A straight line T, which passes through the points A(-3, -5) and B(3, -6) is perpendicular to a line L at the point C(-2, -2). a) Find the gradient of line T. b) Find the equation of line T in the for

Here are the solutions for question 13: 13. A straight line T, which passes through the points A(-3, -5) and B(3, -6) is perpendicular to a line L at the point C(-2, -2). a) Find the gradient of line T. Step 1: Use the formula for the gradient m = (y_2 - y_1)/(x_2 - x_1) with points A(-3, -5) and B(3, -6). Let (x_1, y_1) = (-3, -5) and (x_2, y_2) = (3, -6). m_T = (-6 - (-5))/(3 - (-3)) Step 2: Simplify the expression. m_T = (-6 + 5)/(3 + 3) m_T = (-1)/(6) The gradient of line T is -(1)/(6). b) Find the equation of line T in the form y = mx + c where m and c are constants. Step 1: Use the point-slope form y - y_1 = m(x - x_1) with the gradient m_T = -(1)/(6) and point A(-3, -5). y - (-5) = -(1)/(6)(x - (-3)) y + 5 = -(1)/(6)(x + 3) Step 2: Distribute the gradient and simplify. y + 5 = -(1)/(6)x - (3)/(6) y + 5 = -(1)/(6)x - (1)/(2) Step 3: Isolate y to get the equation in the form y = mx + c. y = -(1)/(6)x - (1)/(2) - 5 y = -(1)/(6)x - (1)/(2) - (10)/(2) y = -(1)/(6)x - (11)/(2) The equation of line T is y = -(1)/(6)x - (11)/(2). c) Given that another line Q is parallel to T and passes through (1, -3) find the equation of line Q in the form ax + by = c. Step 1: Since line Q is parallel to line T, their gradients are equal. m_Q = m_T = -(1)/(6) Step 2: Use the point-slope form y - y_1 = m(x - x_1) with m_Q = -(1)/(6) and the point (1, -3). y - (-3) = -(1)/(6)(x - 1) y + 3 = -(1)/(6)(x - 1) Step 3: Multiply by 6 to eliminate the fraction. 6(y + 3) = -1(x - 1) 6y + 18 = -x + 1 Step 4: Rearrange the equation into the form ax + by = c. x + 6y = 1 - 18 x + 6y = -17 The equation of line Q is x + 6y = -17. d) Find the points of intersection of L and Q. Step 1: Find the gradient of line L. Line T is perpendicular to line L, so m_L · m_T = -1. m_L · (-(1)/(6)) = -1 m_L = 6 Step 2: Find the equation of line L. Line L passes through C(-2, -2) with gradient m_L = 6. y - (-2) = 6(x - (-2)) y + 2 = 6(x + 2) y + 2 = 6x + 12 y = 6x + 10 Step 3: Solve the system of equations for line L and line Q simultaneously. Equation of L: y = 6x + 10 Equation of Q: x + 6y = -17 Substitute the expression for y from line L into the equation for line Q. x + 6(6x + 10) = -17 x + 36x + 60 = -17 37x + 60 = -17 37x = -17 - 60 37x = -77 x = -(77)/(37) Step 4: Substitute the value of x back into the equation for line L to find y. y = 6(-(77)/(37)) + 10 y = -(462)/(37) + (10 × 37)/(37) y = -4