Question

1. a line l passes through points (-2, 3) and (-1, 6) and is perpendicular to a line p at (-1, 6).

(a) find the equation of l.
(2 marks)

(b) find the equation of p in the form ax + by = c, where a, b and c are constants. (2 marks)

(c) given that another line q is parallel to l and passes through point (1, 2), find the x and y intercepts of q.
(3 marks)

(d) find the point of intersection of lines p and q.
(3 marks)

Explanation:

Step1: Find the slope of line L

Use the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let \((x_1,y_1)=(-2,3)\) and \((x_2,y_2)=(-1,6)\).
\(m=\frac{6 - 3}{-1-(-2)}=\frac{3}{1}=3\)

Step2: Use the point - slope form to find the equation of L

The point - slope form is \(y - y_1=m(x - x_1)\). Using the point \((-1,6)\) and \(m = 3\), we have \(y-6=3(x + 1)\).
Expand: \(y-6=3x+3\), then \(y=3x + 9\)

Step3: Find the slope of line P

If two lines are perpendicular, \(m_1\times m_2=-1\). Since \(m_L = 3\), then \(m_P=-\frac{1}{3}\)
Using the point - slope form with point \((-1,6)\) and \(m =-\frac{1}{3}\), \(y - 6=-\frac{1}{3}(x + 1)\)
Multiply through by 3: \(3y-18=-x - 1\), so \(x+3y=17\)

Step4: Find the equation of line Q

Since Q is parallel to L, \(m_Q = 3\). Using the point - slope form with point \((1,2)\), \(y - 2=3(x - 1)\)
Expand: \(y-2=3x-3\), then \(y=3x-1\)
For x - intercept, set \(y = 0\): \(0=3x-1\), \(x=\frac{1}{3}\)
For y - intercept, set \(x = 0\): \(y=-1\)

Step5: Find the intersection of P and Q

We have \(P:x + 3y=17\) and \(Q:y=3x-1\)
Substitute \(y = 3x-1\) into \(x + 3y=17\): \(x+3(3x - 1)=17\)
Expand: \(x + 9x-3=17\), \(10x=20\), \(x = 2\)
Substitute \(x = 2\) into \(y=3x-1\), \(y=3\times2-1=5\)

Answer:

(a) \(y = 3x+9\)
(b) \(x + 3y=17\)
(c) x - intercept: \(\frac{1}{3}\), y - intercept: \(-1\)
(d) \((2,5)\)