -ax - 3 y = -(a)/(3)x - 1 The gradient of this line, m_L, is -(a)/(3). Step 3: Use the condition for perpendicular lines to find the value of a. If two lines are perpendicular, the product of their gradients is -1. m_PQ × m_L = -1 7 × (-(a)/(3)) = -1 -(7a)/(3) = -1 Multiply both sides by -3: 7a = 3 a = (3)/(7) Step 4: Write down the equation of line PQ. We have the gradient m_PQ = 7 and can use either point P(4, 9) or Q(3, 2). Using point P(4, 9) and the point-slope form y - y_1 = m(x - x_1): y - 9 = 7(x - 4) y - 9 = 7x - 28 y = 7x - 28 + 9 y = 7x - 19 The equation can also be written as 7x - y - 19 = 0. The value of a is (3)/(7) and the equation of line PQ is y = 7x - 19. Send me the next one 📸